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I have a Masters in Statistics from Leeds Trinity University.
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In hypothesis testing, we have two hypotheses a null hypothesis a.docxbradburgess22840
In hypothesis testing, we have two hypotheses: a null hypothesis and an alternative hypothesis. The alternative hypothesis is typically what we want to demonstrate (based on the research question). We collect data to see if a certain population value differs from a given value (≠), is less than a given value (<), or is greater than a given value (>). The null hypothesis is typically a baseline or a known standard against which we are testing. For example: If we want to test to see if a majority of voters voted for a certain candidate, then our alternative hypothesis would be that the population proportion who voted for the candidate is greater than 0.50 (i.e. p > 0.50). This is what we want to demonstrate and is the reason for collecting data. The null hypothesis would be that the population proportion who voted for the candidate is 0.50 (i.e. p = 0.50) which would not be a majority. This is the baseline against which we are testing. Note that the alternative hypothesis covers a range of values, but the null hypothesis is just the one value (i.e. equality).
1. A polling group surveyed a city in Scotland regarding residents’ opinions on independence from the UK. It is generally believed that the percentage of ‘Yes’ votes is 50%. The poll wants to find out whether greater than half (> 50%) of the residents will vote ‘Yes.’ The survey polled 2000 residents, of which 1050 responded that they will vote ‘Yes’ on Scotland independence (52.5%). What are the null and alternative hypotheses?
A) Null: the percentage of ‘Yes’ votes is 52.5%; Alternative: the percentage of ‘Yes’ votes is greater than 52.5%
B) Null: the percentage of ‘Yes’ votes is greater than 52.5%; Alternative: the percentage of ‘Yes’ votes is 52.5%
C) Null: the percentage of ‘Yes’ votes is 50%; Alternative: the percentage of ‘Yes’ votes is greater than 50%
D) Null: the percentage of ‘Yes’ votes is greater than 50%; Alternative: the percentage of ‘Yes' votes is 50%
2. For patients with a particular disease, the population proportion of those successfully treated with a standard treatment that has been used for many years is 0.75. A medical research group invents a new treatment that they believe will be more successful, i.e. the population proportion will exceed 0.75. A doctor plans a clinical trial he hopes will prove this claim. A sample of 100 patients with the disease is obtained. Each person is treated with the new treatment and eventually classified as having either been successfully or not successfully treated with the new treatment. Out of 100 patients, 80 (80%) were successfully treated by the new treatment. What are the null and alternative hypotheses?
A) Null: the population proportion of those successfully treated by the new treatment exceeds 0.75 (p > 0.75); Alternative: the population proportion of those successfully treated by the new treatment is 0.75 (p = 0.75)
B) Null: the population proportion of those successfully treated by the new treatment is 0.75 (p = 0.75); Al.
Inferential statistics are often used to compare the differences between the treatment groups. Inferential statistics use measurements from the sample of subjects in the experiment to compare the treatment groups and make generalizations about the larger population of subjects.
Confidence Intervals in the Life Sciences PresentationNamesS.docxmaxinesmith73660
Confidence Intervals in the Life Sciences Presentation
Names
Statistics for the Life Sciences STAT/167
Date
Fahad M. Gohar M.S.A.S
1
Conservation Biology of Bears
Normal Distribution
Standard normal distribution
Confidence Interval
Population Mean
Population Variance
Confidence Level
Point Estimate
Critical Value
Margin of Error
Welcome to the presentation on Confidence Intervals of Conservation Biology on Bears.
The team will define normal distribution and use an example of variables why this is important. A standard and normal distribution is discussed as well as the difference between standard and other normal distributions. Confidence interval will be defined and how it is used in Conservation Biology and Bears. We will learn how a confidence interval helps researchers estimate of population mean and population variance. The presenters defined a point estimate and try to explain how a point estimate found from a confidence interval. Confidence level is defined and a short explanation of confidence level is related to the confidence interval. Lastly, a critical value and margin of error are explained with examples from the Statdisk.
2
Normal Distribution
A normal distribution is one which has the mean, median, and mode are the same and the standard deviations are apart from the mean in the probabilities that go with the empirical rule. Not all data has the measures of central tendency, since some data sets may not have one unique value which occurs more than once. But every data set has a mean and median. The mean is only good with interval and ratio data, while the median can be used with interval, ratio and ordinal data. Mean is used when they're a lot of outliers, and median is used when there are few.
The normal distribution is continuous, and has only two parameters - mean and variance. The mean can be any positive number and variance can be any positive number (can't be negative - the mean and variance), so there are an infinite number of normal distributions. You want your data to represent the population distribution because when you make claims from the distribution of the sample you took, you want it to represent the whole entire population.
Some examples in the business world: Some industries which use normal distributions are pharmaceutical companies. They model the average blood pressure through normal distributions, and can make medicine which will help majority of the people with high blood pressure. A company can also model its average time to create something using the normal distribution. Several statistics can be calculated with the normal distribution, and hypothesis tests can be done with the normal distribution which models the average time.
Our chosen life science is BEARS. The age of the bears can be modeled by normal distributions and it is important to monitor since that tells us the average age of the bear, and can tell us a lot about the population. If the mean is high and the standard deviatio.
STATUse the information below to answer Questions 1 through 4..docxdessiechisomjj4
STAT
Use the information below to answer Questions 1 through 4.
Given a sample size of 36, with sample mean 670.3 and sample standard deviation 114.9, we perform the following hypothesis test.
Null Hypothesis
Alternative Hypothesis
1. What is the test statistic?
2. At a 10% significance level (90% confidence level), what is the critical value in this test? Do we reject the null hypothesis?
3. What are the border values between acceptance and rejection of this hypothesis?
4. What is the power of this test if the assumed true mean were 710 instead of 700?.
Questions 5 through 8 involve rolling of dice.
5. Given a fair, six-sided die, what is the probability of rolling the die twice and getting a “1” each time?
6. What is the probability of getting a “1” on the second roll when you get a “1” on the first roll?
7. The House managed to load the die in such a way that the faces “2” and “4” show up twice as frequently as all other faces. Meanwhile, all the other faces still show up with equal frequency. What is the probability of getting a “1” when rolling this loaded die?
8. Write the probability distribution for this loaded die, showing each outcome and its probability. Also plot a histogram to show the probability distribution.
Use the data in the table to answer Questions 9 through 11.
x
3
1
4
4
5
y
1
-2
3
5
9
9. Determine SSxx, SSxy, and SSyy.
10.
Find the equation of the regression line. What is the predicted value when
11. Is the correlation significant at 1% significance level (99% confidence level)? Why or why not?
Use the data below to answer Questions 12 through 14.
A group of students from three universities were asked to pick their favorite college sport to attend of their choice: The results, in number of students, are listed as follows:
Football
Basketball
Soccer
Maryland
60
70
20
Duke
10
75
15
UCLA
35
65
25
Supposed a student is randomly selected from the group mentioned above.
12. What is the probability that the student is from UCLA or chooses football?
13. What is the probability that the student is from Duke, given that the student chooses basketball?
14. What is the probability that the student is from Maryland and chooses soccer?
Use the information below to answer Questions 15 and 17.
There are 3600 apples in a shipment. The weight of the apples in this shipment is normally distributed. It is found that it a mean weight of 14 ounces with a standard deviation of 2.5 ounces.
