Title: Inference for Means
Course: Question Banks

Q: A nutritional scientist is investigating the effects of ketogenic dieting on lean body mass. They studied 50 college age men in a paired study: each pair matched on starting lean body mass, height, and race and divided into Groups A and B. Group A consumed a ketogenic diet while Group B consumed a traditional western diet.

Which type of hypothesis test should they use?
- test for one population proportion (The response variable (lean body mass) is quantitative, so this is not in the context of proportions.)
- test for a difference in two population proportions (We have a quantitative response variable (lean body mass). This is a matched-pairs study and our hypothesis is based on the difference between paired values from two data sets.)
- test for a difference in two population means (You are correct that the test will involve a mean since the variable (lean body mass) is quantitative. But this is a matched-pairs study and our hypothesis is based on the difference between paired values from two data sets)
- test for one population mean (We have a quantitative response variable (lean body mass). This is a matched-pairs study and our hypothesis is based on the difference between paired values from two data sets.)
A: test for one population mean

Q: In the article “Randomised clinical trial: yoga vs written self-care advice for ulcerative colitis”, researchers were investigating the effect of yoga on ulcerative colitis (Aliment Pharmacol Ther. 2017 April 5). Participants answered pre and post treatment questionnaires regarding their disease-specific quality of life. Here are the results:

“The yoga group had significantly higher disease-specific quality of life compared to the self-care group after 12 weeks (Δ = 14.6; 95% confidence interval = 2.6-26.7; P = 0.018).Twenty-one [ patients in the yoga group (n=39) ] and 12 [ patients in the self-care group(n=38)]... reached a clinical relevant increase in quality of life at week 12 (P = 0.048). “

The researcher wants to calculate the increase in score of disease-specific quality of life to expect for patients who participate regularly in yoga versus written self-care advice.

What is the best method for the researchers to use to answer this question?
- Use the difference in sample means (14.6) in a hypothesis test for a difference in two population means (or treatment effect). (A hypothesis test will not answer the question “how much change?”.)
- Use the difference in sample proportions (21/39 and 12/38) in a hypothesis test for a difference in two population proportions(or treatment effect). (A hypothesis test will not answer the question “how much change?”.)
- Use the difference in sample means (14.6) to calculate a confidence interval for a difference in two population means (or treatment effect). (To estimate the effect between groups, a confidence interval around the mean difference is appropriate to estimate this effect.)
A: Use the difference in sample means (14.6) to calculate a confidence interval for a difference in two population means (or treatment effect).

Q: Imagine that two friends (Heather and Suzanne) are playing a video game where the player goes on quests to earn coins. The average number of coins earned on a quest is 2000. Heather plays quite a bit, and goes on an average of 20 quests a day. Suzanne doesn't play quite as much, and only goes on 5 quests a day.

On a particular day, which person is less likely to record an average earning below 1500 coins?
- Suzanne (with 5 quests a day), because with fewer quests there will be less variability in mean earnings. (Fewer quests will result is more variability in the average number of coins earned.)
- Heather and Suzanne are equally likely to record such an event, because the number of coins earned on any given quest does not depend on the number of quests undertaken that day. (This is a question about the average earnings in a given day. The more quests in the sample, the lower the variability in average coins earned.)
- Heather (with 20 quests a day), because the mean will typically be closer to 1500 coins with more quests. (There is less variability is sample means when the sample size is larger.)
A: Heather (with 20 quests a day), because the mean will typically be closer to 1500 coins with more quests.

Q: Consider sampling weights from shelter domesticated cats in the United States. Assume the mean height of shelter domesticated cats in the United States is μ = 9 pounds and the standard deviation is σ = 3.3 pounds. Suppose we randomly sample 81 cats from this population and compute the mean weight, then repeat this sampling process 5,000 times and record all the means we get.

Which of the following is the best approximation for the standard deviation of the 5,000 sample means?
- 0.047 (Incorrect. It looks like you confused the size of each sample with the number of samples we collected. Or perhaps you forgot to take the square root of n. The formula for the standard deviation of sample means is)
- 3.3 (Incorrect.We would expect a smaller standard deviation of sample means. The formula for the standard deviation of sample means is)
- 0.37 (Correct. The formula for the standard deviation of sample means is)
A: 0.37

Q: Weight of a cat For computing the amount of radioactive iodine-131 to administer, a veterinary student is concerned with obtaining highly accurate readings of cat weight. If the scale manufacturer claims a standard deviation of 0.1 pounds, how many times should she repeatedly weigh the cat so the sampling distribution of mean weight has a standard deviation of 0.01 pounds?

