Question

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.

1. Using the sample of 25 people, what would the 95% confidence interval be for the population mean?

2. What are the null and alternative hypotheses?

3. What is the critical value at 95% confidence?

4. Calculate the test statistic.

5. Find the p-value.

6. What conclusion would be made here at the 95% confidence level?

7. Would my conclusion change if I changed alpha to .01? Show reasoning.

Answer 1

a)

The confidence interval for mean is obtained using the formula,

From the data values,

The t critical value is obtained from t distribution table for significance level = 0.05 and degree of freedom = n -1 = 25 - 1 = 24.

b)

The null and alternative hypotheses are,

c)

The t critical value is obtained from t distribution table for significance level = 0.05 and degree of freedom = n -1 = 25 - 1 = 24 for one sided test.

d)

The t statistic is obtained using the formula,

Where,

e)

The p-value is obtained from t distribution table for degree of freedom = n - 1 = 25 - 1 = 24 and significance level = 0.05. (in excel use function =T.DIST(-0.885,24,TRUE))

f)

Since the P-value is less than significance level = 0.05 at 95% confidence interval, the null hypothesis is not rejected. Hence, it can be concluded that there is no difference in sample and population mean drink per week.

h)

No, the result would be same as 99% confidence interval will be more wider and population mean is already within the 95% confidence interval.