15. How many of apples have weights between 13 ounces and 15 ounces?
16. What is the probability that a randomly selected mango weighs less than 12.5 ounces?
17. A quality inspector randomly selected 100 apples from the shipment.
a. What is the probability that the 100 randomly selected apples have a mean weight less than 12.5 ounces?
b. Do you come up with the same result in Question 16? Why or why not?
18. A pharmaceutical company has developed a screening test for a rare disease that afflicted 2% of the population. Un.
I am Hannah Dennis. Currently associated with excelhomeworkhelp.com as Excel homework helper. After completing my master's from Kean University, USA. I was in search of an opportunity that expands my area of knowledge hence I decided to help students with their homework. I have written several excel homework till date to help students overcome numerous difficulties they face.
In hypothesis testing, we have two hypotheses a null hypothesis a.docxbradburgess22840
In hypothesis testing, we have two hypotheses: a null hypothesis and an alternative hypothesis. The alternative hypothesis is typically what we want to demonstrate (based on the research question). We collect data to see if a certain population value differs from a given value (≠), is less than a given value (<), or is greater than a given value (>). The null hypothesis is typically a baseline or a known standard against which we are testing. For example: If we want to test to see if a majority of voters voted for a certain candidate, then our alternative hypothesis would be that the population proportion who voted for the candidate is greater than 0.50 (i.e. p > 0.50). This is what we want to demonstrate and is the reason for collecting data. The null hypothesis would be that the population proportion who voted for the candidate is 0.50 (i.e. p = 0.50) which would not be a majority. This is the baseline against which we are testing. Note that the alternative hypothesis covers a range of values, but the null hypothesis is just the one value (i.e. equality).
1. A polling group surveyed a city in Scotland regarding residents’ opinions on independence from the UK. It is generally believed that the percentage of ‘Yes’ votes is 50%. The poll wants to find out whether greater than half (> 50%) of the residents will vote ‘Yes.’ The survey polled 2000 residents, of which 1050 responded that they will vote ‘Yes’ on Scotland independence (52.5%). What are the null and alternative hypotheses?
A) Null: the percentage of ‘Yes’ votes is 52.5%; Alternative: the percentage of ‘Yes’ votes is greater than 52.5%
B) Null: the percentage of ‘Yes’ votes is greater than 52.5%; Alternative: the percentage of ‘Yes’ votes is 52.5%
C) Null: the percentage of ‘Yes’ votes is 50%; Alternative: the percentage of ‘Yes’ votes is greater than 50%
D) Null: the percentage of ‘Yes’ votes is greater than 50%; Alternative: the percentage of ‘Yes' votes is 50%
2. For patients with a particular disease, the population proportion of those successfully treated with a standard treatment that has been used for many years is 0.75. A medical research group invents a new treatment that they believe will be more successful, i.e. the population proportion will exceed 0.75. A doctor plans a clinical trial he hopes will prove this claim. A sample of 100 patients with the disease is obtained. Each person is treated with the new treatment and eventually classified as having either been successfully or not successfully treated with the new treatment. Out of 100 patients, 80 (80%) were successfully treated by the new treatment. What are the null and alternative hypotheses?
A) Null: the population proportion of those successfully treated by the new treatment exceeds 0.75 (p > 0.75); Alternative: the population proportion of those successfully treated by the new treatment is 0.75 (p = 0.75)
B) Null: the population proportion of those successfully treated by the new treatment is 0.75 (p = 0.75); Al.
Inferential statistics are often used to compare the differences between the treatment groups. Inferential statistics use measurements from the sample of subjects in the experiment to compare the treatment groups and make generalizations about the larger population of subjects.
Confidence Intervals in the Life Sciences PresentationNamesS.docxmaxinesmith73660
Confidence Intervals in the Life Sciences Presentation
Names
Statistics for the Life Sciences STAT/167
Date
Fahad M. Gohar M.S.A.S
1
Conservation Biology of Bears
Normal Distribution
Standard normal distribution
Confidence Interval
Population Mean
Population Variance
Confidence Level
Point Estimate
Critical Value
Margin of Error
Welcome to the presentation on Confidence Intervals of Conservation Biology on Bears.
The team will define normal distribution and use an example of variables why this is important. A standard and normal distribution is discussed as well as the difference between standard and other normal distributions. Confidence interval will be defined and how it is used in Conservation Biology and Bears. We will learn how a confidence interval helps researchers estimate of population mean and population variance. The presenters defined a point estimate and try to explain how a point estimate found from a confidence interval. Confidence level is defined and a short explanation of confidence level is related to the confidence interval. Lastly, a critical value and margin of error are explained with examples from the Statdisk.
2
Normal Distribution
A normal distribution is one which has the mean, median, and mode are the same and the standard deviations are apart from the mean in the probabilities that go with the empirical rule. Not all data has the measures of central tendency, since some data sets may not have one unique value which occurs more than once. But every data set has a mean and median. The mean is only good with interval and ratio data, while the median can be used with interval, ratio and ordinal data. Mean is used when they're a lot of outliers, and median is used when there are few.
The normal distribution is continuous, and has only two parameters - mean and variance. The mean can be any positive number and variance can be any positive number (can't be negative - the mean and variance), so there are an infinite number of normal distributions. You want your data to represent the population distribution because when you make claims from the distribution of the sample you took, you want it to represent the whole entire population.
Some examples in the business world: Some industries which use normal distributions are pharmaceutical companies. They model the average blood pressure through normal distributions, and can make medicine which will help majority of the people with high blood pressure. A company can also model its average time to create something using the normal distribution. Several statistics can be calculated with the normal distribution, and hypothesis tests can be done with the normal distribution which models the average time.
Our chosen life science is BEARS. The age of the bears can be modeled by normal distributions and it is important to monitor since that tells us the average age of the bear, and can tell us a lot about the population. If the mean is high and the standard deviatio.
STATUse the information below to answer Questions 1 through 4..docxdessiechisomjj4
STAT
Use the information below to answer Questions 1 through 4.
Given a sample size of 36, with sample mean 670.3 and sample standard deviation 114.9, we perform the following hypothesis test.
Null Hypothesis
Alternative Hypothesis
1. What is the test statistic?
2. At a 10% significance level (90% confidence level), what is the critical value in this test? Do we reject the null hypothesis?
3. What are the border values between acceptance and rejection of this hypothesis?
4. What is the power of this test if the assumed true mean were 710 instead of 700?.
Questions 5 through 8 involve rolling of dice.
5. Given a fair, six-sided die, what is the probability of rolling the die twice and getting a “1” each time?
6. What is the probability of getting a “1” on the second roll when you get a “1” on the first roll?
7. The House managed to load the die in such a way that the faces “2” and “4” show up twice as frequently as all other faces. Meanwhile, all the other faces still show up with equal frequency. What is the probability of getting a “1” when rolling this loaded die?
8. Write the probability distribution for this loaded die, showing each outcome and its probability. Also plot a histogram to show the probability distribution.
Use the data in the table to answer Questions 9 through 11.
x
3
1
4
4
5
y
1
-2
3
5
9
9. Determine SSxx, SSxy, and SSyy.
10.
Find the equation of the regression line. What is the predicted value when
11. Is the correlation significant at 1% significance level (99% confidence level)? Why or why not?
Use the data below to answer Questions 12 through 14.
