This document is a PowerPoint presentation about hypothesis testing for two samples and chi-square tests. It covers topics like independent and dependent sample tests, testing differences between proportions, one-tailed and two-tailed tests. Examples are provided to demonstrate how to perform two-sample t-tests, tests of proportions, and chi-square tests using contingency tables with 2 rows and 3 rows. Step-by-step instructions and formulas are given. Key chapters from the textbook are reviewed.
This document provides an outline and overview of topics covered in a course on inductive statistics, including probability distributions, sampling distributions, estimation, and hypothesis testing. Key topics discussed include interval estimation for means and proportions, using t-distributions when sample sizes are small and variances are unknown, and the basics of hypothesis testing such as null and alternative hypotheses. Examples are provided to illustrate concepts like confidence intervals for means, proportions, and hypothesis testing.
This document contains a PowerPoint presentation on inductive statistics covering topics like probability distributions, sampling distributions, estimation, hypothesis testing for means and proportions, and two-sample hypothesis tests. It provides an overview of the chapters that will be covered, examples of hypothesis tests for means and proportions when the population standard deviation is known and unknown, and examples of independent and dependent two-sample hypothesis tests for differences in means and proportions with both large and small sample sizes. Step-by-step explanations are given for conducting hypothesis tests.
The document summarizes key concepts from chapters 6 and 7 of a statistics textbook. Chapter 6 discusses sampling and calculating standard error for infinite and finite populations. Chapter 7 introduces estimation, including interval estimates and point estimates. It provides examples of calculating standard error and confidence intervals. The document also lists SPSS tips for t-tests.
The document describes how to perform a student's t-test to compare two samples. It provides steps for both a matched pairs t-test and an independent samples t-test. For a matched pairs t-test, the steps are: 1) state the null and alternative hypotheses, 2) calculate the differences between pairs, 3) calculate the mean difference, 4) calculate the standard deviation of the differences, 5) calculate the standard error, 6) calculate the t value, 7) determine the degrees of freedom, 8) find the critical t value, and 9) determine if there is a statistically significant difference. For an independent samples t-test, similar steps are followed to calculate means, standard deviations, the difference between
The chi-square test is used to determine if there is a relationship between two categorical variables in two or more independent groups. It can be used when data is arranged in a contingency table with observed and expected frequencies. A sample problem demonstrates how to calculate chi-square by finding the difference between observed and expected counts, squaring these differences, dividing by the expected counts, and summing across all cells. Degrees of freedom and critical values from tables determine whether to reject or fail to reject the null hypothesis of independence. Larger tables can be partitioned into subtables to identify where differences lie. Guidelines are provided for when chi-square or Fisher's exact test should be used based on sample size and expected cell counts.
This document summarizes a seminar on independent and paired t-tests. The t-test assesses whether the means of two groups are statistically different. An independent t-test is used to compare the means of two independent groups, while a paired t-test compares the means of the same variable measured at two different time points. Both tests calculate a critical t-value to compare to the calculated t-value to determine whether to reject the null hypothesis of no difference between the means. The document provides examples of performing independent and paired t-tests using formulas and the statistical software SPSS.
The document outlines the steps for conducting frequency distribution, cross-tabulation, and hypothesis testing analyses. It discusses measuring location, variability, and shape in frequency distributions. Cross-tabulation involves analyzing two or more variables simultaneously through tables. Hypothesis testing follows general steps including formulating hypotheses, selecting a test, determining significance levels, collecting data, and making conclusions.
1) The chi-square test of independence is used to determine if there is a relationship between two categorical variables. It compares observed frequencies to expected frequencies if the null hypothesis of independence is true.
2) A contingency table is constructed with the observed frequencies. Expected frequencies are calculated for each cell based on row and column totals and the grand total.
3) The chi-square statistic is calculated by summing the squared differences between observed and expected frequencies divided by the expected frequency for each cell. This value is then compared to a critical value from the chi-square distribution to determine if the null hypothesis should be rejected.
This document provides an outline and overview of topics covered in a course on inductive statistics, including probability distributions, sampling distributions, estimation, and hypothesis testing. Key topics discussed include interval estimation for means and proportions, using t-distributions when sample sizes are small and variances are unknown, and the basics of hypothesis testing such as null and alternative hypotheses. Examples are provided to illustrate concepts like confidence intervals for means, proportions, and hypothesis testing.
This document contains a PowerPoint presentation on inductive statistics covering topics like probability distributions, sampling distributions, estimation, hypothesis testing for means and proportions, and two-sample hypothesis tests. It provides an overview of the chapters that will be covered, examples of hypothesis tests for means and proportions when the population standard deviation is known and unknown, and examples of independent and dependent two-sample hypothesis tests for differences in means and proportions with both large and small sample sizes. Step-by-step explanations are given for conducting hypothesis tests.
The document summarizes key concepts from chapters 6 and 7 of a statistics textbook. Chapter 6 discusses sampling and calculating standard error for infinite and finite populations. Chapter 7 introduces estimation, including interval estimates and point estimates. It provides examples of calculating standard error and confidence intervals. The document also lists SPSS tips for t-tests.
The document describes how to perform a student's t-test to compare two samples. It provides steps for both a matched pairs t-test and an independent samples t-test. For a matched pairs t-test, the steps are: 1) state the null and alternative hypotheses, 2) calculate the differences between pairs, 3) calculate the mean difference, 4) calculate the standard deviation of the differences, 5) calculate the standard error, 6) calculate the t value, 7) determine the degrees of freedom, 8) find the critical t value, and 9) determine if there is a statistically significant difference. For an independent samples t-test, similar steps are followed to calculate means, standard deviations, the difference between
The chi-square test is used to determine if there is a relationship between two categorical variables in two or more independent groups. It can be used when data is arranged in a contingency table with observed and expected frequencies. A sample problem demonstrates how to calculate chi-square by finding the difference between observed and expected counts, squaring these differences, dividing by the expected counts, and summing across all cells. Degrees of freedom and critical values from tables determine whether to reject or fail to reject the null hypothesis of independence. Larger tables can be partitioned into subtables to identify where differences lie. Guidelines are provided for when chi-square or Fisher's exact test should be used based on sample size and expected cell counts.
This document summarizes a seminar on independent and paired t-tests. The t-test assesses whether the means of two groups are statistically different. An independent t-test is used to compare the means of two independent groups, while a paired t-test compares the means of the same variable measured at two different time points. Both tests calculate a critical t-value to compare to the calculated t-value to determine whether to reject the null hypothesis of no difference between the means. The document provides examples of performing independent and paired t-tests using formulas and the statistical software SPSS.
The document outlines the steps for conducting frequency distribution, cross-tabulation, and hypothesis testing analyses. It discusses measuring location, variability, and shape in frequency distributions. Cross-tabulation involves analyzing two or more variables simultaneously through tables. Hypothesis testing follows general steps including formulating hypotheses, selecting a test, determining significance levels, collecting data, and making conclusions.
1) The chi-square test of independence is used to determine if there is a relationship between two categorical variables. It compares observed frequencies to expected frequencies if the null hypothesis of independence is true.
2) A contingency table is constructed with the observed frequencies. Expected frequencies are calculated for each cell based on row and column totals and the grand total.
3) The chi-square statistic is calculated by summing the squared differences between observed and expected frequencies divided by the expected frequency for each cell. This value is then compared to a critical value from the chi-square distribution to determine if the null hypothesis should be rejected.
Test of significance (t-test, proportion test, chi-square test)Ramnath Takiar
The presentation discusses the concept of test of significance including the test of significance examples of t-test, proportion test and chi-square test.
The chi-square test is used to determine if experimental data fits the results expected from genetic theory. It involves calculating an expected and observed count for each phenotype, then using a chi-square formula to determine how well the observed fits the expected. The result is compared to critical chi-square values from a table based on degrees of freedom and probability to determine if the null hypothesis should be rejected or failed to be rejected.
The Chi Square Test is used to determine if observed data fits a hypothesized distribution. It involves calculating the Chi Square statistic by comparing observed and expected values and interpreting the result using a Chi Square table. The document provides an example using Drosophila genetics to test if two traits are independently assorting. The null hypothesis is that the traits are independently assorting. Expected values are calculated based on this. The Chi Square value is found to be not statistically significant, so the null hypothesis that the traits are independently assorting is not rejected.
The document discusses Chi-Square tests, which are used when assumptions of normality are violated. It provides requirements for Chi-Square tests, including that variables must be independent and samples sufficiently large. The key steps are outlined: determine appropriate test, establish significance level, formulate hypotheses, calculate test statistic using frequencies, determine degrees of freedom, and compare to critical value. An example compares party membership to opinions on gun control to demonstrate a Chi-Square test of independence.
