Aresearcher was intecosted in seeing if there was a signilcant dillere.pdf

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Aresearcher was intecosted in seeing if there was a signilcant dillerense betwoen Ohio Univerity undergraduate and graduble students in lema of the average zumber of hours that they stidy on the weekend. So, he collected data troen two random samples (below) Use aphay m os. x. State tin mal and atemative thpotheces in uaget. ( 2 poins) ti. Det io the coteria be mating a seivion. That in, fre the offoa value ( i panti) E Consva the acorophale lesi usiaile. sHow youR woredi Beate complete try Nor noding the ymianct! (3 pousts) Compute the appropriate test-statistic. SHOW YOUR WOAKI Pleaso complete by NOT pooling the variancel (3 poinss) Calciated Numerator = Cavovaled Denominator = Calculated Final value = d. Evaluate the null typothesis (based on your answen to the above stepe). (Crow One) PEECT FML TO REHET (1 point e. Bined on your evaluation of the cill hypothais, what ia your condusice? (1 point) the cos 2 (3 points).

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EPS 525 – Introduction to Statistics Assignment No. 5 – One-way Analysis of Variance Name: A researcher conducted a study to examine the effects of secure, anxious, and avoidant attachment styles on the physiology of sleep. Participants were selected using a stratified random sampling approach to ensure representation of each of the three styles. The sleep patterns of 30 secure, 30 anxious, and 30 avoidant children were monitored. Of primary importance to the researcher was the overall percentage of time each child spent in deep (delta) sleep. Following is the average amount of time that each child spent in delta sleep, expressed as a percentage of total sleep time (ranging from 0.0 to 100.0). For the attachment styles, 1 = secure, 2 = anxious, and 3 = avoidant. Data Output for this Assignment is found on the last four pages. The gray boxes for your answers will expand as necessary for your responses. 1. (2 points) What would the null hypothesis be for this study? Show/write the appropriate symbols and expression in words. H0: . 2. (2 points) What would the alternative hypothesis be for this study? Show/write the appropriate symbols and expression in words. Ha: 3. Prior to examining whether the group means differ; you need to test the underlying assumptions of the one-way analysis of variance. 3.a. (2 points) Has the assumption of independence been met for this data? FORMCHECKBOX Yes FORMCHECKBOX No (check your answer selection) Indicate how you made your determination. 3.b. (3 points) Has the assumption of normality been met for this data, using an alpha level of .001? FORMCHECKBOX Yes FORMCHECKBOX No (check your answer selection) Indicate how you made your determination. Be sure to include all applicable values and symbols. 3.c. (3 points) Has the assumption of homogeneity of variance been met for this data, using an alpha level of .05? That is, is this assumption met or not met? FORMCHECKBOX Yes FORMCHECKBOX No (check your answer selection) Indicate how you came to your conclusion. Be sure to include all applicable values and symbols. 4. (2 points) The next question that needs to be answered is whether all of the groups are the same in their percentage of time in deep (delta) sleep using an alpha level of .05. If applicable (or indicate why not), use the Welch statistic. What is your conclusion (at this point) from this analysis? Indicate how you came to your conclusion. Be sure to include all applicable values and symbols. 5. (3 points) Calculate the measure of association and interpret its meaning. W T W B MS SS MS K SS + - - = ) 1 ( 2 wWhere SSB = K = MSW = SST = MSW = Therefore, (2 = This means: 6. (1 point) Write the statistical strand for this one-way ANOVA analysis. 7. (4 points) Assuming that you found a significant F, which pairs of groups differ? Indicate which post hoc procedure you used and why. Indicate your findings fr ...

