Question

A drug company that manufactures a diet drug claims that those using the drug for 30 days will lose at least 15 pounds. You sample 30 people who have used the drug and find that the average weight loss was 12 pounds with a standard deviation of 5 pounds. (a) Test the claim at the .05 significance level. (b) No claim was made by the company. They took a sample with the above results and ask you to construct a two-sided 95% confidence interval for the population mean.

MUST BE SOLVED BY HAND, WRITTEN OUT STEP BY STEP. NO EXCEL PROGRAM!

Answer 1

Answer)

As the population s.d is unknown here we will use t distribution to conduct the test and estimate the interval as well

A)

Null hypothesis Ho : u >= 15

Alternate hypothesis Ha : u < 15

Test statistics t = (sample mean - claimed mean)/(s.d/√n)

t = (12-15)/(5/√30)

t = -3.286

Degrees of freedom is = n-1 = 29

For 29 df and -3.286 test statistics p-value from t distribution is 0.0013

As obtained P-Value is less than tha given significance level 0.05

We reject the null hypothesis Ho

So we do not have enough evidence to support the claim.that u >=15

B)

Degrees of freedom is = n-1 = 29

For 29 df and 95% confidence level critical value t from t table is = 2.05

MOE = t*s.d/√n = 2.05*5/√30 = 1.87

Confidence interval is given by

(Mean - moe, mean +moe)

Mean = 12

[10.13, 13.87].

We are 95% confident that the population mean (μ) falls between 10.13 and 13.87.