A group of students from three universities were asked to pick their favorite college sport to attend of their choice: The results, in number of students, are listed as follows:
Football
Basketball
Soccer
Maryland
60
70
20
Duke
10
75
15
UCLA
35
65
25
Supposed a student is randomly selected from the group mentioned above.
12. What is the probability that the student is from UCLA or chooses football?
13. What is the probability that the student is from Duke, given that the student chooses basketball?
14. What is the probability that the student is from Maryland and chooses soccer?
Use the information below to answer Questions 15 and 17.
There are 3600 apples in a shipment. The weight of the apples in this shipment is normally distributed. It is found that it a mean weight of 14 ounces with a standard deviation of 2.5 ounces.
15. How many of apples have weights between 13 ounces and 15 ounces?
16. What is the probability that a randomly selected mango weighs less than 12.5 ounces?
17. A quality inspector randomly selected 100 apples from the shipment.
a. What is the probability that the 100 randomly selected apples have a mean weight less than 12.5 ounces?
b. Do you come up with the same result in Question 16? Why or why not?
18. A pharmaceutical company has developed a screening test for a rare disease that afflicted 2% of the population. Un.
I am James S. I am a Probability Homework Expert at statisticshomeworkhelper.com. I hold a Master's in Statistics, from Melbourne, Australia. I have been helping students with their homework for the past 8 years. I solved homework related to Probability .
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Approximately 20 of 1000 fish in a pond are affected with a skin .docxrossskuddershamus
Approximately 20% of 1000 fish in a pond are affected with a skin disease. A random sample of 20 fish are selected. What is the mean of the sampling distribution for the proportion of your sample that is infected? What is the standard deviation of the sampling distribution for the proportion of your sample that is infected?
Chapter 4, Section 3, Exercise 075
Match the p-values with the appropriate conclusion:
(a) The evidence against the null hypothesis is significant, but only at the 10% level.
(b) The evidence against the null and in favor of the alternative is very strong.
(c) There is not enough evidence to reject the null hypothesis, even at the 10% level.
(d) The result is significant at a 5% level but not at a 1% level.
Chapter 4, Section 3, Exercise 082
Sleep or Caffeine for Memory?
The consumption of caffeine to benefit alertness is a common activity practiced by 90% of adults in North America. Often caffeine is used in order to replace the need for sleep. One recent study1 compares students' ability to recall memorized information after either the consumption of caffeine or a brief sleep. A random sample of 35 adults (between the ages of 18-39 ) were randomly divided into three groups and verbally given a list of 24 words to memorize. During a break, one of the groups takes a nap for an hour and a half, another group is kept awake and then given a caffeine pill an hour prior to testing, and the third group is given a placebo. The response variable of interest is the number of words participants are able to recall following the break. The summary statistics for the three groups are in the table below. We are interested in testing whether there is evidence of a difference in average recall ability between any two of the treatments. Thus we have three possible tests between different pairs of groups: Sleep vs Caffeine, Sleep vs Placebo, and Caffeine vs Placebo.
Group
Sample size
Mean
Standard Deviation
Sleep
12
15.25
3.3
Caffeine
12
12.25
3.5
Placebo
11
13.70
3.0
1 Mednick, Cai, Kanady, and Drummond, "Comparing the benefits of caffeine, naps and placebo on verbal, motor and perceptual memory", Behavioural Brain Research, 193 (2008), 79-86.
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(a) In the test comparing the sleep group to the caffeine group, the p-value is 0.003.
What is the conclusion of the test?
H0.
In the sample , which group had better recall ability?
According to the test results, do you think sleep is really better than caffeine for recall ability?
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(b) In the test comparing the sleep group to the placebo group, the p-value is 0.06.
What is the conclusion of the test, using a 5% significance level?
H0.
What is the conclusion of the test, if we use a 10% significance level?
H0.
How strong is the evidence of a difference in mean recall ability between these two treatments?
Warn.
InstructionDue Date 6 pm on October 28 (Wed)Part IProbability a.docxdirkrplav
InstructionDue Date: 6 pm on October 28 (Wed)
Part IProbability and Sampling Distributions1.Thinking about probability statements. Probability is measure of how likely an event is to occur. Match one of probabilities that follow with each statement of likelihood given (The probability is usually a more exact measure of likelihood than is the verbal statement.)Answer0 0.01 0.3 0.6 0.99 1(a) This event is impossible. It can never occur.(b) This event is certain. It will occur on every trial.(c) This event is very unlikely, but it will occur once in a while in a long sequence of trials.(d) This event will occur more often that not.2. Spill or Spell? Spell-checking software catches "nonword errors" that result in a string of letters that is not a word, as when "the" is typed as "the." When undergraduates are asked to write a 250-word essay (without spell-checking), the number X of nonword errors has the following distribution:Value of X01234Probability0.10.20.30.30.1(a) Check that this distribution satisfies the two requirements for a legitimate assignment of probabilities to individual outcomes.(b) Write the event "at least one nonword error" in term of X (for example, P(X >3)). What is the probability of this event?(c) Describe the event X ≤ 2 in words. What is its probability? 3. Discrete or continuous? For each exercise listed below, decide whether the random variable described is discrete or continuous and explains the sample space.(a) Choose a student in your class at random. Ask how much time that student spent studying during the past 24 hours.(b) In a test of a new package design, you drop a carton of a dozen eggs from a height of 1 foot and count the number of broken eggs.(c) A nutrition researcher feeds a new diet to a young male white rat. The response variable is the weight (in grams) that the rat gains in 8 weeks.4. Tossing Coins(a) The distribution of the count X of heads in a single coin toss will be as follows. Find the mean number of heads and the variance for a single coin toss.Number of Heads (Xi)01mean:Probability (Pi)0.50.5variance:(b) The distribution of the count X of heads in four tosses of a balanced coin was as follows but some missing probabilities. Fill in the blanks and then find the mean number of heads and the variance for the distribution with assumption that the tosses are independent of each other.Number of Heads (Xi)01234mean:Probability (Pi)0.06250.0625variance:(c) Show that the two results of the means (i.e. single toss and four tosses) are related by the addition rule for means. (d) Show that the two results of the variances (i.e. single toss and four tosses) are related by the addition rule for variances (note: It was assumed that the tosses are independent of each other). 5. Generating a sampling distribution. Let's illustrate the idea of a sampling distribution in the case of a very small sample from a very small .
Midterm 2 – Practice Exercises
1
1. The amount of material used in making a custom sail for a sailboat is normally
distributed with a standard deviation of 64 square feet. For a random sample of 15
sails, the mean amount of material used is 912 square feet. Which of the following
represents a 99% confidence interval for the population mean amount of material
used in a custom sail?
A. 912 ± 49.2
B. 912 ± 42.6
C. 912 ± 44.3
D. 912 ± 46.8
2. The number of beverage cans produced each hour from a vending machine is
normally distributed with a standard deviation of 8.6. For a random sample of 12
hours, the average number of beverage cans produced was 326.0. Assume a 99%
confidence interval for the population mean number of beverage cans produced
per hour. Calculate the margin of error of the 99% confidence interval.
A. 1.85
B. 3.60
C. 6.41
D. 10.56
3. The number of beverage cans produced each hour from a vending machine is
normally distributed with a standard deviation of 8.6. For a random sample of 12
hours, the average number of beverage cans produced was 326.0. Assume a 99%
confidence interval for the population mean number of beverage cans produced
per hour. Find the upper confidence limit of the 99% confidence interval.