Which engineer is more likely to come the closest to the true length of the tower rod he or she is weighing?
- 10 times (Incorrect. The formula for the standard deviation of sample means is We know the population standard deviation and also the desired standard deviation of the sampling distribution of the mean.)
- 101 times (Close, but looks like you used n-1 instead of n. The formula for the standard deviation of sample means is We know the population standard deviation and also the desired standard deviation of the sampling distribution of the mean.)
- 100 times (Correct! The formula for the standard deviation of sample means is We know the population standard deviation and also the desired standard deviation of the sampling distribution of the mean.)
A: 100 times

Q: A quality and assurance associate is measuring the diameter on a random sample of 25 mm diameter washers and finds a mean diameter of 24.5 mm. Suppose the population standard deviation is 0.2 mm.

True or false? Finding a random sample with a mean this low in a population with mean 25mm and standard deviation 02mm is very unlikely.
- False (A sample of size 36 with a mean this high is unlikely. The standard deviation of the sampling distribution of means for samples of size 36 is)
- True (A sample of size 36 with a mean this high is unlikely. The standard deviation of the sampling distribution of means for samples of size 36 is . This sample mean is more than 2 standard deviations higher than a mean of 25mm.)
A: True

Q: Suppose we are analysing the timeliness of a direct flights from New York’s JFK International Airport to Los Angelos’ LAX International Airport.

Suppose the flight time is 6.3 hours with a population standard deviation of 1 hour.

Suppose a sample of 49 flights for a particular airline gives a mean flight time of 6.4 hours.

True or false? Finding a random sample with a mean this high in a population with mean 6.3 and standard deviation 1 hour is very unlikely.
- True (A sample of size 49 with mean this high is very likely. The standard deviation of the sampling distribution of means for samples of size 36 is . So our sample mean of 6.4 is less than 1 standard deviation above the mean.)
- False (The standard deviation of the sampling distribution of means for samples of size 49 is . So our sample mean of 6.4 is less than 1 standard deviation above the mean. So this sample mean is fairly common and doesn’t suggest that the flight time for that particular airline is much higher than average.)
A: False

Q: A local grocery store sells a variety of bagels fresh every morning. The mean weight of these bagels is 3 oz with a standard deviation of 0.7 oz.

Suppose we take a random sample of 49 of sesame bagels and record their weights. The mean weight of our sample of 49 slices is 3.1oz.

True or False: based on this sample, it seem likely that sesame bagels weigh more than the other bagels.
- True (A sample of size 49 with mean this high is very likely. The standard deviation of the sampling distribution of means for samples of size 49 is . So our sample mean of 3.1 is just 1 standard deviation above the mean.)
- False (A sample of size 49 with mean this high is very likely. The standard deviation of the sampling distribution of means for samples of size 49 is)
A: False

Q: Suppose that based on a random sample, a 95% confidence interval for the mean hours slept (per day) among graduate students was found to be (6.5, 6.9). What is the margin of error of this confidence interval?
- 6.7 (The center of the confidence interval is 6.7, which therefore must be the sample mean. Recall that the width of the confidence interval is twice the margin of error.)
- .4 (The width of the confidence interval is 0.4. Recall that the width is twice the margin of error.)
- .2 (Indeed, since the width of the confidence interval is 0.4 (6.9 - 6.5), the margin of error must be 0.4 / 2 = 0.2)
A: .2

Q: A diet pill manufacturer finds on average, it’s customers lose 12.3 pounds with a standard error of 6.4 pounds. For a sample of n = 100 people, calculate the 95% confidence interval and interpret your results.
- The 95% confidence interval is (10.32, 14.28). We are 95% certain the confidence interval contains the true population mean. (Incorrect: Looks like you forgot to incorporate the standard error into your calculation.)
- The 95% confidence interval is (5.9, 18.7). We are 95% certain the confidence interval contains the true population mean. (Incorrect: Looks like you forgot to incorporate the critical t value into your calculation.)
- The 95% confidence interval is (-0.37, 24.972). Since the interval includes 0, we conclude at the 95% confidence level that the diet pill may not be effective. (Correct: Since 0 is in the 95% confidence interval, we cannot reject the null hypothesis that the difference in weight is 0.)
A: The 95% confidence interval is (-0.37, 24.972). Since the interval includes 0, we conclude at the 95% confidence level that the diet pill may not be effective.

Q: Suppose the mean hours slept (per day) among a sample of college students was 7.1, and a 95% confidence interval for the population mean hours slept was constructed. Which one of these could it be?
- (7.1, 8.1) (The confidence interval must be centered at the estimate (the sample mean, in this case).)
- (7.0, 7.3) (The confidence interval must be centered at the estimate (the sample mean, in this case).)
- (6.9, 7.3) (The confidence interval must be centered at the sample mean (7.1 in our case), and this is the only interval of the three that satisfies this requirement.)
A: (6.9, 7.3)

Q: In a study of the impact of weekly peer-run review sessions, a teacher compares exam scores between sections which have a mandatory weekly review or no organized weekly review. Of her students, 74 were in the “review group” and 115 as “non-review group.” The difference in mean exam scores out of 100 (review minus non-review) is 13.7 points with a margin of error of 3.1 points with 95% confidence.

