Question

A substance researcher wanted to test the effects of an educational intervention about alcohol use and sexual risk behavior on alcohol consumption among college women that drink. Six college women were randomly selected from the population of students at a local college. Prior to the education intervention, participants were asked to self-report their alcohol consumption (in number of drinks) over the past 30 days. The participants then attended weekly educational sessions over a period of 4 weeks. One month following the conclusion of the educational sessions, participants were again asked to report their alcohol consumption over the past 30 days. Did the educational intervention reduce alcohol consumption among study participants? Below is the data from the study.

Baseline alcohol consumption: 11, 7, 35, 21, 40, 0
Follow up alcohol consumption: 3, 6, 27, 18, 40, 0

a.  State both the null and alternative hypothesis in symbols.
b. Calculate the degrees of freedom (df) and find the t critical value with a significance level of .05.
c. Conduct the appropriate t-test to test your hypothesis (alpha = .05). Show all of your work.
d.  Report your decision.
e.  Interpret your findings.

Answer 1

Paired Sample t test Output

For the score differences we have, mean is Dˉ=3.3333, the sample standard deviation is sD=3.7771, and the sample size is n=6. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: μD =0 Ha: μD >0 This corresponds to a Right-tailed test, for which a t-test for two paired samples be used. (2a) Critical Value Based on the information provided, the significance level is α=0.05, and the degree of freedom is n-1=6-1=5. Therefore the critical value for this Right-tailed test is tc​=2.015. This can be found by either using excel or the t distribution table. (2b) Rejection Region The rejection region for this Right-tailed test is t>2.015 (3)Test Statistics The t-statistic is computed as follows: (4) The p-value The p-value is the probability of obtaining sample results as extreme or more extreme than the sample results obtained, under the assumption that the null hypothesis is true. In this case, the p-value is 0.0415 (5) The Decision about the null hypothesis (a) Using traditional method Since it is observed that t=2.1617 > tc​=2.015, it is then concluded that the null hypothesis is rejected. (b) Using p-value method Using the P-value approach: The p-value is p=0.0415, and since p=0.0415≤0.05, it is concluded that the null hypothesis is rejected. (6) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ1​ is greater than μ2, at the 0.05 significance level. There is enough evidence to claim that the educational intervention reduce alcohol consumption among study participants

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