The document discusses statistical tests such as the t-test and F-test. The t-test is used to compare means of two samples, such as comparing sample means before and after treatment. There are different types of t-tests, including paired samples and independent samples t-tests. The F-test, also called the F-ratio, compares variances between samples and is used in analysis of variance (ANOVA) to test differences between two or more groups. Examples are provided to demonstrate how to perform t-tests and F-tests to analyze data and test hypotheses.
The document discusses the concept of degrees of freedom. It provides definitions from various statistics textbooks and dictionaries. Degrees of freedom refers to the number of independent variables or pieces of information in a sample. The document also gives everyday examples to help explain degrees of freedom, such as dividing up time among tasks. Finally, it discusses statistical applications of degrees of freedom, such as in calculating sample variance, t-tests, ANOVA, and regression analysis.
The Chi-Square (χ2) test is a nonparametric test used to test hypotheses about frequency distributions across categories. It can test for independence and compare variances. The key steps are: 1) Calculate expected frequencies under the null hypothesis, 2) Calculate the differences between observed and expected frequencies, 3) Sum the squared differences divided by the expected values to get the chi-square statistic, 4) Compare this statistic to critical values from the chi-square distribution to determine significance. An example tests if vaccination prevents smallpox attacks, finding the results do not support independence between vaccination status and attacks, suggesting vaccination is effective.
The document discusses the chi-square test, which is used to determine if an observed frequency distribution differs from an expected theoretical distribution. It can be used as a test of independence to determine if two variables are associated, and as a test of goodness of fit to assess how well an expected distribution fits observed data. The steps of the chi-square test are outlined, including calculating the test statistic, determining degrees of freedom, and comparing the statistic to critical values to determine if the null hypothesis can be rejected. An example of a chi-square test of independence is shown to test if perceptions of fairness of performance evaluation methods are independent of each other.
This document provides an overview of common statistical hypothesis tests including:
1) The t-test, which is used to test differences between sample means and the significance of sample means.
2) The chi-square test, which evaluates differences between observed and expected frequencies without distributional assumptions. It is used for goodness of fit tests and tests of independence.
3) Analysis of variance (ANOVA), which tests for differences among two or more means by analyzing variance estimates. It is used to evaluate whether experimental factors have significant effects on outcomes.
The document defines key terms like null and alternative hypotheses, type I and II errors, level of significance, and degrees of freedom. It also outlines applications and procedures
The document discusses chi-square tests, which are used to analyze categorical data when the variables of interest are nominal rather than continuous. It explains the steps to conduct a chi-square test of goodness of fit and independence. An example is provided of a chi-square test of goodness of fit used to analyze the gender of characters on cereal boxes. The null hypothesis is that males and females would appear equally, and it is rejected because the chi-square statistic is greater than the critical value from the chi-square table.
parametric test of difference z test f test one-way_two-way_anova Tess Anoza
The document provides information about z-tests and F-tests. It defines what a z-test and F-test are, explains why and how they are used, and discusses the z-test for one sample and two sample groups as well as the F-test. For the z-test, it provides the formula, steps to use it for one and two sample groups, and an example problem. For the F-test, it defines what it is, when it is used, provides the formula and steps to compute the F-value including using an ANOVA table. It also provides an example problem to demonstrate solving the F-value.
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This document discusses statistical methods for comparing means, including t-tests and analysis of variance (ANOVA). It explains how t-tests can be used to compare two means or paired samples, and how ANOVA can compare two or more means. Key assumptions and procedures are outlined for one-sample t-tests, paired t-tests, independent t-tests with equal and unequal variances, and one-way between-subjects ANOVAs.
The document describes how to perform a paired t-test to determine if there is a statistically significant difference between two related samples. Specifically, it provides the steps to test if the number of fingerprints detected is higher after enhancing with ninhydrin compared to detection with DFO alone. The steps include: setting the null and alternative hypotheses, calculating the differences between paired samples, finding the mean difference, standard deviation, t-value, degrees of freedom, and critical value to determine if the null hypothesis can be rejected in favor of the alternative hypothesis of ninhydrin enhancing detection. The analysis rejects the null hypothesis, concluding ninhydrin treatment leads to more detectable fingerprints.
Hypothesis testing , T test , chi square test, z test Irfan Ullah
- The document discusses hypothesis testing and the p-value approach, which involves specifying the null and alternative hypotheses, calculating a test statistic, determining the p-value, and comparing it to the significance level α to determine whether to reject or accept the null hypothesis.
- It also discusses type I and type II errors, degrees of freedom as the number of independent pieces of information, and chi-square and t-tests as statistical tests.
The document discusses how to use a chi-squared (x2) test to examine differences between observed and expected frequencies of categorical data. It provides guidelines for when a chi-squared test is appropriate, how to perform the calculation, and how to interpret the results. A case study example is presented of a student analyzing questionnaire responses about the 2012 Olympics using a chi-squared test to determine if response frequencies differed significantly between demographic groups.
Tests of significance are statistical methods used to assess evidence for or against claims based on sample data about a population. Every test of significance involves a null hypothesis (H0) and an alternative hypothesis (Ha). H0 represents the theory being tested, while Ha represents what would be concluded if H0 is rejected. A test statistic is computed and compared to a critical value to either reject or fail to reject H0. Type I and Type II errors can occur. Steps in hypothesis testing include stating hypotheses, selecting a significance level and test, determining decision rules, computing statistics, and interpreting the decision. Hypothesis tests are used to answer questions about differences in groups or claims about populations.
This document provides information about the z-test, including its definition, formula, assumptions, and an example. The z-test is used to test if the means of two populations are different based on random samples from those populations. It can be used when sample sizes are large (n>=30) and the population standard deviation is known. An example compares the mean IQ scores of a sample of 30 students to the known population mean of 100 to test a principal's claim that students have above average intelligence. The z-score is calculated and compared to the critical value to determine if the null hypothesis can be rejected.
The document discusses introductory statistics concepts including descriptive statistics, inferential statistics, types of variables, levels of measurement, frequency tables, histograms, and other methods for organizing and presenting data. Chapter 1 covers what statistics is, types of statistics, variables, and levels of measurement. Chapter 2 discusses describing data through frequency tables, bar charts, histograms, and other graphs. The learning goals are to understand descriptive vs inferential statistics and distinguish between different statistical concepts.
This document provides an introduction to statistics, covering key topics such as what statistics is, its functions, applications in business, and subject matter. Statistics is defined as both a set of numerical data and a set of techniques for collecting, organizing, analyzing, and interpreting quantitative data. It serves functions like simplifying complex facts, providing comparisons, and forecasting. Statistics is used widely in business decision making across areas like marketing, finance, and operations. The subject matter of statistics has two parts - descriptive statistics, which summarizes data, and inferential statistics, which makes conclusions about large groups by studying samples.
The document discusses motivation, conflict, and negotiation in the workplace. It examines intrinsic and extrinsic motivation, as well as goal-setting theory and management by objectives as effective motivational models. Two conflict resolution techniques are analyzed: issues clarification and shared values exploration. Integrative bargaining is presented as a negotiation technique that leads to win-win solutions and long-term consistent relationships.
Test of significance (t-test, proportion test, chi-square test)Ramnath Takiar
The presentation discusses the concept of test of significance including the test of significance examples of t-test, proportion test and chi-square test.
The chi-square test is used to determine if experimental data fits the results expected from genetic theory. It involves calculating an expected and observed count for each phenotype, then using a chi-square formula to determine how well the observed fits the expected. The result is compared to critical chi-square values from a table based on degrees of freedom and probability to determine if the null hypothesis should be rejected or failed to be rejected.
The Chi Square Test is used to determine if observed data fits a hypothesized distribution. It involves calculating the Chi Square statistic by comparing observed and expected values and interpreting the result using a Chi Square table. The document provides an example using Drosophila genetics to test if two traits are independently assorting. The null hypothesis is that the traits are independently assorting. Expected values are calculated based on this. The Chi Square value is found to be not statistically significant, so the null hypothesis that the traits are independently assorting is not rejected.
The document discusses Chi-Square tests, which are used when assumptions of normality are violated. It provides requirements for Chi-Square tests, including that variables must be independent and samples sufficiently large. The key steps are outlined: determine appropriate test, establish significance level, formulate hypotheses, calculate test statistic using frequencies, determine degrees of freedom, and compare to critical value. An example compares party membership to opinions on gun control to demonstrate a Chi-Square test of independence.
The document discusses statistical tests such as the t-test and F-test. The t-test is used to compare means of two samples, such as comparing sample means before and after treatment. There are different types of t-tests, including paired samples and independent samples t-tests. The F-test, also called the F-ratio, compares variances between samples and is used in analysis of variance (ANOVA) to test differences between two or more groups. Examples are provided to demonstrate how to perform t-tests and F-tests to analyze data and test hypotheses.