Which gives the best interpretation of what we can conclude about the impact of mandatory weekly peer-run review sessions?
- We are 95% certain that the gain in test score is between 10.6 to 16.8 points for any student that is assigned to a mandatory review group. (Incorrect. The confidence level is not a statement about the likelihood that the interval captures individual test score gains.)
- We expect 95% of the students in the mandatory review group to gain on average 10.6 to 16.8 points on their exam. (Incorrect: The confidence level is not a statement about the percentage of the sample.)
- We are 95% certain that the confidence interval of 10.6 to 16.8 points includes the true mean points gained. (The confidence level is the likelihood that the mean number of points gained by the treatment group falls between 10.6-16.8.)
A: We are 95% certain that the confidence interval of 10.6 to 16.8 points includes the true mean points gained.

Q: In a 2016 article “A century of trends in adult human height”, researchers reanalysed 1472 population studies of more than 18.6 million participants worldwide between 1896 and 1996. They report that “The largest gain in adult height over the past century has occurred in South Korean women and Iranian men, who became 20.2 cm (95% confidence interval 17.5–22.7) and 16.5 cm (13.3–19.7) taller, respectively.”

What is best conclusion about the change in height of South Korean women?

Reference: NCD Risk Factor Collaboration (NCD-RisC). A century of trends in adult human height. July 26. 2016. eLife Sciences. Retrieved from: https://elifesciences.org/content/5/e13410ent

Which of the following is a valid conclusion?
- We are 95% confident South Korean women have gained 19.7cm, the most in height from 1896 to 1996. (Incorrect. The confidence level is not referring to which group gained the most in height. What is the 95% confidence interval referring to?)
- We are 95% confident that our sample of South Korean women from 1896 and 1996 will have a difference in sample means between 17.5 cm to 22.7 cm. (Incorrect.The difference in the sample means for our sample is 20.2 cm and this difference is in the interval. Every confidence interval is built around a sample statistic. In other words, we are 100% confident that this is true.)
- We are 95% confident that basset hounds in 1896 were 17.5 cm taller on average and in 1996 are 22.7 cm taller on average. (Incorrect. The confidence interval is estimating the difference in the population means, not an average value for each population. The 17.5 and 22.7 are the lower and upper bounds of an interval that is estimating the difference in the mean heights for the two populations.)
- We are 95% confident that South Korean women in 1996 are between 17.5 cm to 22.7 cm taller than South Korean women in 1896. (“We are 95% confident that South Korean women in 1996 are between 17.5 cm to 22.7 cm taller than South Korean women in 1896” is a valid conclusion.)
A: We are 95% confident that South Korean women in 1996 are between 17.5 cm to 22.7 cm taller than South Korean women in 1896.

Q: In a recent study, 1,115 males 25 to 35 years of age were randomly chosen and asked about their exercise habits. Based on the study results, the researchers estimated the mean time that a male 25 to 35 years of age spends exercising with 90%, 95%, and 99% confidence intervals. These were (not necessarily in the same order):

For which of the three intervals do you have the most confidence that it captures the population mean (μ)?
- (3, 4) (Notice that (3, 4) isn’t the widest interval. A wider interval would give us more confidence of capturing the population mean.)
- (2.5, 4.5) (Notice that (2.5, 4.5) isn’t the widest interval. A wider interval would give us more confidence of capturing the population mean.)
- (2, 5) (The widest interval gives us the most confidence of capturing the population mean because it covers more of the number line.)
A: (2, 5)

Q: In a recent study, 1,115 males 25 to 35 years of age were randomly chosen and asked about their exercise habits. Based on the study results, the researchers estimated the mean time that a male 25 to 35 years of age spends exercising with 90%, 95%, and 99% confidence intervals. These were (not necessarily in the same order):

The confidence interval in which you have the most confidence that it captures the population mean μ must be:
- the 90% confidence interval (Recall that the confidence level measures our confidence that the interval captures the population mean.)
- the 95% confidence interval (Recall that the confidence percentage measures our confidence that the interval captures the population mean.)
- the 99% confidence interval (The confidence percentage measures our confidence that the interval captures the population mean, so the widest interval must be the one with the largest confidence.)
A: the 99% confidence interval

