Question

A recent national survey found that high school students watched an average (mean) of 6.6 DVDs per month with a population standard deviation of 0.90 hour. The distribution of DVDs watched per month follows the normal distribution. A random sample of 43 college students revealed that the mean number of DVDs watched last month was 6.10. At the 0.05 significance level, can we conclude that college students watch fewer DVDs a month than high school students?

b.	State the decision rule.

Reject H₀ if z > -1.645

Reject H₁ if z < -1.645

Reject H₀ if z < -1.645

Reject H₁ if z > -1.645

Answer 1

Solution :

Given that,

Population mean = = 6.10

Sample mean = = 6.6

Population standard deviation = = 0.90

Sample size = n = 43

Level of significance = = 0.05

This is a left tailed test.

The null and alternative hypothesis is,

Ho: 6.10

Ha: 6.10

The test statistics,

Z =( - )/ (/n)

= ( 6.6 - 6.10 ) / ( 0.90 / 43 )

= 3.643

Critical value of  the significance level is α = 0.05, and the critical value for a left-tailed test is

= -1.645

Since it is observed that |z| = 3.643 > = -1.645, it is then concluded that the null hypothesis is rejected.

Reject Ho if z > -1.645