The document discusses the concept of degrees of freedom. It provides definitions from various statistics textbooks and dictionaries. Degrees of freedom refers to the number of independent variables or pieces of information in a sample. The document also gives everyday examples to help explain degrees of freedom, such as dividing up time among tasks. Finally, it discusses statistical applications of degrees of freedom, such as in calculating sample variance, t-tests, ANOVA, and regression analysis.
The Chi-Square (χ2) test is a nonparametric test used to test hypotheses about frequency distributions across categories. It can test for independence and compare variances. The key steps are: 1) Calculate expected frequencies under the null hypothesis, 2) Calculate the differences between observed and expected frequencies, 3) Sum the squared differences divided by the expected values to get the chi-square statistic, 4) Compare this statistic to critical values from the chi-square distribution to determine significance. An example tests if vaccination prevents smallpox attacks, finding the results do not support independence between vaccination status and attacks, suggesting vaccination is effective.
The document discusses the chi-square test, which is used to determine if an observed frequency distribution differs from an expected theoretical distribution. It can be used as a test of independence to determine if two variables are associated, and as a test of goodness of fit to assess how well an expected distribution fits observed data. The steps of the chi-square test are outlined, including calculating the test statistic, determining degrees of freedom, and comparing the statistic to critical values to determine if the null hypothesis can be rejected. An example of a chi-square test of independence is shown to test if perceptions of fairness of performance evaluation methods are independent of each other.
This document provides an overview of common statistical hypothesis tests including:
1) The t-test, which is used to test differences between sample means and the significance of sample means.
2) The chi-square test, which evaluates differences between observed and expected frequencies without distributional assumptions. It is used for goodness of fit tests and tests of independence.
3) Analysis of variance (ANOVA), which tests for differences among two or more means by analyzing variance estimates. It is used to evaluate whether experimental factors have significant effects on outcomes.
The document defines key terms like null and alternative hypotheses, type I and II errors, level of significance, and degrees of freedom. It also outlines applications and procedures
The document discusses chi-square tests, which are used to analyze categorical data when the variables of interest are nominal rather than continuous. It explains the steps to conduct a chi-square test of goodness of fit and independence. An example is provided of a chi-square test of goodness of fit used to analyze the gender of characters on cereal boxes. The null hypothesis is that males and females would appear equally, and it is rejected because the chi-square statistic is greater than the critical value from the chi-square table.
parametric test of difference z test f test one-way_two-way_anova Tess Anoza
The document provides information about z-tests and F-tests. It defines what a z-test and F-test are, explains why and how they are used, and discusses the z-test for one sample and two sample groups as well as the F-test. For the z-test, it provides the formula, steps to use it for one and two sample groups, and an example problem. For the F-test, it defines what it is, when it is used, provides the formula and steps to compute the F-value including using an ANOVA table. It also provides an example problem to demonstrate solving the F-value.
Statistical tests /certified fixed orthodontic courses by Indian dental academy Indian dental academy
The Indian Dental Academy is the Leader in continuing dental education , training dentists in all aspects of dentistry and offering a wide range of dental certified courses in different formats.
Indian dental academy provides dental crown & Bridge,rotary endodontics,fixed orthodontics,
Dental implants courses.for details pls visit www.indiandentalacademy.com ,or call
0091-9248678078
This document discusses statistical methods for comparing means, including t-tests and analysis of variance (ANOVA). It explains how t-tests can be used to compare two means or paired samples, and how ANOVA can compare two or more means. Key assumptions and procedures are outlined for one-sample t-tests, paired t-tests, independent t-tests with equal and unequal variances, and one-way between-subjects ANOVAs.
The document describes how to perform a paired t-test to determine if there is a statistically significant difference between two related samples. Specifically, it provides the steps to test if the number of fingerprints detected is higher after enhancing with ninhydrin compared to detection with DFO alone. The steps include: setting the null and alternative hypotheses, calculating the differences between paired samples, finding the mean difference, standard deviation, t-value, degrees of freedom, and critical value to determine if the null hypothesis can be rejected in favor of the alternative hypothesis of ninhydrin enhancing detection. The analysis rejects the null hypothesis, concluding ninhydrin treatment leads to more detectable fingerprints.
Hypothesis testing , T test , chi square test, z test Irfan Ullah
- The document discusses hypothesis testing and the p-value approach, which involves specifying the null and alternative hypotheses, calculating a test statistic, determining the p-value, and comparing it to the significance level α to determine whether to reject or accept the null hypothesis.
- It also discusses type I and type II errors, degrees of freedom as the number of independent pieces of information, and chi-square and t-tests as statistical tests.
The document discusses how to use a chi-squared (x2) test to examine differences between observed and expected frequencies of categorical data. It provides guidelines for when a chi-squared test is appropriate, how to perform the calculation, and how to interpret the results. A case study example is presented of a student analyzing questionnaire responses about the 2012 Olympics using a chi-squared test to determine if response frequencies differed significantly between demographic groups.
Tests of significance are statistical methods used to assess evidence for or against claims based on sample data about a population. Every test of significance involves a null hypothesis (H0) and an alternative hypothesis (Ha). H0 represents the theory being tested, while Ha represents what would be concluded if H0 is rejected. A test statistic is computed and compared to a critical value to either reject or fail to reject H0. Type I and Type II errors can occur. Steps in hypothesis testing include stating hypotheses, selecting a significance level and test, determining decision rules, computing statistics, and interpreting the decision. Hypothesis tests are used to answer questions about differences in groups or claims about populations.
This document provides information about the z-test, including its definition, formula, assumptions, and an example. The z-test is used to test if the means of two populations are different based on random samples from those populations. It can be used when sample sizes are large (n>=30) and the population standard deviation is known. An example compares the mean IQ scores of a sample of 30 students to the known population mean of 100 to test a principal's claim that students have above average intelligence. The z-score is calculated and compared to the critical value to determine if the null hypothesis can be rejected.
The document discusses introductory statistics concepts including descriptive statistics, inferential statistics, types of variables, levels of measurement, frequency tables, histograms, and other methods for organizing and presenting data. Chapter 1 covers what statistics is, types of statistics, variables, and levels of measurement. Chapter 2 discusses describing data through frequency tables, bar charts, histograms, and other graphs. The learning goals are to understand descriptive vs inferential statistics and distinguish between different statistical concepts.
This document provides an introduction to statistics, covering key topics such as what statistics is, its functions, applications in business, and subject matter. Statistics is defined as both a set of numerical data and a set of techniques for collecting, organizing, analyzing, and interpreting quantitative data. It serves functions like simplifying complex facts, providing comparisons, and forecasting. Statistics is used widely in business decision making across areas like marketing, finance, and operations. The subject matter of statistics has two parts - descriptive statistics, which summarizes data, and inferential statistics, which makes conclusions about large groups by studying samples.
The document discusses motivation, conflict, and negotiation in the workplace. It examines intrinsic and extrinsic motivation, as well as goal-setting theory and management by objectives as effective motivational models. Two conflict resolution techniques are analyzed: issues clarification and shared values exploration. Integrative bargaining is presented as a negotiation technique that leads to win-win solutions and long-term consistent relationships.
This document discusses denouncing injustice and proclaiming justice according to Judiac Christian doctrine and human experience. It refers to teachings of the church and notes that the Catholic Bishop's Conference of the Philippines is the official organization of the Catholic hierarchy in the Philippines.
This study analyzes the impact of foreign direct investment (FDI) and international trade on labor productivity in 30 Chinese provinces from 1979 to 2006. It finds strong evidence that productivity in a given Chinese region is influenced by surrounding regions, indicating positive spatial autocorrelation. Additionally, FDI and trade are found to have positive and significant impacts on labor productivity and exhibit positive spatial spillovers among provinces. These findings are robust after accounting for spatial effects and alternative specifications.
The document discusses direct marketing and its importance in today's business world. It covers various channels of direct marketing like direct mail, telemarketing, email marketing. It also discusses the planning process for direct marketing which involves idea generation, objective setting, implementation and measuring results. Industrial or business to business marketing is also discussed along with differences between industrial and consumer marketing. The role of different promotional tools like advertising, sales promotion, publicity and direct marketing for business markets is explained.
1. The document discusses different types of chemical bonding including ionic bonding, covalent bonding, and metallic bonding.
2. Ionic bonding involves the transfer of electrons between atoms to form ions with opposite charges that are attracted in a giant lattice structure.
3. Covalent bonding can form either simple molecules held together by shared electron pairs or giant covalent structures with thousands of atoms bonded together.
Learning Objectives
To understand the escalating importance of logistics and supply-chain management as crucial tools for competitiveness.
To learn about materials management and physical distribution.
To learn why international logistics is more complex than domestic logistics.
To see how the transportation infrastructure in host countries often dictates the options open to the manager.
To learn why international inventory management is crucial for success.