Q: In a recent study, 1,115 males 25 to 35 years of age were randomly chosen and asked about their exercise habits. Based on the study results, the researchers estimated the mean time that a male 25 to 35 years of age spends exercising with 90%, 95%, and 99% confidence intervals. These were (not necessarily in the same order):

Which of the three confidence intervals provides the most precise estimation?
- (2, 5) (The narrower the confidence interval, the more precise the estimation it provides of the unknown mean.)
- (2.5, 4.5) (The narrower the confidence interval, the more precise the estimation it provides of the unknown mean.)
- (3, 4) (Indeed, since this is the narrowest confidence interval, it is the one that provides the most precise estimation of the unknown mean.)
A: (3, 4)

Q: In a recent study, 1,115 males 25 to 35 years of age were randomly chosen and asked about their exercise habits. Based on the study results, the researchers estimated the mean time that a male 25 to 35 years of age spends exercising with 90%, 95%, and 99% confidence intervals. These were (not necessarily in the same order):

The confidence interval that provides the most precise estimation, must be:
- the 99% confidence interval (Recall that there is a trade-off between the level of precision and level of confidence.)
- the 95% confidence interval (Recall that there is a trade-off between the level of precision and level of confidence.)
- the 90% confidence interval (Indeed, since there is a trade-off between the level of confidence and precision, the interval that provides the most precise estimation must be the one with the lowest level of confidence.)
A: the 90% confidence interval

Q: A medical researcher wants to estimate μ, the mean weight of babies born to women over the age of 40. The researcher chooses a random sample of 100 pregnant women who are over 40. Using the mean birth weight of the 100 babies in the sample, the researcher calculates the 95% confidence interval for μ. With 95% confidence, the researcher estimates the mean birth weight of all babies born to women who are over the age of 40 to be between 2935 and 3135 grams. The researcher wants to maintain the 95% level of confidence but report a confidence interval with a smaller margin of error. What should she do?
- Redo the study, but this time with a sample size 64. (A smaller sample will increase the margin of error (which is the opposite of what we want).)
- Redo the study and choose a different sample of size 100. (A different sample will probably produce a different sample mean and standard deviation, so the resulting confidence interval may have a different center and a different margin of error. But we do not know if the margin of error will be smaller or larger.)
- Redo the study, but this time with a sample size 225. (Increasing the sample size will decrease the margin of error.)
A: Redo the study, but this time with a sample size 225.

Q: The National Assessment of Educational Progress (NAEP) gives a math assessment every year to 12th graders in the U.S. On the math test, a score above 275 indicates that a student has the skills to balance a checkbook. For a random sample of 500 young men, the mean NAEP math score is 272 with a standard deviation of 78. Do we have evidence to support that claim that young men nationwide have a mean score below 275?

The null and alternative hypotheses are H_0: µ = 275, H_a: µ < 275. The level of significance is 5%.

The t-test statistic is −0.86 with a P-value of 0.20. What is the correct conclusion?
- The evidence suggests that young men nationwide have a mean score less than 275. (This statement says the evidence supports H_a. This is not correct because the P-value (0.20) is greater than the significance level (0.05). So we fail to reject H_0.)
- It is likely that these 500 young men have a mean score less than 275. (This statement is not connected to the hypotheses. We know that this particular sample of 500 young men has a mean score less than 275. It is 272, so it does not make sense to say this is likely. The conclusion in a hypothesis test is never about the sample. It is always a statement about the population.)
- The evidence suggests that young men nationwide have a mean score equal to 275. (This statement says the evidence suggests that H_0 is true. We can never say that H_0 is true, but we can fail to reject it. Since the P-value (0.20) is greater than the significance level (0.05), we fail to reject H_0.)
- We do not have enough evidence to conclude that the mean score is less than 275 for young men nationwide. (This statement says we do not have enough evidence to support H_a. This is correct because the P-value (0.20) is greater than the significance level (0.05). So we fail to reject H_0.)
A: We do not have enough evidence to conclude that the mean score is less than 275 for young men nationwide.

Q: According to the Pet Care Foundation of Los Angeles, the average number of cats per household was 2.19 in 1996. A cat lover living in Portland, OR suspects that Portland has a higher number of cats per household than the national average .

Here are the null and alternative hypotheses for her study: H_0: µ =2.19; H_a: µ > 2.19. What does µ represent in these hypotheses?
- mean number of cats per household. (Incorrect. You are right that the unit of interest is mean number of cats, but we need to be more specific about the population of interest.)
- mean number households with at least 2.19 cats in Portland, OR. (Incorrect. You are correct that our population of interest resides in Portland, but be careful: we are not looking at the mean number of households!)
- mean number of cats per household in Portland, OR. (Correct. The mean we are trying to estimate is the population of Portland cats and comparing that to the national average.)
A: mean number of cats per household in Portland, OR.

