Question

Suppose a study determines that the amount of time that college students on a given campus work out each week changes from university to university. You are intrigued by this and randomly interview 60 BU students. You find that the average weekly gym time is 3 hours, and the standard deviation is 1 hour.

i. Is the sample mean normally distributed? Why or why not?  

ii. Your friend thinks the average gym time for Boston University students is equal to 2.5 hours. Test their hypothesis at a 5 percent significance level. Be sure to clearly state the null hypothesis, the rejection region, and your conclusion.

iii. Calculate the p-value. Offer a range if you can't obtain the exact value.

Suppose now you do know that the population standard deviation is 1.

iv. Find the 90 percent confidence interval for the population mean. State the general formula, fill in the appropriate values, and determine the exact boundaries of the interval. [20 points]

Answer 1

i. As sample size is 60>30, as per central limit theorem distribution of sample mean is normal.

ii. Here claim is that mean is equal to 2.5, so hypothesis is vs

So test statistics is

The t-critical values for a two-tailed test, for a significance level of α=0.05 are

tc​=−2.001 and tc​=2.001

Graphically

As test statistics falls in the rejection region we reject the null hypothesis

iii. P value is TDIST(3.87,59,2)=0.00028

iv. Now as population standard deviation is given we can use z distribution

z value for 90% CI is 1.645 as P(-1.645<z<1.645)=0.90

Margin of Error is

Hence CI is