A. 340.25
B. 325.98
C. 319.59
D. 332.41
4. If we change a 95% confidence interval estimate to a 99% confidence interval
estimate, we can expect
A. the size of the confidence interval to increase
B. the size of the confidence interval to decrease
C. the size of the confidence interval to remain the same
D. the sample size to increase
Midterm 2 – Practice Exercises
2
5. If a sample has 20 observations and a 90% confidence estimate for µ is needed,
the appropriate t‐score is:
A. 2.120
B. 1.746
C. 2.131
D. 1.729
6. We are interested in conducting a study to determine what percentage of voters
would vote for the incumbent member of parliament. What is the minimum size
sample needed to estimate the population proportion with a margin of error of
0.07 or less at 95% confidence?
A. 200
B. 100
C. 58
D. 196
7. The sample size needed to provide a margin of error of 2 or less with a 0.95
confidence coefficient when the population standard deviation equals 11 is
A. 10
B. 11
C. 116
D. 117
8. The manager of the local health club is interested in determining the number of
times members use the weight room per month. She takes a random sample of 15
members and finds that over the course of a month, the average number of visits
was 11.2 with a standard deviation of 3.2. Assuming that the monthly number of
visits is normally distributed, which of the following represents a 95% confidence
interval for the average monthly usage of all health club members?
A. 11.2 ± 1.74
B. 11.2 ± 1.77
C. 11.2 ± 1.62
D. 11.2 ± 1.83
Midterm 2 – Practice Exercises
3
9. The s.
Midterm 2 – Practice Exercises
1
1. The amount of material used in making a custom sail for a sailboat is normally
distributed with a standard deviation of 64 square feet. For a random sample of 15
sails, the mean amount of material used is 912 square feet. Which of the following
represents a 99% confidence interval for the population mean amount of material
used in a custom sail?
A. 912 ± 49.2
B. 912 ± 42.6
C. 912 ± 44.3
D. 912 ± 46.8
2. The number of beverage cans produced each hour from a vending machine is
normally distributed with a standard deviation of 8.6. For a random sample of 12
hours, the average number of beverage cans produced was 326.0. Assume a 99%
confidence interval for the population mean number of beverage cans produced
per hour. Calculate the margin of error of the 99% confidence interval.
A. 1.85
B. 3.60
C. 6.41
D. 10.56
3. The number of beverage cans produced each hour from a vending machine is
normally distributed with a standard deviation of 8.6. For a random sample of 12
hours, the average number of beverage cans produced was 326.0. Assume a 99%
confidence interval for the population mean number of beverage cans produced
per hour. Find the upper confidence limit of the 99% confidence interval.
A. 340.25
B. 325.98
C. 319.59
D. 332.41
4. If we change a 95% confidence interval estimate to a 99% confidence interval
estimate, we can expect
A. the size of the confidence interval to increase
B. the size of the confidence interval to decrease
C. the size of the confidence interval to remain the same
D. the sample size to increase
Midterm 2 – Practice Exercises
2
5. If a sample has 20 observations and a 90% confidence estimate for µ is needed,
the appropriate t‐score is:
A. 2.120
B. 1.746
C. 2.131
D. 1.729
6. We are interested in conducting a study to determine what percentage of voters
would vote for the incumbent member of parliament. What is the minimum size
sample needed to estimate the population proportion with a margin of error of
0.07 or less at 95% confidence?
A. 200
B. 100
C. 58
D. 196
7. The sample size needed to provide a margin of error of 2 or less with a 0.95
confidence coefficient when the population standard deviation equals 11 is
A. 10
B. 11
C. 116
D. 117
8. The manager of the local health club is interested in determining the number of
times members use the weight room per month. She takes a random sample of 15
members and finds that over the course of a month, the average number of visits
was 11.2 with a standard deviation of 3.2. Assuming that the monthly number of
visits is normally distributed, which of the following represents a 95% confidence
interval for the average monthly usage of all health club members?
A. 11.2 ± 1.74
B. 11.2 ± 1.77
C. 11.2 ± 1.62
D. 11.2 ± 1.83
Midterm 2 – Practice Exercises
3
9. The s ...
ELEMENTS OF STATISTICS / TUTORIALOUTLET DOT COMalbert0076
Unit 3 Problem Set NAME: Elements of Statistics--FHSU Virtual College--Spring 2017
REMEMBER, these are assessed preparatory problems related to the content of Unit 3. The Unit 3 Exam will consist of similar types of
Lecture 6 Point and Interval Estimation.pptxshakirRahman10
Point and Interval Estimation:
Objectives:
Apply the basics of inferential statistics in terms of point estimation.
Compute point and estimation of population means and confidence interval.
Interpret the results of point and interval estimation.
Estimation:
Estimating the value of parameter from the sample:
An aspect of inferential statistics.
Why to estimate: Population is large enough so we
can only estimate.
Types of estimation:
Point Estimation:
A specified number value (single value) that is an estimate of a population parameter. The point estimate of the population mean µ is the sample mean.
Interval Estimate:
Range of values to estimate about population parameter.
Confidence Interval Estimation:
Range of values to estimate about population parameter.
May contain the parameter or not (Degree of confidence).
Ranges between two values.
Example:
Age (in years) 4 BScN students: 20<µ < 25 or (22.5 +2.5)
FORMULA:
Point estimate (x) + Critical Value x Standard Error.
Confidence Interval is a particular interval of estimate.
Given that sample size is large, the 95% of the sample means taken from same population and same sample size will fall in + 1.96 SD of the population mean.
Three commonly used Confidence Intervals are 90%, 95% (by default) , and 99%.
Why not too small or too large confidence intervals?
Too wide: 99.9% Interval too broad
Too narrow: 80 % More uncertainty to have population mean.
The 99% of the sample means taken from same population and
same sample size will fall in + 2.575 SD of the population mean.
Interpretation:
99% probability that interval will enclose population parameter and 1% chance that it will not have population parameter.
Level of confidence: The level of certainty that the interval will have the true population mean.
Chances of Error: Chances that the interval will not cater the true parameter.
Sum of level of confidence and chances of error =100%
Question 1. [12 marks]Market research has indicated that custo.docxIRESH3
Question 1. [12 marks]
Market research has indicated that customers are likely to bypass Roma tomatoes that weigh less than 70 grams. A produce company produces Roma tomatoes that average 74.0 grams with a standard deviation of 3.2 grams.
(a) [2 marks] Assuming that the normal distribution is a reasonable model for the
weights of these tomatoes, what proportion of Roma tomatoes are currently undersize (less than 70g)?
(b) [2 marks] How much must a Roma tomato weigh to be among the heaviest 10%?
(c ) [2 marks] The aim of the current research is to reduce the proportion of
undersized tomatoes to no more than 2%. One way of reducing this proportion is to reduce the standard deviation. If the average size of the tomatoes remains 74.0 grams, what must the target standard deviation be to achieve the 2% goal?
(d) [3 marks] The company claims that the goal of 2% undersized tomatoes is
reached.To test this, a random sample of 25 tomatoes is taken. What is the
distribution of undersized tomatoes in this sample if the company's claim is true?
Explain your reasoning.
Question 2:
In an article in Marketing Science, Silk and Berndt investigate the output of advertising agencies. They describe ad agency output by finding the shares of dollar billing volume coming from various media categories such as network television, spot television, newspaper, radio, and so forth.