Q: For a statistics class, a student is flipping a coin to test determine whether the coin is fair or not. She flips the coin 4 times in a row and obtains 3 heads and 1 tail. Confident that she has an unfair coin, she stops her experiment and reports that she has an unfair coin.

Is her reasoning correct?
- Yes, her reasoning is correct. If the coin was fair, it would have given her 2 heads and 2 tails. (A fair coin when tossed 4 times could have many different outcomes, up to chance. Though this scenario is perhaps more likely, it does not have to happen all the time!)
- No, her reasoning is incorrect. The sample size was too small to detect if the coin is fair or unfair. (The sample is too small to be able to conclude with high confidence whether the coin is fair or unfair.)
A: No, her reasoning is incorrect. The sample size was too small to detect if the coin is fair or unfair.

Q: The American Society for the Prevention of Cruelty to Animals (ASPCA) collects statistics on pets in the United States. One of the measurements taken by ASPCA is the age of shelter animals.Suppose that the ASPCA finds in a particular shelter, the average age of a shelter cat is 4 years old. Suppose also that, in 2017, a random sample of 150 shelter cats have mean age of 2.5 years old, with a standard deviation of 0.9 years. At the 5% significance level, we conduct a one-sided T-test to see if the mean age in 2017 is significantly less than 4 years old. Statistical software tells us that the p-value = 0.014.

Which of the following is the most appropriate conclusion?
- This suggests that the mean age for cats in U.S. shelters is lower than 4 years old (Incorrect.The null hypothesis involves the mean age for that particular shelter’s cats, not the overall population of U.S. cats.)
- This suggests that the mean age for cats at that particular shelter is 4 years old. (Incorrect. The null hypothesis is that the mean age for that shelter’s cats is 4 years old. If this is true, then there is only a 1.4% chance that we will see a mean age of 2.5 years old or less. This is highly unlikely, suggesting the mean age is lower.)
- This suggests that the mean age for cats at that particular shelter is lower than 4 years old. (Correct. The null hypothesis is that the mean age for that shelter’s cats is 4 years old. If this is true, then there is only a 1.4% chance that we will see a mean age of 2.5 years old or less.)
A: This suggests that the mean age for cats at that particular shelter is lower than 4 years old.

Q: A publishing company wanted to test whether typing speed differs when using word processor A or word processor B. A random sample of 25 typists was selected and the typing speeds (in words per minute) were recorded for each typist when using word processor A and then when using word processor B. (Which word processor is used first is determined for each secretary by a coin flip).
- (H_0: μ = 0) and (H_a: μ < 0) (The problem states: "A publishing company wanted to test whether typing speed differs when using word processor A or word processor B.")
- (H_0: μ = 0) and (H_a: μ > 0) (The problem states: "A publishing company wanted to test whether typing speed differs when using word processor A or word processor B.")
- None of the above. This is not a matched pairs situation. (This is a matched pairs situation. Each of the 25 typists was measured twice; once using word processor A and once using word processor B.)
- (H_0: μ = 0) and (H_a: μ ≠ 0) (Since we would like to test whether the typing speeds differ, the appropriate hypotheses are: (H_0: μ = 0) and (H_a: μ ≠ 0).)
A: (H_0: μ = 0) and (H_a: μ ≠ 0)

Q: Suppose we want to assess the effect of music on concentration. We measure “concentration” by recording the time it takes a person to complete a simple word puzzle. Each person does a puzzle with and without music. A coin flip determines the order of the treatments. Our subjects are a sample of 18 volunteers from a statistics class.

We test the hypotheses (at the 5% level):

H_0 : μ = 0

H_a : μ > 0

where µ is the mean of the difference in times (“no music” minus “music”).

The sample mean is 4 seconds with a standard deviation of 9 seconds. The sample is not random, but treatments are randomly ordered. In addition, the distribution of times is not strongly skewed, so we decide to use the t-test despite the small sample size. The t-test statistic is approximately 1.89 with a P-value of approximately 0.038.