Hcm611 1001 B 01 P1 T2 Ip Carl Wills.DocxTom Wills
This document discusses organizational structures for Regency Health Care Group and compares traditional and patient-focused structures. A traditional structure focuses on specialization, top-down hierarchy, and departmentalization to maximize coordination and efficiency. A patient-focused structure emphasizes cross-training, decentralization, flexibility, and patient care teams to improve integration and care. The document analyzes the pros and cons of each approach and argues that a patient-focused structure provides better care.
This document provides a summary of key concepts from Chapter 3 of a statistics textbook, including:
- How to calculate measures of central tendency like the mean, median, mode, and weighted mean
- The characteristics and properties of each measure
- How the positions of the mean, median and mode relate to the shape of the distribution
- How to calculate the mean, median and mode for grouped data
- What the geometric mean represents and how it is calculated
Correlation and regression analysis are statistical methods used to determine if a relationship exists between variables and describe the nature of that relationship. A scatter plot graphs the independent and dependent variables and allows visualization of any trends in the data. The correlation coefficient measures the strength and direction of the linear relationship between variables, ranging from -1 to 1. Regression finds the linear "best fit" line that minimizes the residuals and can be used to predict dependent variable values.
The document discusses Toyota's recalls of millions of vehicles in 2009-2010 due to issues with accelerators getting stuck. This damaged Toyota's reputation for quality and reliability. Rapid global expansion may have compromised quality systems as production moved overseas. Solutions included overhauling quality processes, communicating directly with customers, and regaining trust through a public relations campaign. The recalls significantly hurt Toyota's financial performance in the short term through lost sales and costs. Marketing would play a role in rebuilding Toyota's brand image and regaining customer confidence.
This document provides a summary of key concepts from chapters on simple regression and correlation analysis. It defines regression analysis as determining the nature and strength of relationships between variables. Scatter plots are used to visualize these relationships. The regression line estimates the relationship between an independent and dependent variable. Correlation analysis describes the degree of linear relationship between variables using the coefficient of determination and coefficient of correlation. Examples are provided to demonstrate calculating the regression equation and correlation coefficient.
Sexualharassment 150206205318-conversion-gate02Tyler Kah Jun
Sexual harassment is bullying, coercion, or inappropriate promises of rewards in exchange for sexual favors. It can include verbal harassment that makes people uncomfortable, sending sexually explicit emails, or unwanted physical contact like pinching or fondling. Employers can help prevent sexual harassment by establishing clear policies, monitoring the workplace, training supervisors, and making sure employees dress appropriately.
This document provides information about statistical tests that can be used to make inferences when comparing two samples or populations. Specifically, it discusses:
- Tests for comparing two proportions, means, variances or standard deviations from independent and dependent samples using z-tests, t-tests and F-tests.
- The assumptions and procedures for each test, including how to determine critical values and calculate test statistics.
- Examples of how to perform hypothesis tests and construct confidence intervals for various statistical comparisons between two samples or populations using a TI calculator.
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Chapter 9: Inferences from Two Samples
9.4: Two Variances or Standard Deviations
This document discusses hypothesis testing of claims about population standard deviations and variances. It provides the steps to conduct hypothesis tests of such claims using the chi-square distribution. The chi-square test can be used to test if a sample variance or standard deviation is significantly different from a claimed population variance or standard deviation. Examples show how to identify the null and alternative hypotheses, calculate the test statistic, find the critical value, and make a decision to reject or fail to reject the null hypothesis based on the test statistic and critical value.
1) The document discusses statistical inference and hypothesis testing. It covers topics like point and interval estimation, confidence intervals, hypothesis testing steps and terminology, tests for population means and proportions, and chi-square tests for independence.
2) An example calculates a 95% confidence interval for the mean hours students work per week based on sample data.
3) The final section discusses contingency tables and chi-square tests, providing an example to test if hand dominance and gender are associated using a contingency table. It shows calculating expected frequencies and the chi-square test statistic to evaluate the null hypothesis of independence.
t test for single mean, t test for means of independent samples, t test for means of dependent sample ( Paired t test). Case study / Examples for hands on experience of how SPSS can be used for different hypothesis testing - t test.
1. The document discusses hypothesis testing of claims about population parameters such as proportions, means, standard deviations, and variances from one or two samples.
2. Key concepts include hypothesis tests using z-tests, t-tests, and chi-square tests. Confidence intervals are also constructed for parameters.
3. Two examples are provided to demonstrate hypothesis testing of claims about two population proportions using z-tests. The null hypothesis is rejected in one example but not the other.
This document provides an outline and overview of Chapter 9 from a statistics textbook. The chapter covers hypothesis testing for single populations, including:
- Establishing null and alternative hypotheses
- Understanding Type I and Type II errors
- Testing hypotheses about single population means when the standard deviation is known or unknown
- Testing hypotheses about single population proportions and variances
- Solving for Type II errors
The chapter teaches students how to implement the HTAB (Hypothesis, Test Statistic, Accept/Reject regions, Boundaries, Conclusion) system to scientifically test hypotheses using statistical techniques like z-tests and t-tests. Key concepts covered include one-tailed and two-tailed tests, critical values, p
Nonparametric Test Chi-Square Test for Independence Th.docxpauline234567
Nonparametric Test
Chi-Square Test for Independence
The test is used to determine whether two categorical variables are independent.
Notation for the Chi-Square Test for Independence (Please note that the notation varies
depending on the text)
O represents the observed frequency of an outcome
E represents the expected frequency of an outcome
r represents the number of rows in the contingency table
c represents the number of columns in the contingency table
n represents the total number of trials
Test Statistic
2
2
1 1
O E
E
df r c
The Chi-Square test is a hypothesis test. There are seven steps for a hypothesis test.
1. State the null hypothesis
2. State the alternative hypothesis
3. State the level of significance
4. State the test statistic
5. Calculate
6. Statistical Conclusion
7. Experimental Conclusion
Example
A university is interested to know if the choice of major has a relationship to gender. A
random sample of 200 incoming freshmen students was taken (100 male and 100
female). There major and gender were recorded. The results are shown in the
contingency table below.
Major Female Male
Math 5 15
Nursing 44 10
English 10 10
Pre-Med 17 20
History 4 5
Education 15 20
Undecided 5 20
To determine if there is a relationship between the gender of a freshmen student and
thei declared major perform the hypothesis test (Use level of significance 0.05 ) .
Step 1: Null Hypothesis
0 : Gender and Major of Freshmen students are independentH
Step 2: Alternative Hypothesis
: Gender and Major of Freshmen students are not independentAH
Step 3: Level of Significance
0.05
Step 4: Test Statistic
2
2
1 1
O E
E
df r c
Step 5: Calculations
There are several calculations for this test. We have to find the expected frequency for
each cell in the contingency table. The expected frequency is the probability under the
null hypothesis times the total frequency for the given row. Here the probability under
the null hypothesis is .5, as the probability of being male and female is equal.
rE pn
Major Female Male
Math 1 .5 20 10E 2 .5 20 10E
Nursing 3 .5 54 27E 4 .5 54 27E
English 5 .5 20 10E 6 .5 20 10E
Pre-Med 7 .5 37 18.5E 8 .5 37 18.5E
History 9 .5 9 4.5E 10 .5 9 4.5E
Education 11 .5 35 17.5E 12 .5 35 17.5E
Undecided 13 .5 25 12.5E 14 .5 25 12.5E
Know calculate the test statistic.
2
2
2 2 2 2
2
2 2 2 2
2 2 2 2
2 2
2
5 10 15 10 44 27 10 27
10 10 27 27
10 10 10 10 17 18.5 20 18.5
10 10 18.5 18.5
4 4.5 5 4.5 15 17.5 20 17.5
4.5 4.5 17.5 17.5
5 12.5 20 12.5
12.5 12.5
2.5 2.5 10.7 10.7 0 0 .1216
obs
ob.
RM&IPR M4.pdfResearch Methodolgy & Intellectual Property Rights Series 4T.D. Shashikala
This PPT is prepared for VTU-Karnataka, Mtech/PhD Research Methodology syllabus based on C.R. Kothari, Gaurav Garg, Research Methodology: Methods and Techniques, New
Age International, 4th Edition, 2018
The document discusses hypothesis testing methods, including:
1. The p-value method which rejects the null hypothesis if the p-value is less than the significance level.
2. The critical value method which uses rejection regions determined by the significance level and critical values of the test statistic's distribution.
3. Examples are provided to demonstrate one-tailed and two-tailed hypothesis tests for population means when the population standard deviation is known or unknown.
Statistics practice for finalBe sure to review the following.docxdessiechisomjj4
Statistics practice for final
Be sure to review the following and have this information handy when taking Final GHA:
· Calculating z alpha/2 and t alpha/2 on Tables II and IV
· Find sample size for estimating population mean. Formula 8.3 p. 321 OCR.