Suppose that a random sample of 400 U.S. advertising agencies gives an average percentage share of billing volume from network television equal to 7.46 percent with a standard deviation of 1.42 percent. Further, suppose that a random sample of 400 U.S. advertising agencies gives an average percentage share of billing volume from spot television commercials equal to 12.44 percent with a standard deviation of 1.55 percent.
Using the sample information, does it appear that the mean percentage share of billing volume from spot television commercials for the U.S. advertising agencies is greater than the mean percentage share of billing volume from network television? Explain.
Module #3: Sampling Distributions, Estimates, and Hypothesis Testing
Question 3:
[3] Identify which of these types of sampling is used: random, systematic, convenience, stratified, or cluster.
a) The instructor of this course observed at a Walnut Creek Police sobriety checkpoint at which every fifth driver was stopped and interviewed. Some drivers were arrested.
b) The instructor of this course observed professional wine tasters working at a winery in Napa Valley, CA. Assume that a taste test involved three different wines randomly selected from each of five different wineries.
c) The U.S. Department of Corrections collects data about returning prisoners by randomly selecting five federal prisons and surveying all of the prisoners in each of the prisons.
d) In a Gallup poll, 1003 adults were called after their telephone numbers were randomly generated by a computer, and 20% of t ...
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Approximately 20 of 1000 fish in a pond are affected with a skin .docxrossskuddershamus
Approximately 20% of 1000 fish in a pond are affected with a skin disease. A random sample of 20 fish are selected. What is the mean of the sampling distribution for the proportion of your sample that is infected? What is the standard deviation of the sampling distribution for the proportion of your sample that is infected?
Chapter 4, Section 3, Exercise 075
Match the p-values with the appropriate conclusion:
(a) The evidence against the null hypothesis is significant, but only at the 10% level.
(b) The evidence against the null and in favor of the alternative is very strong.
(c) There is not enough evidence to reject the null hypothesis, even at the 10% level.
(d) The result is significant at a 5% level but not at a 1% level.
Chapter 4, Section 3, Exercise 082
Sleep or Caffeine for Memory?
The consumption of caffeine to benefit alertness is a common activity practiced by 90% of adults in North America. Often caffeine is used in order to replace the need for sleep. One recent study1 compares students' ability to recall memorized information after either the consumption of caffeine or a brief sleep. A random sample of 35 adults (between the ages of 18-39 ) were randomly divided into three groups and verbally given a list of 24 words to memorize. During a break, one of the groups takes a nap for an hour and a half, another group is kept awake and then given a caffeine pill an hour prior to testing, and the third group is given a placebo. The response variable of interest is the number of words participants are able to recall following the break. The summary statistics for the three groups are in the table below. We are interested in testing whether there is evidence of a difference in average recall ability between any two of the treatments. Thus we have three possible tests between different pairs of groups: Sleep vs Caffeine, Sleep vs Placebo, and Caffeine vs Placebo.
Group
Sample size
Mean
Standard Deviation
Sleep
12
15.25
3.3
Caffeine
12
12.25
3.5
Placebo
11
13.70
3.0
1 Mednick, Cai, Kanady, and Drummond, "Comparing the benefits of caffeine, naps and placebo on verbal, motor and perceptual memory", Behavioural Brain Research, 193 (2008), 79-86.
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(a) In the test comparing the sleep group to the caffeine group, the p-value is 0.003.
What is the conclusion of the test?
H0.
In the sample , which group had better recall ability?
According to the test results, do you think sleep is really better than caffeine for recall ability?
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(b) In the test comparing the sleep group to the placebo group, the p-value is 0.06.
What is the conclusion of the test, using a 5% significance level?
H0.
What is the conclusion of the test, if we use a 10% significance level?
H0.
How strong is the evidence of a difference in mean recall ability between these two treatments?
Warn.
InstructionDue Date 6 pm on October 28 (Wed)Part IProbability a.docxdirkrplav
InstructionDue Date: 6 pm on October 28 (Wed)
Part IProbability and Sampling Distributions1.Thinking about probability statements. Probability is measure of how likely an event is to occur. Match one of probabilities that follow with each statement of likelihood given (The probability is usually a more exact measure of likelihood than is the verbal statement.)Answer0 0.01 0.3 0.6 0.99 1(a) This event is impossible. It can never occur.(b) This event is certain. It will occur on every trial.(c) This event is very unlikely, but it will occur once in a while in a long sequence of trials.(d) This event will occur more often that not.2. Spill or Spell? Spell-checking software catches "nonword errors" that result in a string of letters that is not a word, as when "the" is typed as "the." When undergraduates are asked to write a 250-word essay (without spell-checking), the number X of nonword errors has the following distribution:Value of X01234Probability0.10.20.30.30.1(a) Check that this distribution satisfies the two requirements for a legitimate assignment of probabilities to individual outcomes.(b) Write the event "at least one nonword error" in term of X (for example, P(X >3)). What is the probability of this event?(c) Describe the event X ≤ 2 in words. What is its probability? 3. Discrete or continuous? For each exercise listed below, decide whether the random variable described is discrete or continuous and explains the sample space.(a) Choose a student in your class at random. Ask how much time that student spent studying during the past 24 hours.(b) In a test of a new package design, you drop a carton of a dozen eggs from a height of 1 foot and count the number of broken eggs.(c) A nutrition researcher feeds a new diet to a young male white rat. The response variable is the weight (in grams) that the rat gains in 8 weeks.4. Tossing Coins(a) The distribution of the count X of heads in a single coin toss will be as follows. Find the mean number of heads and the variance for a single coin toss.Number of Heads (Xi)01mean:Probability (Pi)0.50.5variance:(b) The distribution of the count X of heads in four tosses of a balanced coin was as follows but some missing probabilities. Fill in the blanks and then find the mean number of heads and the variance for the distribution with assumption that the tosses are independent of each other.Number of Heads (Xi)01234mean:Probability (Pi)0.06250.0625variance:(c) Show that the two results of the means (i.e. single toss and four tosses) are related by the addition rule for means. (d) Show that the two results of the variances (i.e. single toss and four tosses) are related by the addition rule for variances (note: It was assumed that the tosses are independent of each other). 5. Generating a sampling distribution. Let's illustrate the idea of a sampling distribution in the case of a very small sample from a very small .
Midterm 2 – Practice Exercises
1
1. The amount of material used in making a custom sail for a sailboat is normally
distributed with a standard deviation of 64 square feet. For a random sample of 15
sails, the mean amount of material used is 912 square feet. Which of the following
represents a 99% confidence interval for the population mean amount of material
used in a custom sail?
A. 912 ± 49.2
B. 912 ± 42.6
C. 912 ± 44.3
D. 912 ± 46.8
2. The number of beverage cans produced each hour from a vending machine is
normally distributed with a standard deviation of 8.6. For a random sample of 12
hours, the average number of beverage cans produced was 326.0. Assume a 99%
confidence interval for the population mean number of beverage cans produced
per hour. Calculate the margin of error of the 99% confidence interval.
A. 1.85
B. 3.60
C. 6.41
D. 10.56
3. The number of beverage cans produced each hour from a vending machine is
normally distributed with a standard deviation of 8.6. For a random sample of 12
hours, the average number of beverage cans produced was 326.0. Assume a 99%
confidence interval for the population mean number of beverage cans produced
per hour. Find the upper confidence limit of the 99% confidence interval.