What can we conclude?
- For college students, music is associated with statistically significant improvements in concentration as measured by time to complete a simple word puzzle. (You are correct that the P-value supports the alternative hypothesis since it is less than the significance level of 0.05, but the phrase “is associated with” is a weak description of the effect. This experiment randomized the treatment order. In addition, the matched pairs design controlled the effects of individual characteristics, such as natural ability with word puzzles. So we can use stronger cause-and-effect language.)
- College students listening to music performed significantly faster on word puzzles than students who were not listening to music. (This statement says we compared two separate groups: those who listened to music and those who did not. But this is a matched-pairs design with a single sample of subjects.)
- Music is not associated with statistically significant improvements in concentration as measured by time to complete a simple word puzzle. Completion times only increased 4 seconds on average. (You are right that the times only increased by 4 seconds. But the improvement is statistically significant. The P-value is less than the significance level of 0.05. So the results support the alternative hypothesis.)
- For college students, music produces statistically significant improvements in concentration as measured by time to complete a simple word puzzle. (The P-value supports the alternative hypothesis since it is less than the significance level of 0.05. The word “produces” implies a cause-and-effect relationship. This is appropriate because this experiment is well-designed. We randomized the treatment order. In addition, the matched pairs design controlled the effects of individual characteristics, such as natural ability with word puzzles.)
A: For college students, music produces statistically significant improvements in concentration as measured by time to complete a simple word puzzle.

Q: Idexx is a commonly used laboratory for processing pet lab work. Sensitivity is the proportion of given the disease is present, the test returns a positive result. Specificity is the proportion of given the disease is not present, the test returns a negative result. Their SNAP FIV/FeLV combo test has a published accuracy of 98.6% sensitivity and 98.2% specificity for FeLV screening.

In this context, which is the best way to interpret the percentage 100-98.2% = 1.8%?
- Type II error rate (A type II error occurs when we fail to reject the null hypothesis even though the alternative hypothesis is true. In this context, a type II error would occur if we got a negative test result even though the disease is known to be present.)
- Type I error rate (A type I error occurs when we reject the null hypothesis even though it is true. In this situation, a type I error occurs when we get a positive test result even though the disease is not present. In other words, given the disease is not present, the test result will either be negative (98.2% of the time) or falsely positive (1.8% of the time))
A: Type I error rate

Q: Idexx is a commonly used laboratory for processing pet lab work. Sensitivity is the proportion of given the disease is present, the test returns a positive result. Specificity is the proportion of given the disease is not present, the test returns a negative result. Their SNAP FIV/FeLV combo test has a published accuracy of 98.6% sensitivity and 98.2% specificity for FeLV screening.

In this context, which is the best way to interpret the percentage 100-98.6% = 1.4%?
- Type I error rate (A type I error occurs when we reject the null hypothesis even though it is true. In this situation, a type I error occurs when we get a positive test result even though the disease is not present.)
- Type II error rate (A type II error occurs when we fail to reject the null hypothesis even though the alternative hypothesis is true. In this context, a type II error would occur if we got a negative test result even though the disease is known to be present. In other words, given the disease is present, the test result will either be correctly positive (98.6% of the time) or falsely negative (1.4% of the time))
A: Type II error rate

Q: The American Society for the Prevention of Cruelty to Animals (ASPCA) collects statistics on pets in the United States. One of the measurements taken by ASPCA is the age of shelter animals.Suppose that the ASPCA finds in a particular shelter, the average age of a shelter cat is 4 years old. Suppose also that, in 2017, a random sample of 150 shelter cats have mean age of 2.5 years old, with a standard deviation of 0.9 years. At the 5% significance level, we conduct a one-sided T-test to see if the mean age in 2017 is significantly less than 4 years old. Statistical software tells us that the p-value = 0.014.

Which of the following is the most appropriate conclusion?
- There is a 1.4% chance that a random sample of 150 shelter cats will have a mean age of 4 years or less if the sample mean is 2.5 years old. (Incorrect.The null hypothesis is that the mean age for that shelter’s cats is 4 years old. If this is true, then there is only a 1.4% chance that we will see a mean age of 2.5 years old or less.)
- There is a 1.4% chance that mean age for all shelter cats is 2.5 years old. (Incorrect. The null hypothesis is that the mean age for that shelter’s cats is 4 years old. If this is true, then there is only a 1.4% chance that we will see a mean age of 2.5 years old or less.)
- There is a 1.4% chance that a random sample of 150 shelter cats will have a mean age of 2.5 years old or less if the mean age for a cat is 4 years old. (Correct. The null hypothesis is that the mean age for that shelter’s cats is 4 years old. If this is true, then there is only a 1.4% chance that we will see a mean age of 2.5 years old or less.)
A: There is a 1.4% chance that a random sample of 150 shelter cats will have a mean age of 2.5 years old or less if the mean age for a cat is 4 years old.

