Question

A medical researcher is investigating the effect of drinking coffee on systolic blood pressure. The researcher assumes the average systolic blood pressure is 120 mmHg. For a random sample of 200 patients, the researcher takes two measurements of systolic blood pressure. The first systolic blood pressure measurement is taken during a week when the patients drink no coffee, and the second systolic blood pressure measurement is taken during a week when the patients drink at least two cups of coffee. The medical researcher wonders whether there is a significant difference between the blood pressure measurements.

Which of the following is the correct null and alternative hypothesis for the medical researcher’s study?

1. H₀: µ = 120; H_a: µ ≠ 120
2. H₀: µ = 0; H_a: µ ≠ 0
3. H₀: µ = 0; H_a: µ ≠ 120

Question 2

In a fictional study, suppose that a psychologist is studying the effect of daily meditation on resting heart rate. The psychologist believes the patients who not meditate have a higher resting heart rate. For a random sample of 45 pairs of identical twins, the psychologist randomly assigns one twin to one of two treatments. One twin in each pair meditates daily for one week, while the other twin does not meditate. At the end of the week, the psychologist measures the resting heart rate of each twin. Assume the mean resting heart rate is 80 heart beats per minute.

The psychologist conducts a T-test for the mean of the differences in resting heart rate of patients who do not meditate minus resting heart rate of patients who do meditate.

Which of the following is the correct null and alternative hypothesis for the psychologist’s study?

1. H₀: µ = 80; H_a: µ > 80
2. H₀: µ = 0; H_a: µ ≠ 0
3. H₀: µ = 0; H_a: µ > 0

Question 3

Facebook friends: According to Facebook’s self-reported statistics, the average Facebook user has 130 Facebook friends. For a statistics project a student at Contra Costa College (CCC) tests the hypothesis that CCC students will average more than 130 Facebook friends. She randomly selects 3 classes from the schedule of classes and distributes a survey in these classes. Her sample contains 45 students.

From her survey data she calculates that the mean number of Facebook friends for her sample is: ¯x= 138.7 with a standard deviation of: s=79.3.

She chooses a 5% level of significance. What can she conclude from her data?

1. Nothing. The conditions for use of a t-model are not met. She cannot trust that the p-value is accurate for this reason.
2. We cannot conclude that the average number of Facebook friends for CCC students is greater than 130. The sample mean of 138.7 is not significantly greater than 130.
3. Her data supports her claim. The average number of Facebook friends for CCC students is significantly greater than 130.

Question 4

According to a 2014 research study of national student engagement in the U.S., the average college student spends 17 hours per week studying. A professor believes that students at her college study less than 17 hours per week. The professor distributes a survey to a random sample of 80 students enrolled at the college.

From her survey data the professor calculates that the mean number of hours per week spent studying for her sample is: ¯x= 15.6 hours per week with a standard deviation of s = 4.5 hours per week.

The professor chooses a 5% level of significance. What can she conclude from her data?

1. The data supports the professor’s claim. The average number of hours per week spent studying for students at her college is less than 17 hours per week.
2. The professor cannot conclude that the average number of hours per week spent studying for students at her college is less than 17 hours per week. The sample mean of 15.6 is not significantly less than 17.
3. Nothing. The conditions for use of a t-model are not met. The professor cannot trust that the p-value is accurate for this reason.

Answer 1

1. The correct hypothesis would be:

H0: u=120 vs Ha: u not equal 120

2. The correct hypothesis is,

H0: u=0 vs Ha: u>0

3. We are testing,

H0: u= 130 vs H1: u>130

From the given date, test statistic is:

138.7-130/(79.3/√45) = 0.736

p value of this one sided t test is,

P(t44>0.736) = 0.2328

Since the p-value of this test>0.05, we have insufficient evidence to reject H0 and support the claim that u>130  

4. We are testing,

H0: u=17 vs H1: u<17

From the data given, test statistic,

t = 15.6-17/(4.5/√80) = -2.783

P-VALUE of this test is, P(t79<-2.783) = 0.00337

Since the p-value of this test<significance level of 0.05, we have sufficient evidence to reject H0 and support the claim.