· Stating H0 and H1 claims about variation and about the mean. Chapter 9 OCR.
· Type I and Type II errors p. 345 OCR.
· Confidence Interval for difference between two population means. Chapter 10 OCR p. 428
· Pooled sample standard deviation. Chapter 10 OCR p. 397
· What do Chi-Square tests measure? How are their degrees of freedom calculated? Chapter 12 OCR
· Find F test statistic using One-Way ANOVA.xls Be sure to enable editing and change values to match your problem. One-Way ANOVA.xls
· Find equation of regression line used to predict. To do on Excel, go to a blank worksheet, enter x values in one column and their matching y values in another column. Click Insert – Select Scatterplot. Right click any one of the points (diamonds) on the graph. Left click “Add a Trendline.” Check “Display Equation on Chart” box. Regression equation will appear on chart. Try it here with No. 20 below.
Practice Problems
Chapter 8 Final Review
1) In which of the following situations is it reasonable to use the z-interval
procedure to obtain a confidence interval for the population mean?
Assume that the population standard deviation is known.
A. n = 10, the data contain no outliers, the variable under consideration is
not normally distributed.
B. n = 10, the variable under consideration is normally distributed.
C. n = 18, the data contain no outliers, the variable under consideration is
far from being normally distributed.
D. n = 18, the data contain outliers, the variable under consideration is
normally distributed.
Find the necessary sample size.
2) The weekly earnings of students in one age group are normally
distributed with a standard deviation of 10 dollars. A researcher wishes to
estimate the mean weekly earnings of students in this age group. Find the
sample size needed to assure with 95 percent confidence that the sample
mean will not differ from the population mean by more than 2 dollars.
Find the specified t-value.
3) For a t-curve with df = 6, find the two t-values that divide the area under
the curve into a middle 0.99 area and two outside areas of 0.005.
Provide an appropriate response.
4) Under what conditions would you choose to use the t-interval procedure
instead of the z-interval procedure in order to obtain a confidence
interval for a population mean? What conditions must be satisfied in
order to use the t-interval procedure?
CHAPTER 8 Answers
1) B
2) 97
3) -3.707, 3.707
4) When the population standard deviation is unknown, the t-interval procedure is used instead of the
z-interval procedure. The t-interval procedure works provided that the population is normally
distributed or the.
Final Exam ReviewChapter 10Know the three ideas of s.docxlmelaine
Final Exam Review
Chapter 10
Know the three ideas of sampling.
• Examine a part of the whole: A sample can give
information about the population.
vA parameter is a number used in a model of the population.
vA statistic is a number that is calculated from the sample data.
vThe sample to sample differences are called the sampling
variability (or sampling error).
• Randomize to make the sample representative.
• The sample size is what matters.
• In a simple random sample (SRS), every possible group of
n individuals has an equal chance of being our sample.
Chapter 13
• Know the general rules of probability and how to apply them.
• The General Addition Rule says that :
P(A) or P(B) = P(A) + P(B) – P(A and B).
• The General Multiplication Rule says that :
P(A and B) = P(A) x P(B|A).
•Know that the conditional probability of an event B given the
event A is P(B|A) = P(A and B)/P(A).
•Know how to define and use independence:
Events A and B are independent if P(A|B) = P(A) or P(A and B) =
P(A) × P(B)
Chapter 14
• The expected value of a (discrete) random variable is:
• The variance for a random variable is:
• Rules combining RVs:
E(X ± c) = E(X) ± c Var(X ± c) = Var(X)
E(aX) = aE(X) Var(aX) = a2Var(X)
( ) ( )E X x P xµ = = ⋅∑
( ) ( ) ( )22 Var X x P xσ µ= = − ⋅∑
Chapter 15 :Geometric Probability Model
for Bernoulli Trials: GEOM(n, p)
•p = probability of success
•q = 1 – p = probability of failure
•X = number of trials until the first success occur
•Expected value:
•Standard deviation:
P X = x( ) = qx−1p
E X( ) = µ = 1
p
σ =
q
p2
Chapter 15: Binomial Probability Model for
Bernoulli Trials: BINOM(n, p)
•x = number of trials
•p = probability of success
•q = 1 – p = probability of failure
•X = number of successes in n trials
P X = x( ) = nCx pxqn−x, nCx =
n!
x! n − x( )!
Mean: µ = np
Standard deviation: σ = npq
Chapter 15:Poisson for Small p
• For rare events (small p), np may be less than 10.
• Use the Poisson instead of the Normal model.
• l = np mean number of successes
• X = number of successes
•
•
• Good approximation if n ³ 20 with p ≤ 0.05
or n ≤ 100 with p ≤ 0.10
( )
!
ll-
==
xe
P X x
x
( ) , ( )l l= =E X SD X
Chapter 16: One-Proportion Z-Interval
• Conditions met, find level C confidence interval for p
• Confidence interval is
• Standard deviation estimated by
• z* specifies number of SEs needed for C% of random
samples to yield confidence intervals that capture the
true parameter. Use table below to get z*
• Interpretation : we are 95% confident that the interval contains the
true proportion of X in the population
p̂ ± z * ×SE(p̂)
SE(p̂) =
p̂q̂
n
Chapter 16: One-Proportion Z-Interval
•Sampling Distribution for Proportions is Normal.
• Mean is p.
•
σ (p̂) = SD(p̂) =
pq
n
Chapter 16: One-Proportion Z-Interval
Sample Size and Standard Deviation
•
• Larger sample size → Smaller standard deviaaon
σ
( )=SD y
n
ˆ( )=
pq
SD p
n
Chapter 16: One-Proporti ...
The document discusses the history and procedures for conducting the Mann-Whitney U test. It originated from the work of Frank Wilcoxon, Henry Mann, and Donald Whitney in the late 19th/early 20th century. The test is a nonparametric alternative to the independent t-test that can be used to compare two independent groups when the dependent variable is either ordinal or continuous. It works by ranking the data from both groups together and comparing the sums of the ranks for each group to determine if they are significantly different, indicating differences in central tendency. Examples are provided demonstrating how to calculate the test statistic U and conduct statistical inferences.
The chi-square test is used to determine if an observed frequency distribution differs from an expected theoretical distribution. It can test for independence and goodness of fit. Karl Pearson introduced the chi-square test to compare observed and expected frequencies across categories. The test calculates a chi-square statistic and compares it to a critical value to determine if the null hypothesis that the distributions are the same can be rejected. Examples demonstrated how to calculate expected frequencies, the chi-square statistic, degrees of freedom, and compare to critical values to test independence between variables and goodness of fit to theoretical distributions.
This document discusses testing differences between two dependent samples using matched pairs. It provides examples of how to:
1) Calculate the differences between matched pairs and find the mean and standard deviation of the differences.
2) Use a t-test to determine if the mean difference is statistically significant and construct a 90% confidence interval for the true mean difference between two dependent samples.
3) Apply these methods to an example comparing cholesterol levels before and after a mineral supplement, testing the claim that the supplement changes cholesterol levels.
My significant event was losing my father at an early age. So if y.docxrosemarybdodson23141
My significant event was losing my father at an early age. So if you are good at psychology describe how that can affect a young woman without a dad at an early age.
Assignment 1.1: Significant Event Draft
Due Week 4 Note: Students will turn in this assignment twice: first as a draft (Week 4 Assignment) and then as a final paper (Week 8 Assignment). Students will submit first draft of the assignment and then submit the final draft after receiving feedback from the professor and making the recommended changes. They will be graded separately for each submission of the assignment.
Select a significant event (either positive or negative) that occurred before you reached adulthood and that has had a life-shaping effect on your life. In this assignment, you will use what you’ve learned in Weeks 1 thru 4 of this course and base your paper on your readings, along with information from library research, to discuss psychological aspects of your event.
Write a two to three (2-3) page paper in which you:
1.Briefly describe your significant event (one to two [1-2] sentences).
2.Describe your event in terms of at least two (2) different perspectives in psychology (e.g., behavioral, cognitive, psychodynamic, cultural/social, etc.)
3.Determine what learning (e.g., classical, operant, etc.) may have resulted from your event, or arisen because of your event.
4.Discuss why the memory of the event you described may or may not be completely accurate.
5.Use at least two (2) quality academic resources in this assignment. Note: One of these resources may be your textbook. Articles from professional journals are certainly a high quality resource. Magazine and newspaper articles are also accepted for this assignment. Articles published on the Internet may also be suitable, if they originate with credible persons or organizations. Please note that articles from Wikipedia, ask.com, and the like are not suitable.
Your assignment must follow these formatting requirements:
•Be typed, double spaced, using Times New Roman font (size 12), with one-inch margins on all sides; references must follow APA format. Check with your professor for any additional instructions.