Q: A publishing company wanted to test whether typing speeds differ when using word processor A or word processor B. The typing speeds (in words per minute) are recorded for a random sample of 25 typists using processor A, and for another (different) random sample of 25 typists using word processor B. Which "case" does this study fall into?
- Comparing two means—matched pairs (While it is true that this example calls for comparing two means (the mean typing speeds using the two word processors), note that each of the samples was chosen randomly.)
- Comparing two means—independent samples (Indeed, this example calls for comparing two means (the mean typing speeds using the two word processors), and since each of the two samples was chosen randomly, they are independent.)
A: Comparing two means—independent samples

Q: Gabapentin and Tinnitus In the article Short-Term Effects of Gabapentin on Subjective Tinnitus in Acoustic Trauma Patients, researchers collected data in a double-blind, randomized clinical trial (Iran J Otorhinolaryngol. 2017 Mar; 29(91): 95–100.).

Methods: “In this double-blind, randomized clinical trial, 103 patients with tinnitus due to acoustic trauma who were referred to the ENT clinic of Taleghani Hospital during 2014 were randomized to the gabapentin (300 mg bid, n=55) or control (n=48) groups. The two groups were then compared before and after 6 weeks of treatment using a visual analog scale (VAS). At least a 30% reduction in VAS was considered a response to treatment.”

Results: “Differences between the two groups regarding sex, age, duration of disease, and audiometry results was not significant (P>0.05). After 6 weeks’ treatment, the VAS significantly decreased in both groups (P<0.001), but the reduction was significantly greater in the gabapentin group compared with control (P<0.001). Forty-nine patients (89%) in the gabapentin group and 28 control patients (58.3%) responded to treatment (≥30% reduction in VAS), with the difference between the two groups being statistically significant (P<0.001).“

Which gives the best interpretation of what we can conclude about the impact of gabapentin on subjective tinnitus?
- At the 0.05 level, we fail to reject H_0; the differences between the two groups are not statistically significant. (The lack of differences between the groups regarding sex, age, duration of disease, and audiometry results were not the key focus of the study. The lack of differences is used to justify the control and treatment groups are comparable.)
- At the 0.05 level, we fail to reject H_0. Although at the 95% confidence level we see a statistically significant difference between the proportion with decreased VAS, since both groups exhibited statistically significant decreases, no conclusions can be drawn. (Although it is interesting that the control also showed a statistically significant reduction in VAS, the statistic of concern here is the difference between the two groups. In addition, the statistical significance of one does not negate the statistical significance of the other.)
- At the 0.05 level, we reject H_0; the study results suggest that Gabapentin may reduce VAS as “Differences between the two groups regarding sex, age, duration of disease, and audiometry results was not significant (P>0.05)” (The hypothesis we are testing is about gabapentin’s effect not on the study design.)
- 3.At the 0.05 level, we reject H_0; the study results suggest that Gabapentin may reduce VAS as “the reduction was significantly greater in the gabapentin group compared with control (P<0.001).“ (This study is an experiment, so we are attempting to establish or debunk a cause-and-effect relationship. Since the control and treatment group difference was statistically significant, we have evidence supporting that Gabapentin may reduce VAS.)
A: 3.At the 0.05 level, we reject H_0; the study results suggest that Gabapentin may reduce VAS as “the reduction was significantly greater in the gabapentin group compared with control (P<0.001).“

Q: In a 2010 article “Survival and the Development of Azotemia after Treatment of Hyperthyroid Cats”, researchers studied 300 hyperthyroid cats (Journal of Veterinary Internal Medicine, Vol 24, No. 24, July 2010). Two laboratory bloodwork markers for chronic kidney disease are urine protein: creatinine ratio (UPC) and urine albumin:creatinine ratio. The article states “UPC decreased … with treatment (pretreatment 0.32 [0.01–3.40], posttreatment 0.18 [0.05–3.65]; P < .001, n = 42), while UAC did not (pretreatment 33.9 [3.13–3,256.47] mg/g, posttreatment 21.3 [2.47–1,539.40] mg/g; P= .105, n = 42).” Consider the set of hypotheses:

H_0: mean UPC pre-treatment and post-treatment are the same

H_A: mean UPC is reduced post-treatment.

What conclusion can we draw from this study?
- The data provide sufficient evidence to conclude that there is no difference in the mean UPC before or after treatment at the 0.001 significance level. (Since the p-value is less than 0.001, there is enough evidence to reject the null hypothesis.)
- The data do not provide sufficient evidence to conclude mean UPC is lower post-treatment than it was pre-treatment at the 0.001 significance level. (Since the p-value is less than 0.001, there is enough evidence to reject the null hypothesis.)
- We accept the null hypothesis, and conclude that mean UPC is lower post-treatment than it was pre-treatment at the 0.001 significance level (You are right that we can conclude that mean UPC is lower post-treatment than it was pre-treatment at the 0.001 significance level. However, this is the alternative hypothesis statement. Also, in hypothesis testing we never accept the null hypothesis.)
- The data provide sufficient evidence to reject the null hypothesis and to conclude that the mean UPC is lower post-treatment than it was pre-treatment at the 0.001 significance level. (Since the p-value is less than 0.001, there is enough evidence to reject the null hypothesis.)
A: The data provide sufficient evidence to reject the null hypothesis and to conclude that the mean UPC is lower post-treatment than it was pre-treatment at the 0.001 significance level.

