Question

Question 13 Dean Halverson recently read that full-time college students study 20 hours Type numbers in the boxes. each week. She decides to do a study at her university to see if there is 10 points evidence that students study an average of less than 20 hours each week. A random sample of 32 students were asked to keep a diary of their activities over a period of several weeks. It was found that he average number of hours that the 32 students studied each week was 19.1 hours. The sample standard deviation of 4.2 hours. Find the p-value. The p-value should be rounded to 4-decimal places. Question 14 A medical researcher is studying the effects of a drug on blood pressure. Type numbers in the boxes. Subjects in the study have their blood pressure taken at the beginning of the 10 points study. After being on the medication for 4 weeks, their blood pressure is taken again. The change in blood pressure is recorded and used in doing the hypothesis test. Change: Final Blood Pressure - Initial Blood Pressure The researcher wants to know if there is evidence that the drug affects blood pressure. At the end of 4 weeks, 35 subjects in the study had an average change in blood pressure of -2.4 with a standard deviation of 5.2. Find the p-value for the hypothesis test. Your answer should be rounded to 4 decimal places.

Answer 1

13)

Below are the null and alternative Hypothesis,
Null Hypothesis: μ = 20
Alternative Hypothesis: μ < 20

Rejection Region
This is left tailed test, for α = 0.05 and df = 31
Critical value of t is -1.696.
Hence reject H0 if t < -1.696

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (19.1 - 20)/(4.2/sqrt(32))
t = -1.212

P-value Approach
P-value = 0.1173

14)

Below are the null and alternative Hypothesis,
Null Hypothesis: μ = 0
Alternative Hypothesis: μ ≠ 0

Rejection Region
This is two tailed test, for α = 0.05 and df = 34
Critical value of t are -2.032 and 2.032.
Hence reject H0 if t < -2.032 or t > 2.032

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (-2.4 - 0)/(5.2/sqrt(35))
t = -2.73

P-value Approach
P-value = 0.0100