A. 340.25
B. 325.98
C. 319.59
D. 332.41
4. If we change a 95% confidence interval estimate to a 99% confidence interval
estimate, we can expect
A. the size of the confidence interval to increase
B. the size of the confidence interval to decrease
C. the size of the confidence interval to remain the same
D. the sample size to increase
Midterm 2 – Practice Exercises
2
5. If a sample has 20 observations and a 90% confidence estimate for µ is needed,
the appropriate t‐score is:
A. 2.120
B. 1.746
C. 2.131
D. 1.729
6. We are interested in conducting a study to determine what percentage of voters
would vote for the incumbent member of parliament. What is the minimum size
sample needed to estimate the population proportion with a margin of error of
0.07 or less at 95% confidence?
A. 200
B. 100
C. 58
D. 196
7. The sample size needed to provide a margin of error of 2 or less with a 0.95
confidence coefficient when the population standard deviation equals 11 is
A. 10
B. 11
C. 116
D. 117
8. The manager of the local health club is interested in determining the number of
times members use the weight room per month. She takes a random sample of 15
members and finds that over the course of a month, the average number of visits
was 11.2 with a standard deviation of 3.2. Assuming that the monthly number of
visits is normally distributed, which of the following represents a 95% confidence
interval for the average monthly usage of all health club members?
A. 11.2 ± 1.74
B. 11.2 ± 1.77
C. 11.2 ± 1.62
D. 11.2 ± 1.83
Midterm 2 – Practice Exercises
3
9. The s.
Midterm 2 – Practice Exercises
1
1. The amount of material used in making a custom sail for a sailboat is normally
distributed with a standard deviation of 64 square feet. For a random sample of 15
sails, the mean amount of material used is 912 square feet. Which of the following
represents a 99% confidence interval for the population mean amount of material
used in a custom sail?
A. 912 ± 49.2
B. 912 ± 42.6
C. 912 ± 44.3
D. 912 ± 46.8
2. The number of beverage cans produced each hour from a vending machine is
normally distributed with a standard deviation of 8.6. For a random sample of 12
hours, the average number of beverage cans produced was 326.0. Assume a 99%
confidence interval for the population mean number of beverage cans produced
per hour. Calculate the margin of error of the 99% confidence interval.
A. 1.85
B. 3.60
C. 6.41
D. 10.56
3. The number of beverage cans produced each hour from a vending machine is
normally distributed with a standard deviation of 8.6. For a random sample of 12
hours, the average number of beverage cans produced was 326.0. Assume a 99%
confidence interval for the population mean number of beverage cans produced
per hour. Find the upper confidence limit of the 99% confidence interval.
A. 340.25
B. 325.98
C. 319.59
D. 332.41
4. If we change a 95% confidence interval estimate to a 99% confidence interval
estimate, we can expect
A. the size of the confidence interval to increase
B. the size of the confidence interval to decrease
C. the size of the confidence interval to remain the same
D. the sample size to increase
Midterm 2 – Practice Exercises
2
5. If a sample has 20 observations and a 90% confidence estimate for µ is needed,
the appropriate t‐score is:
A. 2.120
B. 1.746
C. 2.131
D. 1.729
6. We are interested in conducting a study to determine what percentage of voters
would vote for the incumbent member of parliament. What is the minimum size
sample needed to estimate the population proportion with a margin of error of
0.07 or less at 95% confidence?
A. 200
B. 100
C. 58
D. 196
7. The sample size needed to provide a margin of error of 2 or less with a 0.95
confidence coefficient when the population standard deviation equals 11 is
A. 10
B. 11
C. 116
D. 117
8. The manager of the local health club is interested in determining the number of
times members use the weight room per month. She takes a random sample of 15
members and finds that over the course of a month, the average number of visits
was 11.2 with a standard deviation of 3.2. Assuming that the monthly number of
visits is normally distributed, which of the following represents a 95% confidence
interval for the average monthly usage of all health club members?
A. 11.2 ± 1.74
B. 11.2 ± 1.77
C. 11.2 ± 1.62
D. 11.2 ± 1.83
Midterm 2 – Practice Exercises
3
9. The s ...
ELEMENTS OF STATISTICS / TUTORIALOUTLET DOT COMalbert0076
Unit 3 Problem Set NAME: Elements of Statistics--FHSU Virtual College--Spring 2017
REMEMBER, these are assessed preparatory problems related to the content of Unit 3. The Unit 3 Exam will consist of similar types of
Lecture 6 Point and Interval Estimation.pptxshakirRahman10
Point and Interval Estimation:
Objectives:
Apply the basics of inferential statistics in terms of point estimation.
Compute point and estimation of population means and confidence interval.
Interpret the results of point and interval estimation.
Estimation:
Estimating the value of parameter from the sample:
An aspect of inferential statistics.
Why to estimate: Population is large enough so we
can only estimate.
Types of estimation:
Point Estimation:
A specified number value (single value) that is an estimate of a population parameter. The point estimate of the population mean µ is the sample mean.
Interval Estimate:
Range of values to estimate about population parameter.
Confidence Interval Estimation:
Range of values to estimate about population parameter.
May contain the parameter or not (Degree of confidence).
Ranges between two values.
Example:
Age (in years) 4 BScN students: 20<µ < 25 or (22.5 +2.5)
FORMULA:
Point estimate (x) + Critical Value x Standard Error.
Confidence Interval is a particular interval of estimate.
Given that sample size is large, the 95% of the sample means taken from same population and same sample size will fall in + 1.96 SD of the population mean.
Three commonly used Confidence Intervals are 90%, 95% (by default) , and 99%.
Why not too small or too large confidence intervals?
Too wide: 99.9% Interval too broad
Too narrow: 80 % More uncertainty to have population mean.
The 99% of the sample means taken from same population and
same sample size will fall in + 2.575 SD of the population mean.
Interpretation:
99% probability that interval will enclose population parameter and 1% chance that it will not have population parameter.
Level of confidence: The level of certainty that the interval will have the true population mean.
Chances of Error: Chances that the interval will not cater the true parameter.
Sum of level of confidence and chances of error =100%
Question 1. [12 marks]Market research has indicated that custo.docxIRESH3
Question 1. [12 marks]
Market research has indicated that customers are likely to bypass Roma tomatoes that weigh less than 70 grams. A produce company produces Roma tomatoes that average 74.0 grams with a standard deviation of 3.2 grams.
(a) [2 marks] Assuming that the normal distribution is a reasonable model for the
weights of these tomatoes, what proportion of Roma tomatoes are currently undersize (less than 70g)?
(b) [2 marks] How much must a Roma tomato weigh to be among the heaviest 10%?
(c ) [2 marks] The aim of the current research is to reduce the proportion of
undersized tomatoes to no more than 2%. One way of reducing this proportion is to reduce the standard deviation. If the average size of the tomatoes remains 74.0 grams, what must the target standard deviation be to achieve the 2% goal?
(d) [3 marks] The company claims that the goal of 2% undersized tomatoes is
reached.To test this, a random sample of 25 tomatoes is taken. What is the
distribution of undersized tomatoes in this sample if the company's claim is true?
Explain your reasoning.
Question 2:
In an article in Marketing Science, Silk and Berndt investigate the output of advertising agencies. They describe ad agency output by finding the shares of dollar billing volume coming from various media categories such as network television, spot television, newspaper, radio, and so forth.