Q: In a study of the impact of weekly peer-run review sessions, a teacher compares exam scores between sections which have a mandatory weekly review or no organized weekly review. Of her students, 74 were in the “review group” and 115 as “non-review group.” The difference in mean exam scores out of 100 (review minus non-review) is 13.7 points with a margin of error of 3.1 points with 95% confidence.

Which gives the best interpretation of what we can conclude about the impact of mandatory weekly peer-run review sessions?
- At the 5% level, the differences between the two groups was not statistically significant. (The lack of differences between the groups regarding sex, age, duration of disease, and audiometry results were not the key focus of the study. The lack of differences is used to justify the control and treatment groups are comparable.)
- There is a 95% chance that if a student attends weekly review sessions, their score will increase by 10.6 to 16.8 points than if they did not attend. (95% confidence does not tell us that there is a 95% chance.)
- With such a large margin of error, this study does not suggest that there is a difference in mean exam scores when we compare the two groups. (The interval does not contain zero, so the interval does suggest that there is a difference in mean exam score for the two groups.)
- Mandatory weekly peer-run review sessions are associated with higher test scores. When review session mandated students are compared to non-mandated students, we are 95% confident that the review session mandated students will score 10.6 to 16.8 points more than if they were not required to attend. (This study is an observational study, so the use of the phrase “is associated with” is appropriate. The use of the phrase “95% confident” with a specific confidence interval is also appropriate.)
A: Mandatory weekly peer-run review sessions are associated with higher test scores. When review session mandated students are compared to non-mandated students, we are 95% confident that the review session mandated students will score 10.6 to 16.8 points more than if they were not required to attend.

Q: In a 2016 article “A century of trends in adult human height”, researchers reanalysed 1472 population studies of more than 18.6 million participants worldwide between 1896 and 1996. They report that “The largest gain in adult height over the past century has occurred in South Korean women and Iranian men, who became 20.2 cm (95% confidence interval 17.5–22.7) and 16.5 cm (13.3–19.7) taller, respectively.”

What is best conclusion about the change in height of South Korean women?

Reference: NCD Risk Factor Collaboration (NCD-RisC). A century of trends in adult human height. July 26. 2016. eLife Sciences. Retrieved from: https://elifesciences.org/content/5/e13410ent

Which of the following is a valid conclusion?
- We are 95% confident South Korean women have gained the most in height from 1896 to 1996. (The confidence level is not referring to which group gained the most in height. What is the 95% confidence interval referring to?)
- We are 95% confident that our sample of South Korean women from 1896 and 1996 will have a difference in sample means between 17.5 cm to 22.7 cm. (The difference in the sample means for our sample is 20.2 cm and this difference is in the interval. Every confidence interval is built around a sample statistic. In other words, we are 100% confident that this is true.)
- We are 95% confident that basset hounds in 1896 were 17.5 cm taller on average and in 1996 are 22.7 cm taller on average. (The confidence interval is estimating the difference in the population means, not an average value for each population. The 17.5 and 22.7 are the lower and upper bounds of an interval that is estimating the difference in the mean heights for the two populations.)
- We are 95% confident that South Korean women in 1996 are between 17.5 cm to 22.7 cm taller than South Korean women in 1896. (“We are 95% confident that South Korean women in 1996 are between 17.5 cm to 22.7 cm taller than South Korean women in 1896” is a valid conclusion.)
A: We are 95% confident that South Korean women in 1996 are between 17.5 cm to 22.7 cm taller than South Korean women in 1896.

Q: In a study of the impact of weekly mentoring sessions, a university randomly assigns mandatory weekly sessions to a sample of 200 students taking the same large section math class. The researcher then compares final exam scores of these students. Of the students, 99 were in the “mentor group” and 101 as “non-mentor group.” The difference in mean exam scores out of 100 (review minus non-review) is 6.7 points with a standard error of 1.1 points. .

Calculate the 95% confidence interval.
- (4.73, 8.67) (It looks like you forgot to include the standard error (SE). The formula is (difference of means) ± (critical value)(SE).)
- (4.53, 8.87)
- (4.90, 8.50) (It looks like you calculated (difference of means) ± (z crit value)(SE). Recall, when the population standard deviation is unknown and instead the standard error is used as an estimate, a z-distribution is not appropriate.)
- (4.53, 8.87) (The formula is (difference of means) ± (critical value)(SE).)
A: (4.53, 8.87)