A recent national survey found that high school students watched an average (mean) of 6.8 DVDs per month with a population standard deviation of 1.8 DVDs. The distribution of DVDs watched per month follows the normal distribution. A random sample of 36 college students revealed that the mean number of DVDs watched last month was 6.2. At the .05 significance level, can we conclude that college students watch fewer DVDs a month than high school students? value:
10.00 points
Required information
a. State the null hypothesis and the alternate hypothesis.
b. State the decision rule.
c. Compute the value of the test statistic. (Negative amount should be indicated by a minus sign. Round your answer to 1 decimal place.)
d. What is your decision regarding H0?
e. What is the p-value? (Round your answer to 4 decimal places.)

A)Null hypothesis; μ ≥ 6.8

Alternative hypothesis; μ < 6.8

B) The decision rule is that we reject the null hypothesis for all z-scores less than - 1.65

C)test statistic: z = -2

D) We reject Null hypothesis H0.

E) P-value ≈ 0.0228

Explanation:

We are given;

Sample size; n = 36

Population mean; μ = 6.8

Sample mean; x¯ = 6.2

Standard deviation; σ = 1.8

A) Let's state the hypotheses;

Null hypothesis; μ ≥ 6.8

Alternative hypothesis; μ < 6.8

B) The significance level is 0.05.

Let us find out the z-score that corresponds to a probability of 1 - 0.5 - 0.05 = 0.45 in the normal probability table attached.

From the table attached, we can that the corresponding z-score is approximately 1.6 + 0.05 = 1.65

The alternative hypothesis contains less than symbol which means this test is left tailed.

Thus, the rejection region includes all z-scores that are smaller than -1.65. Which means we reject the null hypothesis for all z-scores less than - 1.65

C) Formula for the test statistic is;

z = (x¯ - μ)/(σ/√n)

z = (6.2 - 6.8)/(1.8/√36)

z = -0.6/0.3

z = -2

D) The test statistic gotten is z = -2

This is less than the z-score for the rejection the rejection region is answer B above.

Thus, we reject the null hypothesis.

E) from online p-value from z-score calculator attached , using; z = -2; significance value = 0.05, one tailed hypothesis, we have;

P-value = 0.02275

To 4 decimal places gives;

P-value ≈ 0.0228