•Include a cover page containing the title of the assignment, the student’s name, the professor’s name, the course title, and the date. The cover page and the reference page are not included in the required page length.
The specific course learning outcomes associated with this assignment are:
•Relate psychological concepts to real-world situations.
•Describe the major theories of learning, memory, cognition, consciousness, development, and social psychology.
•Use technology and information resources to research issues in psychology.
Part 1 of 16 -
Question 1 of 23
1.0 Points
A company operates four machines during three shifts each day. From production records, the data in the table below were collected. At the .05 level of significance test to determine if the number of breakdowns is independent of the sh.
Categorical Data and Statistical AnalysisMichael770443
In this presentation, we will introduce two tests and hypothesis testing based on it, and different non-parametric methods such as the Kolmogorov-Smirnov test, the Wilcoxon’s signed-rank test, the Mann-Whitney U test, and the Kruskal-Wallis test.
This document discusses statistical tests for comparing two samples, specifically Fisher's F test, Student's t-test, and Wilcoxon rank-sum test. It provides an example comparing ozone concentrations between two market gardens (Gardens A and B) using these tests. The F test showed the variances between gardens were not significantly different. The t-test and Wilcoxon test both found the mean ozone concentration was significantly higher in Garden B than Garden A.
The document discusses sampling and hypothesis testing. It defines key concepts like population, sample, parameter, statistic, sampling distribution, null hypothesis, alternative hypothesis, type I and type II errors. It explains different sampling methods and how to test hypotheses about population means using z-tests. Examples are provided to illustrate hypothesis testing for single and two population means. The summary tests hypotheses about population means using z-scores and critical values at given significance levels.
The document provides an overview of topics to be covered in Chapter 16 on time series and forecasting, including using trend equations to forecast future periods and develop seasonally adjusted forecasts, determining and interpreting seasonal indexes, and deseasonalizing data using a seasonal index. It also includes examples of calculating seasonal indices and adjusting sales data to remove seasonal variation. The document is a lecture outline and review for a class on international business taught by Dr. Ning Ding at Hanze University of Applied Sciences Groningen.
Here are the steps to solve this problem:
1) Code the year as t = 1 for 1999, t = 2 for 2000, etc.
2) Calculate the sums: Σt = 15, ΣY = 211.9, Σt2 = 30, ΣtY = 332.5
3) b = (ΣtY - ΣtΣY/n) / (Σt2 - Σt2/n) = 6.55
4) a = Y - bX = 29.4 - 6.55(1) = 22.85
5) Ŷ = 22.85 + 6.55t
To estimate vending sales
This document provides an overview of simple linear regression and correlation. It discusses key concepts such as dependent and independent variables, scatter diagrams, regression analysis, the least-squares estimating equation, and the coefficients of determination and correlation. Scatter diagrams are used to determine the nature and strength of relationships between variables. Regression analysis finds relationships of association but not necessarily of cause and effect. The least-squares estimating equation models the dependent variable as a function of the independent variable.
This document provides an overview of central tendency measures that will be covered in Chapter 3-A, including the mean, mode, and median for both ungrouped and grouped data. It also includes examples of calculating the mean, weighted mean, and mode. The document reviews key concepts such as the difference between parameters and statistics. Overall, the document previews and reviews important concepts related to measures of central tendency that will be covered in the upcoming chapter.
This document provides an overview and summary of topics covered in a research methods course. It discusses reviewing concepts from prior lectures, including different types of research and variables. Today's lecture will cover instrumentation, validity and reliability, and threats to internal validity. Instrumentation discusses how to collect and measure data. Validity and reliability refer to the accuracy and consistency of measurements. Threats to internal validity could interfere with determining the true effect of independent variables on dependent variables.
This document provides an overview of content covered in Statistics 2, including a review of chapter 5 on sampling distributions. It includes examples of questions from quizzes on topics like the normal distribution and binomial approximation. The document also provides tips on using SPSS for descriptive statistics, such as inputting and defining variable data, and analyzing frequencies.
This document summarizes a course on research methods and techniques. It outlines the structure and requirements of the course, including reading a textbook and attending lectures. It discusses different types of research and variables. The document covers defining research problems, formulating hypotheses, research ethics, and instrumentation. Self-check exercises are provided to help students understand key concepts.
The document outlines chapters from a statistics textbook, covering topics such as describing and exploring data through frequency tables, histograms, measures of central tendency, dispersion, correlation, and time series analysis. It also discusses deseasonalizing time series data to study trends and uses an example of predicting quarterly sales figures after removing seasonal fluctuations. The later chapters focus on time series forecasting through techniques like determining a seasonal index and forming a least squares regression line to predict future values.
The document provides an overview of key concepts in probability and statistics, including:
- Definitions of probability distributions, random variables, and expected value
- Explanations and examples of the binomial, Poisson, and normal distributions
- How to calculate probabilities and combine them with monetary values for decision making
The document discusses techniques for analyzing time series data and seasonal trends, including calculating moving averages, determining linear and nonlinear trends, seasonal indexes, and deseasonalizing data. It provides examples of computing seasonal indexes using quarterly sales data and removing seasonal variation to study underlying trends. Key steps include organizing the data, taking moving averages, calculating specific seasonal indexes, and adjusting values using seasonal factors.
This document contains summaries and examples of key concepts in regression analysis and correlation from Chapter 12, including:
- Regression analysis is used to estimate relationships between variables and predict future values of dependent variables based on independent variables.
- Correlation analysis describes the strength of the linear relationship between two variables from 0 to 1.
- The least squares method is used to fit a regression line that minimizes the squared errors between observed and predicted values.
This document provides instructions for using Excel to conduct descriptive statistical analyses on earnings data. It includes tips for using Excel functions like automatic numbering, summing values, and ranking data. Students are asked to calculate the average earnings for different quarters and years for an individual named M.T. Based on the results, the fourth quarter had the widest range of earnings between quarters, and the third year had the widest range of earnings between years. This suggests M.T.'s earnings varied the most during the fourth quarter and third year.
The document discusses developing a research topic and proposal. It covers generating ideas, refining a topic, writing research questions and objectives, reviewing literature, and drafting a proposal. Key points include choosing a feasible and significant topic, focusing the study with clear and logical objectives, using theory to guide the research, and convincing the audience and organizing ideas in the proposal. The document provides examples and exercises to help develop a strong research topic and questions.
HOW TO START UP A COMPANY A STEP-BY-STEP GUIDE.pdf46adnanshahzad
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This PowerPoint compilation offers a comprehensive overview of 20 leading innovation management frameworks and methodologies, selected for their broad applicability across various industries and organizational contexts. These frameworks are valuable resources for a wide range of users, including business professionals, educators, and consultants.
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Lesson 06 chapter 9 two samples test and Chapter 11 chi square test
1. Inductive Statistics Dr. Ning DING [email_address] I.007 IBS, Hanze You’d better use the full-screen mode to view this PPT file.
