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 1.0. The distribution of number of movies watched per month follows the normal distribution. A random sample of 49 college students revealed that the mean number of movies watched last month was 6.5. 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. A. H0: μ ≥ 7.2; H1: μ < 7.2

B. H0: μ = 7.2; H1: μ ≠ 7.2

C. H0: μ > 7.2; H1: μ = 7.2

D. H0: μ ≤ 7.2; H1: μ > 7.2

2. State the decision rule.

A. Reject H1 if z < –1.645

B. Reject H0 if z > –1.645

C.Reject H1 if z > –1.645

D. 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.)

Value of the test statistic______________?

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

Answer 1

Answer:

1. Option A

2. Option D) Reject null hypothesis if z < –1.645

3. z = -4.90

4. P-value = 0.00001

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 7.2

Sample mean, $\bar{x}$ = 6.5

Sample size, n = 49

Alpha, α = 0.05

Population standard deviation, σ = 1.0

a) First, we design the null and the alternate hypothesis

$H_{0}: \mu \geq 7.2\text{ movies per month}\\H_A: \mu < 7.2\text{ movies per month}$

Option A) is the correct answer

We use one-tailed(left) z test to perform this hypothesis.

c) Formula:

$z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Putting all the values, we have

$z_{stat} = \displaystyle\frac{6.5 - 7.2}{\frac{1}{\sqrt{49}} } = -4.90$

b) Now, $z_{critical} \text{ at 0.05 level of significance } = -1.645$

Rejection rule:

If calculated z-statistic is less than the critical z-value, we reject the null hypothesis.

Option D) Reject null hypothesis if z < –1.645

d) Now, we calculate the p-value with the help of standard normal table.

P-value = 0.00001

Since the p-value is lower than the significance level, we fail to accept the null hypothesis and reject it.We accept the alternate hypothesis.

We conclude that the college students watch fewer movies a month than high school students.