Question

In: Statistics and Probability

The average “moviegoer” sees 8.5 movies a year. A moviegoer is defined as a person who...

The average “moviegoer” sees 8.5 movies a year. A moviegoer is defined as a person who sees at least one movie in a theater in a 12-month period. A random sample of 40 moviegoers from a large university revealed that the average number of movies seen per person was 9.6. The population standard deviation is 3.2 movies. At the 0.05 level of significance, can it be concluded that this represents a difference from the national average?

STEP 1. State the null and alternate hypothesis

The hypotheses are  (Enter an UPPER CASE Letter Only.)

STEP 2. State the critical value(s). Enter the appropriate letter.

z =

STEP 3. Calculate the test value

z =

STEP 4. Make the decision by rejecting or not rejecting the null hypothesis. Since the test value falls in the non-rejection region, we do not reject the null hypothesis.

Conclusion 1. Reject the null hypothesis. At the α = 0.05 significance level there is enough evidence to conclude that the average number of movies seen by people each year is not different from 8.5.

Conclusion 2. Reject the null hypothesis. At the α = 0.05 significance level there is enough evidence to conclude that the average number of movies seen by people each year is different from 8.5.

Conclusion 3. Do not reject the null hypothesis. At the α = 0.05 significance level there is enough evidence to conclude that the average number of movies seen by people each year is different from 8.5.

Conclusion 4. Do not reject the null hypothesis. At the α = 0.05 significance level there is enough evidence to conclude that the average number of movies seen by people each year is 8.5.

(Enter a number only from the list 1, 2, 3, or 4)

Solutions

Expert Solution

Solution:

Given:

Mean =

Sample size = n = 40

Sample mean =

Population standard deviation =

level of significance = 0.05

We have to test if it can be concluded that this represents a difference from the national average.

STEP 1. State the null and alternate hypothesis

Since hypothesis statement is non-directional, this is two tailed test.

thus

Vs

STEP 2. State the critical value(s).

level of significance = 0.05

Since this is two tailed, we find : Area =

Look in z table for area = 0.0250 or its closest area and find z value

Area 0.0250 corresponds to -1.9 and 0.06

thus z critical value = -1.96

Since this is two tailed test, we have two z critical values: ( -1.96 , 1.96)

STEP 3. Calculate the test value

STEP 4. Make the decision by rejecting or not rejecting the null hypothesis.

Since z = 2.17 > z critical value = 1.96, we reject H0.

Since the test value falls in the rejection region, we reject the null hypothesis.

Conclusion:

Conclusion 2. Reject the null hypothesis. At the α = 0.05 significance level there is enough evidence to conclude that the average number of movies seen by people each year is different from 8.5.


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