2. Table of Contents Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row
3. Chapter 9 Testing Hypotheses: Two-Sample Tests: Basics-Independent σ is known: σ is unknown: H 0 H 1 n <30 & σ is unknown Two-tailed test One-tailed test
4. Chapter 9 Testing Hypotheses: Two-Sample Tests: Practice Ch 9 No. 9-9 P.466 Step 1: Formulate hypotheses 9-9 Step 2: Find the Pooled Estimate of σ 2 Step 3: Calculate the standard error Step 4: Visualize and Find the t scores One-tailed Test df = 16 area=0.10 t=1.746 Review: Chapter 5 Chapter 6 Chapter 7 Chapter 8 Testing Hypothesis ~Test for Mean * when σ is known * when σ is unknown AND n=<30 ~Test for Proportion Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test Gender Mean Standard Deviation Sample size Female 12.8 1.0667 10 Male 11.625 1.4107 8
5. Chapter 9 Testing Hypotheses: Two-Sample Tests: Dependent Samples 9.2 Dependent Samples Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row
6. Chapter 9 Testing Hypotheses: Two-Sample Tests: Dependent Samples 9.2 Dependent Samples Ch 9 Example P.468 Will the participant lose more than 17 pounds after the weight-reducing program? The survey data is: Step 1: Formulate Hypotheses One-tailed Test Example: Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row
7. Chapter 9 Testing Hypotheses: Two-Sample Tests: Dependent Samples 9.2 Dependent Samples Ch 9 Example P.468 Step 2: Calculate the estimated standard deviation of the population difference Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row
8. Chapter 9 Testing Hypotheses: Two-Sample Tests: Dependent Samples 9.2 Dependent Samples Ch 9 Example P.468 Step 3: Find the Standard Error of the population difference Step 4: Calculate the t value Step 5: Visualize and get the t values df = 10-1=9 area = 0.10 t=1.833 One-tailed Test reject H 0 significant difference Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row
9. Chapter 9 Testing Hypotheses: Two-Sample Tests: Practice Ch 9 No. 9-15 P.474 Step 4: Visalize and Calculate the t values t=1.895 9-15 Step 3: Find the Standard Error of the population difference Step 1: Formulate Hypotheses Step 2: Calculate the estimated standard deviation of the population difference df=7 area=0.10 reject H 0 sig difference Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row One-tailed Test
10. Chapter 9 Testing Hypotheses: Two-Sample Tests: Proportion Example: Ch 9 Example P.476 You are testing whether the two drugs cause different blood-pressure levels. The data is as below: Step 1: Formulate Hypotheses Step 3: Calculate the Standard Error of Proportion Difference Step 2: Calculate the Estimated Proportion Difference Two-tailed Test Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row
11. Step 4: Visualize and get the z values -1.96 +1.96 Accept H 0 No significant difference Chapter 9 Testing Hypotheses: Two-Sample Tests: Proportion Ch 9 Example P.476 Two-tailed Test Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row
12. Chapter 9 Testing Hypotheses: Two-Sample Tests: Proportion Step 1: Formulate Hypotheses Step 3: Calculate the Standard Error of Proportion Difference Step 2: Calculate the Estimated Proportion Difference One-tailed Test Example: Ch 9 Example P.480 You are testing whether personal-appearance method causes fewer tax mistakes than mail method. The data is as below: Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row
13. Step 4: Visualize and get the z values One-tailed α =0.15 P=0.35 z= -1.04 Accept H 0 No significant difference Chapter 9 Testing Hypotheses: Two-Sample Tests: Proportion Ch 9 Example P.480 One-tailed Test -1.04 Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row
14. Chapter 9 Testing Hypotheses: Two-Sample Tests: Practice 9-20 Ch 9 No. 9-20 P.483 Step 1: Formulate Hypotheses Step 3: Calculate the Standard Error of Proportion Difference Step 2: Calculate the Estimated Proportion Difference Step 4: Visualize and get the z values z= -1.28 reject H 0 sig difference Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row
15. Chapter 9 Testing Hypotheses: Two-Sample Tests: Practice 9-21 Step 1: Formulate Hypotheses Step 3: Calculate the Standard Error of Proportion Difference Step 2: Calculate the Estimated Proportion Difference Ch 9 No. 9-21 P.483 Step 4: Visualize and get the z values accept H 0 no sig difference Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row
16. Chapter 11 Chi-Square Test-Basics Contingency Table (Cross break table) Rows * Columns == 2*2 table Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row Male Female Total Junior high school 40 60 100 Senior high school 60 40 100 Total 100 100 200
17. Chapter 11 Chi-Square Test-Basics Contingency Table (Cross break table) Rows * Columns == 2*2 table Expected and Observed Values Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row Male Female Total Junior high school 40 60 100 Senior high school 60 40 100 Total 100 100 200 Male Female Total Junior high school 40 (50) 60 (50) 100 Senior high school 60 (50) 40 (50) 100 Total 100 100 200
18. Chapter 11 Chi-Square Test-Basics Contingency Table (Cross break table) 1*1=1 2*2=4 Degree of freedom= (row-1)*(column-1) 2*2 table 3*3 table Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row Male Female Total Junior high school 40 60 100 Senior high school 60 40 100 Total 100 100 200 Social Science Nature Science Sports Junior high school 40 60 20 Senior high school 60 40 40 University 50 30 120
19. Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row Chapter 11 Chi-Square Test–Calculate χ 2 for 2-row Table Step 1: Calculate the expected values Ch 11 Example P.570 Example: Our employees’ attitude toward job-performance reviews. There are two review methods, the present one or the new one. Is the attitude dependent on geography? The survey looks like below:
20. Chapter 11 Chi-Square Test–Calculate χ 2 for 2-row Table Step 2: Calculate the χ 2 2.7644 Ch 11 Example P.570 Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row
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22. Chapter 11 Chi-Square Test–Calculate χ 2 for 3-row Table For a national health insurance program, you believes that lengths of stays in hospitals are dependent on the types of health insurance that people have. The random data from the survey is as below: Ch 11 Example P.575 Step 1: Formulate the hypotheses H 0 : length of stay and insurance types are independent H 1 : length of stay depends on insurance types α=0.01 Example: Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row Days in Hospital Cost Cover <5 5-10 >10 Total <25% 40 75 65 180 25-50% 30 45 75 150 >50% 40 100 190 330 Total 110 220 330 660
23. Chapter 11 Chi-Square Test–Calculate χ 2 for 3-row Table For a national health insurance program, you believes that lengths of stays in hospitals are dependent on the types of health insurance that people have. The random data from the survey is as below: Example: Ch 11 Example P.575 Step 2: Calculate the Expected Frequency For Any Cell RT CT (30) (60) Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row Days in Hospital Cost Cover <5 5-10 >10 Total <25% 40 75 65 180 25-50% 30 45 75 150 >50% 40 100 190 330 Total 110 220 330 660
24. Chapter 11 Chi-Square Test–Calculate χ 2 for 3-row Table Ch 11 Example P.575 Step 3: Calculate the Chi-Square Example: For a national health insurance program, you believes that lengths of stays in hospitals are dependent on the types of health insurance that people have. The random data from the survey is as below: 3.33 Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row Days in Hospital Cost Cover <5 5-10 >10 Total <25% 40 (30) 75 (60) 65 (90) 180 25-50% 30 (25) 45 (50) 75 (75) 150 >50% 40 (55) 100 (110) 190 (165) 330 Total 110 220 330 660
25. Chapter 11 Chi-Square Test–Calculate χ 2 for 3-row Table Ch 11 Example P.575 Step 3: Calculate the Chi-Square Example: For a national health insurance program, you believes that lengths of stays in hospitals are dependent on the types of health insurance that people have. The random data from the survey is as below: 3.33 3.75 6.94 1.00 4.09 0.50 0.91 0.00 3.79 =3.33+3.75+6.94+1.00+0.50+0.00+4.09+0.91+3.79 =24.32 Chi-Square χ 2 =24.32 Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row Days in Hospital Cost Cover <5 5-10 >10 Total <25% 40 (30) 75 (60) 65 (90) 180 25-50% 30 (25) 45 (50) 75 (75) 150 >50% 40 (55) 100 (110) 190 (165) 330 Total 110 220 330 660
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27. Chapter 11 Chi-Square Test–Calculate χ 2 for 3-row Table: Practice Ch 11 No. 11-9/10 P.582 11-9/10 Step 1: Formulate the hypotheses Step 2: Calculate the Expected Frequency For Any Cell Step 3: Calculate the Chi-Square Step 4: Find the critical Chi-Square Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row Economy Weekly Chip Sales High Medium Low Total At peak 20 7 3 30 At Through 30 40 30 100 Rising 20 8 2 30 Falling 30 5 5 40 Total 100 60 40 200
28. Chapter 11 Chi-Square Test–Calculate χ 2 for 3-row Table: Practice Ch 11 No. 11-9/10 P.582 11-9/10 Step 1: Formulate the hypotheses H 0 : Sales and economy are independent H 1 : sales depends on economy α=0.10 Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row Economy Weekly Chip Sales High Medium Low Total At peak 20 7 3 30 At Through 30 40 30 100 Rising 20 8 2 30 Falling 30 5 5 40 Total 100 60 40 200
29. Chapter 11 Chi-Square Test–Calculate χ 2 for 3-row Table: Practice Ch 11 No. 11-9/10 P.582 11-9/10 Step 2: Calculate the Expected Frequency For Any Cell Review: Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row Economy Weekly Chip Sales High Medium Low Total At peak 20 7 3 30 At Through 30 40 30 100 Rising 20 8 2 30 Falling 30 5 5 40 Total 100 60 40 200 Economy Weekly Chip Sales High Medium Low Total At peak 20 (15) 7 (9) 3 (6) 30 At Through 30 (50) 40 (30) 30 (20) 100 Rising 20 (15) 8 (9) 2 (6) 30 Falling 30 (20) 5 (12) 5 (8) 40 Total 100 60 40 200
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31. Table of Contents Review: Chapter 8 Testing Hypothesis Chapter 9: Testing Hypotheses: Two-Sample Tests ~Basics ~Independent Sample Test ~Large Samples ~Small Samples ~Dependent Sample Test ~Differences between Proportions ~Two-tailed Test ~One-tailed Test Chapter 11: Chi-Square ~Basics ~Contingency Table 2-row ~Contingency Table 3-row
32. The Normal Distribution SPSS Tips The data can be downloaded from: Blackboard – Inductive Statsitics STA2—SPSS-- Week 6 Chi-Square.sav
33. The Normal Distribution SPSS Tips APPLE shop in Groningen wants to know whether our IBS students have an intention to buy an iPad and whether their interest depends on their nationalities. They have interviewed 64 Year TWO students and the data can be downloaded from Blackboard—STA2—SPSS –Chi square.sav.
