Question 5

An urban planner is researching commute times in the SanFrancisco Bay Area to find out if commute times have increased. Inwhich of the following situations could the urban planner use ahypothesis test for a population mean? Check all that apply.

1. The urban planner asks a simple random sample of 110 commutersin the San Francisco Bay Area if they believe their commute timehas increased in the past year. The urban planner will compute theproportion of commuters who believe their commute time hasincreased in the past year.
2. The urban planner collects travel times from a random sample of125 commuters in the San Francisco Bay Area. A traffic study fromlast year claimed that the average commute time in the SanFrancisco Bay Area is 45 minutes. The urban planner will see ifthere is evidence the average commute time is greater than 45minutes.
3. The urban planner asks a random sample of 100 commuters in theSan Francisco Bay Area to record travel times on a Tuesday morning.One year later, the urban planner asks the same 100 commuters torecord travel times on a Tuesday morning. The urban planner willsee if the difference in commute times shows an increase.

Question 6

The Food and Drug Administration (FDA) is a U.S. governmentagency that regulates (you guessed it) food and drugs for consumersafety. One thing the FDA regulates is the allowable insect partsin various foods. You may be surprised to know that much of theprocessed food we eat contains insect parts. An example is flour.When wheat is ground into flour, insects that were in the wheat areground up as well.

The mean number of insect parts allowed in 100 grams (about 3ounces) of wheat flour is 75. If the FDA finds more than thisnumber, they conduct further tests to determine if the flour is toocontaminated by insect parts to be fit for human consumption.

The null hypothesis is that the mean number of insect parts per100 grams is 75. The alternative hypothesis is that the mean numberof insect parts per 100 grams is greater than 75.

Is the following a Type I error or a Type II error orneither?

The test fails to show that the mean number of insect parts isgreater than 75 per 100 grams when it is.

1. Type I error
2. Type II error
3. Neither

Question 7

Child Health and Development Studies (CHDS) has been collectingdata about expectant mothers in Oakland, CA since 1959. One of themeasurements taken by CHDS is the age of first time expectantmothers. Suppose that CHDS finds the average age for a first timemother is 26 years old. Suppose also that, in 2015, a random sampleof 50 expectant mothers have mean age of 26.5 years old, with astandard deviation of 1.9 years. At the 5% significance level, weconduct a one-sided T-test to see if the mean age in 2015 issignificantly greater than 26 years old. Statistical software tellsus that the p-value = 0.034.

Which of the following is the most appropriate conclusion?

1. There is a 3.4% chance that a random sample of 50 expectantmothers will have a mean age of 26.5 years old or greater if themean age for a first time mother is 26 years old.
2. There is a 3.4% chance that mean age for all expectant mothersis 26 years old in 2015.
3. There is a 3.4% chance that mean age for all expectant mothersis 26.5 years old in 2015.
4. There is 3.4% chance that the population of expectant motherswill have a mean age of 26.5 years old or greater in 2015 if themean age for all expectant mothers was 26 years old in 1959.

Question 8

Child Health and Development Studies (CHDS) has been collectingdata about expectant mothers in Oakland, CA since 1959. One of themeasurements taken by CHDS is the weight increase (in pounds) forexpectant mothers in the second trimester. In a fictitious study,suppose that CHDS finds the average weight increase in the secondtrimester is 14 pounds. Suppose also that, in 2015, a random sampleof 40 expectant mothers have mean weight increase of 16 pounds inthe second trimester, with a standard deviation of 6 pounds. At the5% significance level, we can conduct a one-sided T-test to see ifthe mean weight increase in 2015 is greater than 14 pounds.Statistical software tells us that the p-value = 0.021.

Which of the following is the most appropriate conclusion?

1. There is a 2.1% chance that a random sample of 40 expectantmothers will have a mean weight increase of 16 pounds or greater ifthe mean second trimester weight gain for all expectant mothers is14 pounds.
2. There is a 2.1% chance that mean second trimester weight gainfor all expectant mothers is 14 pounds in 2015.
3. There is a 2.1% chance that mean second trimester weight gainfor all expectant mothers is 16 pounds in 2015.
4. There is 2.1% chance that the population of expectant motherswill have a mean weight increase of 16 pounds or greater in 2015 ifthe mean second trimester weight gain for all expectant mothers was14 pounds in 1959.

Question 9

A researcher conducts an experiment on human memory and recruits15 people to participate in her study. She performs the experimentand analyzes the results. She uses a t-test for a mean and obtainsa p-value of 0.17.

Which of the following is a reasonable interpretation of herresults?

1. This suggests that her experimental treatment has no effect onmemory.
2. If there is a treatment effect, the sample size was too smallto detect it.
3. She should reject the null hypothesis.
4. There is evidence of a small effect on memory by herexperimental treatment.

Question 10

A criminal investigator conducts a study on the accuracy offingerprint matching and recruits a random sample of 35 people toparticipate. Since this is a random sample of people, we don’texpect the fingerprints to match the comparison print. In thegeneral population, a score of 80 indicates no match. Scoresgreater than 80 indicate a match. If the mean score suggests amatch, then the fingerprint matching criteria are not accurate.

The null hypothesis is that the mean match score is 80. Thealternative hypothesis is that the mean match score is greater than80.

The criminal investigator chooses a 5% level of significance.She performs the experiment and analyzes the results. She uses at-test for a mean and obtains a p-value of 0.04.

Which of the following is a reasonable interpretation of herresults?

1. This suggests that there is evidence that the mean match scoreis greater than 80. This suggests that the fingerprint matchingcriteria are not accurate.
2. If there is a treatment effect, the sample size was too smallto detect it. This suggests that we need a larger sample todetermine if the fingerprint matching criteria are notaccurate.
3. She cannot reject the null hypothesis. This suggests that thefingerprint matching criteria could be accurate.
4. This suggests that there is evidence that the mean match scoreis equal to 80. This suggests that the fingerprint matchingcriteria is accurate.

Question 11

A group of 42 college students from a certain liberal artscollege were randomly sampled and asked about the number ofalcoholic drinks they have in a typical week. The purpose of thisstudy was to compare the drinking habits of the students at thecollege to the drinking habits of college students in general. Inparticular, the dean of students, who initiated this study, wouldlike to check whether the mean number of alcoholic drinks thatstudents at his college in a typical week differs from the mean ofU.S. college students in general, which is estimated to be4.73.

The group of 42 students in the study reported an average of5.31 drinks per with a standard deviation of 3.93 drinks.

Find the p-value for the hypothesis test.

The p-value should be rounded to 4-decimal places.

Question 12

Commute times in the U.S. are heavily skewed to the right. Weselect a random sample of 240 people from the 2000 U.S. Census whoreported a non-zero commute time.

In this sample the mean commute time is 28.9 minutes with astandard deviation of 19.0 minutes. Can we conclude from this datathat the mean commute time in the U.S. is less than half an hour?Conduct a hypothesis test at the 5% level of significance.

What is the p-value for this hypothesis test?

Your answer should be rounded to 4 decimal places.