Question

A recent national survey found that high school students watched an average (mean) of 7.2 movies per month with a population standard deviation of 0.7. The distribution of number of movies watched per month follows the normal distribution. A random sample of 47 college students revealed that the mean number of movies watched last month was 6.2. At the 0.05 significance level, can we conclude that college students watch fewer movies a month than high school students?

1. State the null hypothesis and the alternate hypothesis.

• H₀: μ ≥ 7.2; H₁: μ < 7.2

• H₀: μ = 7.2; H₁: μ ≠ 7.2

• H₀: μ > 7.2; H₁: μ = 7.2

• H₀: μ ≤ 7.2; H₁: μ > 7.2

2. State the decision rule.

• Reject H₁ if z < –1.645

• Reject H₀ if z > –1.645

• Reject H₁ if z > –1.645

• Reject H₀ if z < –1.645

3. Compute the value of the test statistic. (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.)

4. What is your decision regarding H₀?

• Reject H₀

• Do not reject H₀

5. What is the p-value? (Round your answer to 4 decimal places.)

Answer 1

Part a

H0: μ ≥ 7.2; H1: μ < 7.2

Part b

α = 0.05

Test is lower tailed.

So, Z = -1.645

Reject H0 if z < –1.645

Part c

The test statistic formula is given as below:

Z = (Xbar - µ)/[σ/sqrt(n)]

From given data, we have

µ = 7.2

Xbar = 6.2

σ = 0.7

n = 47

α = 0.05

Critical value = -1.645

(by using z-table or excel)

Z = (6.2 – 7.2)/[0.7/sqrt(47)]

Z = -9.7938

Part d

Test statistic is less than critical value, so we reject the null hypothesis.

Reject H0

Part e

P-value = 0.0000

(by using Z-table)

P-value < α = 0.05

So, we reject the null hypothesis

There is sufficient evidence to conclude that college students watch fewer movies a month than high school students.