Question

Intelligence of the general population has a mean of µ = 100 and a standard deviation of σ = 15. A researcher wonders if people who exercise more than 30 min each day have an average mean IQ different from 100. He randomly selects n = 81 people and enrolls them in an exercise program for six months. After, he has each individual take an IQ test and finds the average IQ of the sample is M = 102.

a) Calculate the SE (standard Error)

b) Calculate a 95% confidence interval for the population mean of exercisers. Make sure to use the standard error you calculated in the previous question. Remember to round each calculation to the nearest hundredths digit (X.xx)

c )Using the values you obtained in the previous question, interpret/describe what the confidence interval is telling you.

d) Calculate the Z-test value.

e) What is the p-value for your z-test statistic?

f) Given your Z-test value you calculated in the previous question, what decision should you make regarding the null hypothesis? Justify your answer.

Use an α level = .05 for this problem.

g) What conclusion should you make regarding the effect of exercise on intelligence?

Group of answer choices

Answer 1

a) Standard error= Standard deviation/√n = 15/√81= 1.667

b) 95% CI for u is given as:

Sample mean +- z0.025*Standard error

= 102 +- 1.96*1.667

=(98.73, 105.27)

c) We are 95% confident that the true mean IQ of people who exercise for more than 30 mins each day lies between 98.73 and 105.27

d) z test: (102-100)/1.667= 1.20

e) p-value of this one sided z test is: P(z>1.20) = 1- P(z<=1.2) = 0.1151 (from z tables)

f) Since p-value of this z test >significance level of 0.05, we have insufficient evidence to Reject H0. So we conclude that u=100

g) So we fail to conclude that u>100 at the 5% level of significance.