Suppose that a random sample of 400 U.S. advertising agencies gives an average percentage share of billing volume from network television equal to 7.46 percent with a standard deviation of 1.42 percent. Further, suppose that a random sample of 400 U.S. advertising agencies gives an average percentage share of billing volume from spot television commercials equal to 12.44 percent with a standard deviation of 1.55 percent.
Using the sample information, does it appear that the mean percentage share of billing volume from spot television commercials for the U.S. advertising agencies is greater than the mean percentage share of billing volume from network television? Explain.
Module #3: Sampling Distributions, Estimates, and Hypothesis Testing
Question 3:
[3] Identify which of these types of sampling is used: random, systematic, convenience, stratified, or cluster.
a) The instructor of this course observed at a Walnut Creek Police sobriety checkpoint at which every fifth driver was stopped and interviewed. Some drivers were arrested.
b) The instructor of this course observed professional wine tasters working at a winery in Napa Valley, CA. Assume that a taste test involved three different wines randomly selected from each of five different wineries.
c) The U.S. Department of Corrections collects data about returning prisoners by randomly selecting five federal prisons and surveying all of the prisoners in each of the prisons.
d) In a Gallup poll, 1003 adults were called after their telephone numbers were randomly generated by a computer, and 20% of t ...
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🔥 From probability hiccups to data dilemmas, our tips will guide you through. 🎯 No more stressing, just A+ grades and confidence to spare.
📌 What's in Store:
✅ Expert Techniques: Learn from the pros.
✅ 24/7 Availability: We match your schedule.
✅ Step-by-Step: Clear, concise problem-solving.
✅ Boosted Performance: Watch your grades rise.
🔗 Visit StatisticsHomeworkHelper.com and breeze through your stats assignments! 🌟
Are complex statistics problems leaving you puzzled? Look no further! Introducing StatisticsHomeworkHelper.com, your ultimate destination for conquering statistics challenges with ease.
🔍 Unparalleled Expertise: Our team of experienced statisticians is ready to tackle any problem thrown their way. From basic concepts to advanced analyses, we've got you covered.
📈 Step-by-Step Guidance: Say goodbye to confusion! Our detailed solutions break down even the trickiest questions into manageable steps, helping you grasp the concepts along the way.
⏱️ Time-Saving Assistance: Don't waste hours struggling over a single problem. Our efficient solutions give you more time to focus on other important tasks.
🌐 Anytime, Anywhere: Access our platform 24/7 from the comfort of your home. Whether it's a late-night study session or a last-minute assignment, we're always here to help.
🎓 Excelling Made Easy: Boost your grades and gain a deeper understanding of statistics. With StatisticsHomeworkHelper.com, excelling in your studies has never been more achievable.
🚀 Try Us Today: Visit our website and experience the power of a dedicated statistics homework solver. Let's turn those daunting problems into confident victories!
📢 Spread the word and tag friends who could use a statistics study companion. Together, let's conquer statistics! 📊📚
Get ready to unravel the mysteries of statistics with ease. Visit us at StatisticsHomeworkHelper.com and never fear your homework again! 🎉
At our company, we understand that statistics homework help can be a challenging subject, and we are here to help. Whether you are struggling with probability or any other aspect of statistics, our experienced tutors are ready to guide you through the material and help you to achieve your academic goals. Contact us today to learn more about our statistics homework help services.
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2. 3.102 How Important is Regular Exercise? In a recent poll of 1000 American adults, the
number saying that exercise is an important part of daily life was 753. Use StatKey or other
technology to find and interpret at 90% confidence interval for the proportion of American
adults who think exercise is an important part of daily life.
Solution
Using StatKey or other technology, we produce a bootstrap distribution such as the figure
shown below. For a 90% confidence interval, we find the 5%-tile and 95%-tile points in this
distribution to be 0.730 and 0.774. We are 90% confident that the percent of American
adults who think exercise is an important part of daily life is between 73.0% and 77.4%.
3.104 Comparing Methods for Having Dogs Identify Cancer in People Exercise 2.17 on
page 55 describes a study in which scientists train dogs to smell cancer. Researchers collected
breath and stool samples from patients with cancer as well as from healthy people. A
trained dog was given five samples, randomly displayed, in each test, one from a patient with
cancer and four from health volunteers. The results are displayed in the table below.
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3. Breath Test Stool Test Total
Dog selects cancer 33 37 70
Dog does not select cancer 3 1 4
Total 36 38 74
Solution
The dog got pˆB = 33/36 = 0.917 or 91.7% of the breath samples correct and pˆS = 37/38 =
0.974 or 97.4% of the stool samples correct. (A remarkably high percentage in both cases!) We
create a bootstrap distribution for the difference in proportions using StatKey or other
technology (as in the figure below) and then find the middle 90% of values. Using the figure,
the 90% confidence interval for pB − pS is -0.14 to 0.025. We are 90% confident that the
difference between the proportion correct for breath samples and the proportion correct for
stool samples for all similar tests we might
Use StatKey or other technology to use a bootstrap distribution to find and interpret a 90%
confidence interval for the difference in the proportion of time the dog correctly picks out the
cancer sample between the two types of samples. Is it plausible that there is no difference in
the effectiveness in the two types of methods (breath or stool)?
give this dog is between -0.14 and 0.025. Since a difference of zero represents no difference, and
zero is in the interval of plausible values, it is plausible that there is no difference in the
effectiveness of breath vs stool samples in having this dog detect cancer.
3.105 Average Tip for a Waitress Data 2.12 on page 119 describes information from a sample
of 157 restaurant bills collected at the First Crush bistro. The data is available in Restaurant
Tips. Create a bootstrap distribution using this data and find and interpret a 95% confidence
interval for the average tip left at this restaurant. Find the confidence interval two ways: using the
standard error and using percentiles. Compare your results.
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4. x¯ ± 2SE
3.85 ± 2(0.19)
3.85 ± 0.38
3.47 to 4.23.
For this bootstrap distribution, the 95% confidence interval using the 2.5%-tile and
97.5%-tile is
3.47 to 4.23. We see that the results (rounding to two decimal places) are the same. We
are 95% confident that the average tip left at this restaurant is between $3.47 and $4.23.
3.116 Small Sample Size and Outliers As we have seen, bootstrap distributions are generally
symmetric and bell-shpaed and centered at the value of the original sample statistic. However,
strange things can happen when the sample size is small and there is an outlier present. Use
StatKey or other technology to create a bootstrap distribution for the standard deviation based
on the following data:
8 10 7 12 13 8 10 50
Solution
Using one bootstrap distribution (as shown below), the standard error is SE = 0.19. The
mean tip from the original sample is x¯= 3.85, so a 95% confidence interval using the
standard error is
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5. clusters of dots represent the number of times the outlier is included in the bootstrap sample
(with the cluster on the left containing statistics from samples in which the outlier was not
included, the next one containing statistics from samples that included the outlier once, the
next one containing statistics from samples that included the outlier twice, and so on.)
4.17 Beer and Mosquitoes Does consuming beer attract mosquitoes? A study done in
Burkino Faso, Africa, about the spread of malaria investigated the connection between beer
consumption and mosquito attraction. In the experiment, 25 volunteers consumed a liter of
beer while 18 vol- unteers consumed a liter of water. The volunteers were assigned to the two
groups randomly. The attractiveness to mosquitos of each volunteer was tested twice: before
the beer or water and after. Mosquitoes were released and caught in traps as they
approached the volunteers.