39. The Normal Distribution SPSS Tips Now you get the output! But how to interpret it? important
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Editor's Notes
Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don't qualify for favorable rates, then Y , which equals 10 − X , is a binomial random variable with n = 10 and q = 1 − p = 0.30. We are interested in finding P ( X ≥ 4). We can't use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewrite P ( X ≥ 4) as a probability statement in terms of Y : P ( X ≥ 4) = P (− X ≤ −4) = P (10 − X ≤ 10 − 4) = P ( Y ≤ 6) Now it's just a matter of looking up the probability in the right place on our cumulative binomial table. To find P ( Y ≤ 6), we: Find n = 10 in the first column on the left. Find the column containing p = 0.30 . Find the 6 in the second column on the left, since we want to find F (6) = P ( Y ≤ 6). Now, all we need to do is read the probability value where the p = 0.30 column and the ( n = 10, y = 6) row intersect. What do you get? Do you need a hint? The cumulative binomial probability table tells us that P ( Y ≤ 6) = P ( X ≥ 4) = 0.9894. That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.
Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don't qualify for favorable rates, then Y , which equals 10 − X , is a binomial random variable with n = 10 and q = 1 − p = 0.30. We are interested in finding P ( X ≥ 4). We can't use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewrite P ( X ≥ 4) as a probability statement in terms of Y : P ( X ≥ 4) = P (− X ≤ −4) = P (10 − X ≤ 10 − 4) = P ( Y ≤ 6) Now it's just a matter of looking up the probability in the right place on our cumulative binomial table. To find P ( Y ≤ 6), we: Find n = 10 in the first column on the left. Find the column containing p = 0.30 . Find the 6 in the second column on the left, since we want to find F (6) = P ( Y ≤ 6). Now, all we need to do is read the probability value where the p = 0.30 column and the ( n = 10, y = 6) row intersect. What do you get? Do you need a hint? The cumulative binomial probability table tells us that P ( Y ≤ 6) = P ( X ≥ 4) = 0.9894. That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.
Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don't qualify for favorable rates, then Y , which equals 10 − X , is a binomial random variable with n = 10 and q = 1 − p = 0.30. We are interested in finding P ( X ≥ 4). We can't use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewrite P ( X ≥ 4) as a probability statement in terms of Y : P ( X ≥ 4) = P (− X ≤ −4) = P (10 − X ≤ 10 − 4) = P ( Y ≤ 6) Now it's just a matter of looking up the probability in the right place on our cumulative binomial table. To find P ( Y ≤ 6), we: Find n = 10 in the first column on the left. Find the column containing p = 0.30 . Find the 6 in the second column on the left, since we want to find F (6) = P ( Y ≤ 6). Now, all we need to do is read the probability value where the p = 0.30 column and the ( n = 10, y = 6) row intersect. What do you get? Do you need a hint? The cumulative binomial probability table tells us that P ( Y ≤ 6) = P ( X ≥ 4) = 0.9894. That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.
Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don't qualify for favorable rates, then Y , which equals 10 − X , is a binomial random variable with n = 10 and q = 1 − p = 0.30. We are interested in finding P ( X ≥ 4). We can't use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewrite P ( X ≥ 4) as a probability statement in terms of Y : P ( X ≥ 4) = P (− X ≤ −4) = P (10 − X ≤ 10 − 4) = P ( Y ≤ 6) Now it's just a matter of looking up the probability in the right place on our cumulative binomial table. To find P ( Y ≤ 6), we: Find n = 10 in the first column on the left. Find the column containing p = 0.30 . Find the 6 in the second column on the left, since we want to find F (6) = P ( Y ≤ 6). Now, all we need to do is read the probability value where the p = 0.30 column and the ( n = 10, y = 6) row intersect. What do you get? Do you need a hint? The cumulative binomial probability table tells us that P ( Y ≤ 6) = P ( X ≥ 4) = 0.9894. That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.
Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don't qualify for favorable rates, then Y , which equals 10 − X , is a binomial random variable with n = 10 and q = 1 − p = 0.30. We are interested in finding P ( X ≥ 4). We can't use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewrite P ( X ≥ 4) as a probability statement in terms of Y : P ( X ≥ 4) = P (− X ≤ −4) = P (10 − X ≤ 10 − 4) = P ( Y ≤ 6) Now it's just a matter of looking up the probability in the right place on our cumulative binomial table. To find P ( Y ≤ 6), we: Find n = 10 in the first column on the left. Find the column containing p = 0.30 . Find the 6 in the second column on the left, since we want to find F (6) = P ( Y ≤ 6). Now, all we need to do is read the probability value where the p = 0.30 column and the ( n = 10, y = 6) row intersect. What do you get? Do you need a hint? The cumulative binomial probability table tells us that P ( Y ≤ 6) = P ( X ≥ 4) = 0.9894. That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.
Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don't qualify for favorable rates, then Y , which equals 10 − X , is a binomial random variable with n = 10 and q = 1 − p = 0.30. We are interested in finding P ( X ≥ 4). We can't use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewrite P ( X ≥ 4) as a probability statement in terms of Y : P ( X ≥ 4) = P (− X ≤ −4) = P (10 − X ≤ 10 − 4) = P ( Y ≤ 6) Now it's just a matter of looking up the probability in the right place on our cumulative binomial table. To find P ( Y ≤ 6), we: Find n = 10 in the first column on the left. Find the column containing p = 0.30 . Find the 6 in the second column on the left, since we want to find F (6) = P ( Y ≤ 6). Now, all we need to do is read the probability value where the p = 0.30 column and the ( n = 10, y = 6) row intersect. What do you get? Do you need a hint? The cumulative binomial probability table tells us that P ( Y ≤ 6) = P ( X ≥ 4) = 0.9894. That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.
Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don't qualify for favorable rates, then Y , which equals 10 − X , is a binomial random variable with n = 10 and q = 1 − p = 0.30. We are interested in finding P ( X ≥ 4). We can't use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewrite P ( X ≥ 4) as a probability statement in terms of Y : P ( X ≥ 4) = P (− X ≤ −4) = P (10 − X ≤ 10 − 4) = P ( Y ≤ 6) Now it's just a matter of looking up the probability in the right place on our cumulative binomial table. To find P ( Y ≤ 6), we: Find n = 10 in the first column on the left. Find the column containing p = 0.30 . Find the 6 in the second column on the left, since we want to find F (6) = P ( Y ≤ 6). Now, all we need to do is read the probability value where the p = 0.30 column and the ( n = 10, y = 6) row intersect. What do you get? Do you need a hint? The cumulative binomial probability table tells us that P ( Y ≤ 6) = P ( X ≥ 4) = 0.9894. That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.
Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don't qualify for favorable rates, then Y , which equals 10 − X , is a binomial random variable with n = 10 and q = 1 − p = 0.30. We are interested in finding P ( X ≥ 4). We can't use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewrite P ( X ≥ 4) as a probability statement in terms of Y : P ( X ≥ 4) = P (− X ≤ −4) = P (10 − X ≤ 10 − 4) = P ( Y ≤ 6) Now it's just a matter of looking up the probability in the right place on our cumulative binomial table. To find P ( Y ≤ 6), we: Find n = 10 in the first column on the left. Find the column containing p = 0.30 . Find the 6 in the second column on the left, since we want to find F (6) = P ( Y ≤ 6). Now, all we need to do is read the probability value where the p = 0.30 column and the ( n = 10, y = 6) row intersect. What do you get? Do you need a hint? The cumulative binomial probability table tells us that P ( Y ≤ 6) = P ( X ≥ 4) = 0.9894. That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.
Solution. If we let X denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with n = 10 and p = 0.70. And, if we let Y denote the number of subscribers who don't qualify for favorable rates, then Y , which equals 10 − X , is a binomial random variable with n = 10 and q = 1 − p = 0.30. We are interested in finding P ( X ≥ 4). We can't use the cumulative binomial tables, because they only go up to p = 0.50. The good news is that we can rewrite P ( X ≥ 4) as a probability statement in terms of Y : P ( X ≥ 4) = P (− X ≤ −4) = P (10 − X ≤ 10 − 4) = P ( Y ≤ 6) Now it's just a matter of looking up the probability in the right place on our cumulative binomial table. To find P ( Y ≤ 6), we: Find n = 10 in the first column on the left. Find the column containing p = 0.30 . Find the 6 in the second column on the left, since we want to find F (6) = P ( Y ≤ 6). Now, all we need to do is read the probability value where the p = 0.30 column and the ( n = 10, y = 6) row intersect. What do you get? Do you need a hint? The cumulative binomial probability table tells us that P ( Y ≤ 6) = P ( X ≥ 4) = 0.9894. That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.