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A computer supplies retailer wishes to open a store in one of two Pittsburgh suburbs, Monroeville and Greentree. To decide where to locate its store, the retailer conducts an e-mail survey of randomly selected households in each suburb, in order to measure the average yearly expenditure per household on computer supplies. 13 households from Monroeville responded to the survey: these households spent an average of $81, with a standard deviation of $20. 12 households from Greentree responded to the survey: these households spent an average of $74, with a standard deviation of $22. a) Let 1 and 2 represent the mean expenditures per household in Monroeville and Greentree respectively. Compute the 99% confidence interval for 1 - 2. b) Is the sample evidence sufficient to conclude which suburb spends more? To answer this question, perform a two-sided hypothesis test at = .05. State the alternative and null hypotheses, compute the observed significance level (p-value), and state the conclusion of your test. (For parts a and b, you should assume that both populations are approximately normally distributed, with equal variances.) In a much larger, follow-up survey, 563 households from Monroeville and 587 households from Greentree were polled. The Monroeville households spent an average of $76, with a standard deviation of $19, and the Greentree households spent an average of $80, with a standard deviation of $22. c) Let µ1 and µ2 represent the mean expenditures per household in Monroeville and Greentree respectively. Using the sample evidence from the follow-up survey, compute the 99% confidence interval for µ1 - µ2. d) Is the sample evidence from the follow-up survey sufficient to conclude which suburb spends more? To answer this question, perform a hypothesis test at α = .05. State the alternative and null hypotheses, compute the observed significance level (p-value), and state the conclusion of your test. 1. How are independent samples t-tests and One-way ANOVAs similar? How do the two tests differ? (Points : 12) 2. What is the F-ratio? That is, what makes up the top and bottom portions of the F-ratio? (Points : 10) 3. A social psychologist conducted a study examining the effectiveness of two persuasion techniques. She found the following information in her ANOVA table: SS between = 45.55 SS within = 30.22 SS total = 75.77 Using eta squared, calculate the effect size for her study? What does this mean? (Points : 10) 4. A researcher is interested in how caffeine affects stress. One group gets no caffeine and the other gets 3 cups of coffee. Complete the ANOVA table ANOVA Summary Table Source SS df MS F A-group 150 1 S/A (error) 25 18 Total 175 19 Fill-in the missing A-group MS value: (Points : 5) 5. A researcher is interested in how caffeine affects stress. One group gets no caffeine and the other gets 3 cups of coffee. Complete the ANOVA table ANOVA Summary Table Source SS df MS F A-group 150 1 .

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Chapter 11: Chi-Square Tests and ANOVA 359 Chapter 11: Chi-Square and ANOVA Tests This chapter presents material on three more hypothesis tests. One is used to determine significant relationship between two qualitative variables, the second is used to determine if the sample data has a particular distribution, and the last is used to determine significant relationships between means of 3 or more samples. Section 11.1: Chi-Square Test for Independence Remember, qualitative data is where you collect data on individuals that are categories or names. Then you would count how many of the individuals had particular qualities. An example is that there is a theory that there is a relationship between breastfeeding and autism. To determine if there is a relationship, researchers could collect the time period that a mother breastfed her child and if that child was diagnosed with autism. Then you would have a table containing this information. Now you want to know if each cell is independent of each other cell. Remember, independence says that one event does not affect another event. Here it means that having autism is independent of being breastfed. What you really want is to see if they are not independent. In other words, does one affect the other? If you were to do a hypothesis test, this is your alternative hypothesis and the null hypothesis is that they are independent. There is a hypothesis test for this and it is called the Chi-Square Test for Independence. Technically it should be called the Chi-Square Test for Dependence, but for historical reasons it is known as the test for independence. Just as with previous hypothesis tests, all the steps are the same except for the assumptions and the test statistic. Hypothesis Test for Chi-Square Test 1. State the null and alternative hypotheses and the level of significance Ho : the two variables are independent (this means that the one variable is not affected by the other) HA : the two variables are dependent (this means that the one variable is affected by the other) Also, state your α level here. 2. State and check the assumptions for the hypothesis test a. A random sample is taken. b. Expected frequencies for each cell are greater than or equal to 5 (The expected frequencies, E, will be calculated later, and this assumption means E ≥ 5 ). 3. Find the test statistic and p-value Finding the test statistic involves several steps. First the data is collected and counted, and then it is organized into a table (in a table each entry is called a cell). These values are known as the observed frequencies, which the symbol for an observed frequency is O. Each table is made up of rows and columns. Then each row is totaled to give a row total and each column is totaled to give a column total. Chapter 11: Chi-Squared Tests and ANOVA 360 The null hypothesis is that the variables are independent. Using the multiplication.