Describe the shape of the distribution. Is it appropriate to construct a confidence interval
from this distribution? Explain why the distribution might have the shape it does.
Solution
The bootstrap distribution for the standard deviations (shown below) has at least four
completely separate clusters of dots. It is not at all symmetric and bell-shaped so it would
not be appropriate to use this bootstrap distribution to find a confidence interval for the
standard deviation.The
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6. For the beer group, the total number of mosquitoes caught in the traps before consumption was
434 and the total was 590 after consumption. For the water group, the total was 337 before and
345 after.
(a)Define the relevant parameter(s) and state the null and alternative hypotheses for a test to
see if, after consumption, the average number of mosquitoes is higher for the volunteers
who drank beer.
(b)Compute the average number of mosquitoes per volunteer before consumption for each
group and compare the results. Are the two sample means different? Do you expect that this
difference is just the result of random chance?
(c)Compute the average number of mosquitoes per volunteer after consumption for each group
and compare the results. Are the two sample means different? Do you expect that this
difference is just the result of random chance?
(d)If the difference in part (c) is unlikely to happen by random chance, what can we conclude
about beer consumption and mosquitoes?
(e)If the difference in part (c) is statistically significant, do we have evidence that beer
consumption increases mosquito attraction? Why or why not?
Solution
(a)We define µb to be the mean number of mosquitoes attracted after drinking beer and µw to be
the mean number of mosquitoes attracted after drinking water. The hypotheses are:
H0 : µb = µw Ha : µb > µw
(b)The sample mean number of mosquitoes attracted per participant before consumption for
the beer group is 434/25 = 17.36 and is 337/18 = 18.72 for the water group. These sample means
are
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7. slightly different, but the small difference could be attributed to random chance.
(c)The sample mean number of mosquitoes attracted per participant after consumption is
590/25
= 23.60 for the beer group and is 345/18 = 19.17 for the water group. This difference is
larger than the difference in means before consumption. It is less likely to be due just to
random chance.
(d)The mean number of mosquitoes attracted when drinking beer is higher than when
drink- ing water.
(e)Since this was an experiment, a statistically significant difference would provide evidence
that beer consumption increases mosquito attraction.
4.18 Guilty Verdicts in Court Cases A reporter on cnn.com stated in July 2010 that 95%
of all court cases that go to trial result in a guilty verdict. To test the accuracy of this claim,
we collect a random sample of 2000 court cases that went to trial and record the proportion that
resulted in a guilty verdict.
(a)What is/are the relevant parameter(s)? What sample statistic(s) is/are used to conduct
the test?
(b)Stat the null and alternative hypotheses.
(c)We assess evidence by considering how likely our sample results are when H0 is true.
What does that mean in this case?
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8. (b) The hypotheses are: H0 : p = 0.95
HA : p /= 0.95
(c) How likely is the observed sample proportion when we select a sample of
size 2000 from a population with p = 0.95?
For exercises 4.21 to 4.25, describe tests we might conduct based on Data 2.3, introduced on
page 66. This dataset, stored in ICUAdmissions, contains information about a sample of pa-
tients admitted to a hospital Intensive Care Unit (ICU). For each of the research questions
below, define any relevant parameters and state appropriate null and alternative hypotheses.
4.21 Is there evidence that mean heart rate is higher in male ICU patients than in female
ICU patients?
Solution
We define µm to be mean heart rate for males being admitted to an ICU and µf to be mean
heart rate for females being admitted to an ICU. The hypotheses are:
H0 : µm = µf HA : µm > µf
Solution
(a) The parameter is p, the proportion of all court cases going to trial that end in a guilty
verdict. The sample statistic is pˆ,the proportion of guilty verdicts in the sample of 2000 cases.
22.Is there a difference in the proportion who receive CPR based on whether the
patient’s race is white or black?
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9. Solution
We define pw to be the proportion of white ICU patients who receive CPR and pb to be the
proportion of black ICU patients who receive CPR. The hypotheses are:
H0 : pw = pb HA : pw /= pb
23.Is there a positive linear association between systolic blood pressure and heart rate?
Solution
We define ρ to be the correlation between systolic blood pressure and heart rate for
patients ad- mitted to an ICU. The hypotheses are:
H0 : ρ = 0
HA : ρ > 0
Note: The hypotheses could also be written in terms of β, the slope of a regression line to
predict one of these variables using the other.
24.Is either gender over-representative in patients to the ICU or is the gender breakdown
about equal?
Solution
Notice that this is a test for a single proportion. We define p to be the proportion of ICU
patients
who are female. (We could also have defined p to be the proportion who are male. The
test will work fine either way.) The hypotheses are:
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10. H0 : p = 0.5
HA : p /= 0.5
Also accepted: We define pm to be the proportion of ICU patients who are male and pf to
be the proportion of ICU patients who are female. The hypotheses are:
H0 : pm = pf HA : pm =
/ pf
25.Is the average age of ICU patients at this hospital greater than 50?
Solution
We define µ to be the mean age of ICU patients. The hypotheses are:
H0 : µ = 50
HA : µ > 50
For exercises 4.30, 4.32, and 4.36, indicate whether the analysis involves a statistical test. If it
does involve a statistical test, state the population parameter(s) of interest and null
hypotheses.
4.30 Polling 1000 people in a large community to determine the average number of hours a day
people watch television
Solution
This analysis does not involve a test because there is no claim of interest. We would likely
use a confidence interval to estimate the average.
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11. 4.32 Utilizing the census of a community, which includes information about all residents of the
community, to determine if there is evidence for the claim that the percentage of people in the
community living in a mobile home is greater than 10%.
Solution
This analysis does not include a test because from the information in a census, we can find
exactly the true population proportion.
4.36 Using the complete voting record of a county to see if there is evidence that more than
50% of the eligible voters in the county voted in the last election.
Solution
This analysis does not include a statistical test. Since we have all the information for the popula-
tion, we can compute the proportion who voted exactly and see if it is greater than 50%.
4.40 Euchre One of the authors and some statistician friends have an ongoing series of Euchre
games that will stop when one of the two teams is deemed to be statistically significantly better
than the other team. Euchre is a card game and each game results in a win for one team and a
loss for the other. Only two teams are competing in this series, which we’ll call Team A and
Team B.
(a)Define the parameter(s) of interest.
(b)What are the null and alternative hypotheses if the goal is to determine if either team is
statistically significantly better than the other at winning Euchre?
(c)What sample statistic(s) would they need to measure as the games go on?
(d)Could the winner be determined after one or two games? Why or why not?
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12. Solution
(a)The population of interest is all Euchre games that could be played between these two
teams. The parameter of interest is the proportion of games that a certain team would win,
say p =the proportion of all possible games that team A wins. (We could also just as easily
have used team B.)
(b)We are testing to see whether this proportion is either significantly higher or lower than 0.5.
The hypotheses are:
H0 : p = 0.5
HA : p /= 0.5
(c)The sample statistic is the proportion of games played so far that team A has won. We
could choose to look at the proportion of wins for either team, but must be consistent
defining
the population parameter and calculating the sample statistic. We also need to keep track
of the sample size (number of games played).
(d) No. Even if the two teams are equal (p = 0.5), it is quite possible that one team
could win the first two games just by random chance. Therefore, even if one team wins the
first two games, we would not have conclusive evidence that that team is better.
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