A researcher is testing the effectiveness of a new herbal supplement that claims to improve physical fitness. A sample of n = 16 college students is obtained and each student takes the supplement daily for six weeks. At the end of the 6-week period, each student is given a standardized fitness test and the average score for the sample is M = 39. For the general population of college students, the distribution of test scores is normal with a mean of µ = 35 and a standard deviation of σ = 12. Do students taking the supplement have significantly better fitness scores? Use a one-tailed test with α = .05.

Answer:There is insufficient evidence to support students taking the supplement have significantly better fitness scoresStep-by-step explanation:Sample size = n = 16The average score for the sample is M = 39µ = 35σ = 12We are supposed to find Do students taking the supplement have significantly better fitness scoresUse a one-tailed test with α = .05.[tex]H_0:\mu = 35\\H_a:\mu>35[/tex][tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex][tex]z=\frac{39-35}{\frac{12}{\sqrt{16}}}[/tex][tex]z=1.33[/tex]Refer the z table for p value P(z>1.33)=1-P(z<1.33)=1-0.908=0.092α = .05p value > α So, we failed to reject null hypothesis.So, there is insufficient evidence to support students taking the supplement have significantly better fitness scores