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Descriptive Statistics Formula Sheet Sample Population Characteristic statistic Parameter raw scores x, y, . . . . . X, Y, . . . . . mean (central tendency) M = ∑ x n μ = ∑ X N range (interval/ratio data) highest minus lowest value highest minus lowest value deviation (distance from mean) Deviation = (x − M ) Deviation = (X − μ ) average deviation (average distance from mean) ∑(x − M ) n = 0 ∑(X − μ ) N sum of the squares (SS) (computational formula) SS = ∑ x 2 − (∑ x)2 n SS = ∑ X2 − (∑ X)2 N variance ( average deviation2 or standard deviation 2 ) (computational formula) s2 = ∑ x2 − (∑ x)2 n n − 1 = SS df σ2 = ∑ X2 − (∑ X)2 N N standard deviation (average deviation or distance from mean) (computational formula) s = √∑ x 2 − (∑ x)2 n n − 1 σ = √∑ X 2 − (∑ X)2 N N Z scores (standard scores) mean = 0 standard deviation = ± 1.0 Z = x − M s = deviation stand. dev. X = M + Zs Z = X − μ σ X = μ + Zσ Area Under the Normal Curve -1s to +1s = 68.3% -2s to +2s = 95.4% -3s to +3s = 99.7% Using Z Score Table for Normal Distribution (Note: see graph and table in A-23) for percentiles (proportion or %) below X for positive Z scores – use body column for negative Z scores – use tail column for proportions or percentage above X for positive Z scores – use tail column for negative Z scores – use body column to discover percentage / proportion between two X values 1. Convert each X to Z score 2. Find appropriate area (body or tail) for each Z score 3. Subtract or add areas as appropriate 4. Change area to % (area × 100 = %) Regression lines (central tendency line for all points; used for predictions only) formula uses raw scores b = slope a = y-intercept y = bx + a (plug in x to predict y) b = ∑ xy − (∑ x)(∑ y) n ∑ x2 − (∑ x)2 n a = My - bMx where My is mean of y and Mx is mean of x SEest (measures accuracy of predictions; same properties as standard deviation) Pearson Correlation Coefficient (used to measure relationship; uses Z scores) r = ∑ xy− (∑ x)(∑ y) n √(∑ x2− (∑ x)2 n )(∑ y2− (∑ y)2 n ) r = degree x & 𝑦 𝑣𝑎𝑟𝑦 𝑡𝑜𝑔𝑒𝑡ℎ𝑒𝑟 degree x & 𝑦 𝑣𝑎𝑟𝑦 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑒𝑙𝑦 r 2 = estimate or % of accuracy of predictions PSYC 2317 Mark W. Tengler, M.S. Assignment #9 Hypothesis Testing 9.1 Briefly explain in your own words the advantage of using an alpha level (α) = .01 versus an α = .05. In general, what is the disadvantage of using a smaller alpha level? 9.2 Discuss in your own words the errors that can be made in hypothesis testing. a. What is a type I error? Why might it occur? b. What is a type II error? How does it happen? 9.3 The term error is used in two different ways in the context of a hypothesis test. First, there is the concept of sta

