Question

3. It is known that the average person has 9.5 alcoholic drinks per week. Suppose we take a sample of 25 people and find that the average amount of alcoholic drinks per week they have is 7.8 with a sample standard deviation of 9.6 drinks. We are interested in testing if the average amount of alcoholic drinks per week is actually lower than 9.5 now. a. Using the sample of 25 people, what would the 95% confidence interval be for the population mean? b. What are the null and alternative hypotheses? c. What is the critical value at 95% confidence? d. Calculate the test statistic. e. Find the p-value. f. What conclusion would be made here at the 95% confidence level? g. Would my conclusion change if I changed alpha to .01? Show reasoning.

Answer 1

a)

= 7.8, =9.5, n=25, s=9.6

formula for confidence interval is

where tc is the t critical value for C =95% with df =n-1 =25-1 =24

tc =2.064

7.8 − 3.963 < < 7.8 + 3.963

thus we get confidence interval as (3.837, 11.763)

b)

Ho: 9.5

Ha: < 9.5

c)

now calculate t critical value for left tailed test with = 0.05

Critical value =−1.711

d)

Formula for test statistics is

test statistics = −0.885

e)

Calculate P-Value for left tailed test with df= n-1 = 25-1= 24

We get

P-value = 0.1924

f)

Rule to take decision is

Reject Ho if (p-value) ( )

Here (p-value= 0.1924) > ( =0.05 )

Hence failed to reject the null hypothesis(Ho).

g)

if   = 0.01

Rule to take decision is

Reject Ho if (p-value) ( )

Here (p-value= 0.1924) > ( =0.01 )

Hence failed to reject the null hypothesis(Ho).