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Answer Sheet ISDS 361B HW5 YOUR FULL NAME: Answer Optional for Partial Credit 1. [12 pts.] Forecast for Saturday in week 6: No. of Periods of Data Collected = No. of Periods in Season = MSE = MAD = MAPE = LAD = 2. [18 pts.] a. Forecast (exponential smoothing): No. of Periods of Data Collected = Smoothing Constant (alpha) = Initial Forecast Value = b. Forecast (weighted MA): MSE = MAD = MAPE = LAD = c. I would recommend weighted MA (True/ False): 3. [12 pts.] Forecast for year 8: MSE = MAD = MAPE = LAD = 4. [18 pts.] Forecast for Summer Year 5: No. of Periods of Data Collected = No. of Periods in Season = Coefficients: Intercept = Period = 1 = 2 = Page 1 of 2 1) What is the critical F value for a sample of 7 observations in the numerator and 6 in the denominator? Use a two-tailed test and the 0.1 significance level. (Round your answer to 2 decimal places.) F 2) Arbitron Media Research Inc. conducted a study of the iPod listening habits of men and women. One facet of the study involved the mean listening time. It was discovered that the mean listening time for men was 29 minutes per day. The standard deviation of the sample of the 10 men studied was 8 minutes per day. The mean listening time for the 12 women studied was also 29 minutes, but the standard deviation of the sample was 15 minutes. At the .10 significance level, can we conclude that there is a difference in the variation in the listening times for men and women? (Round your answer to 3 decimal places.) The test statistic is . Decision 3) The following is sample information. Test the hypothesis that the treatment means are equal. Use the 0.01 significance level. Treatment 1 Treatment 2 Treatment 3 7 4 4 4 5 7 6 5 6 6 4 5 (a) State the null hypothesis and the alternate hypothesis. H0 H1 (b) What is the decision rule? (Round your answer to 2 decimal places.) H0 if the test statistic is greater than . (c&d) Compute SST, SSE, and SS total and complete an ANOVA table. (Round SS, MS and F values to 3 decimal places.) Source SS df MS F Treatments Error Total (e) State your decision regarding the null hypothesis. H0 4) A senior accounting major at Midsouth State University has job offers from four CPA firms. To explore the offers further, she asked a sample of recent trainees how many months each worked for the firm before receiving a raise in salary. The sample information is submitted to MINITAB with the following results: Analysis of Variance Source DF SS MS F Factor 5 36.39 7.28 1.92 Error 12 45.54 3.80 Total 17 81.93 Reject if F > . (Round your answer to 2 .

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36086 Topic: Discussion3 Number of Pages: 2 (Double Spaced) Number of sources: 1 Writing Style: APA Type of document: Essay Academic Level:Master Category: Psychology Language Style: English (U.S.) Order Instructions: Attached I will upload the instructions Reference/Module Learning Objectives •Explain what the x2 goodness-of-fit test is and what it does. •Calculate a x2 goodness-of-fit test. •List the assumptions of the x2 goodness-of-fit test. •Calculate the x2 test of independence. •Interpret the x2 test of independence. •Explain the assumptions of the x2 test of independence. The Chi-Square (x2) Goodness-of-Fit test: What It Is and What It Does The chi-square (x2) goodness-of-fit test is used for comparing categorical information against what we would expect based on previous knowledge. As such, it tests what are called observed frequencies (the frequency with which participants fall into a category) against expected frequencies (the frequency expected in a category if the sample data represent the population). It is a nondirectional test, meaning that the alternative hypothesis is neither one-tailed nor two-tailed. The alternative hypothesis for a x2 goodness-of-fit test is that the observed data do not fit the expected frequencies for the population, and the null hypothesis is that they do fit the expected frequencies for the population. There is no conventional way to write these hypotheses in symbols, as we have done with the previous statistical tests. To illustrate the x2 goodness-of-fit test, let's look at a situation in which its use would be appropriate. chi-square (x2) goodness-of-fit test A nonparametric inferential procedure that determines how well an observed frequency distribution fits an expected distribution. observed frequencies The frequency with which participants fall into a category. expected frequencies The frequency expected in a category if the sample data represent the population. Calculations for the x2 Goodness-of-Fit Test Suppose that a researcher is interested in determining whether the teenage pregnancy rate at a particular high school is different from the rate statewide. Assume that the rate statewide is 17%. A random sample of 80 female students is selected from the target high school. Seven of the students are either pregnant now or have been pregnant previously. The χ2goodness-of-fit test measures the observed frequencies against the expected frequencies. The observed and expected frequencies are presented in Table 21.1. TABLE 21.1Observed and expected frequencies for χ2 goodness-of-fit example FREQUENCIES PREGNANT NOT PREGNANT Observed 7 73 Expected 14 66 As can be seen in the table, the observed frequencies represent the number of high school females in the sample of 80 who were pregnant versus not pregnant. The expected frequencies represent what we would expect based on chance, given what is known about the population. In this case, we would expect 17% of the females to be pregnant .

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