Question

A local Internet provider wants to test the claim that the average time a family spends online on a Saturday is at least 7 hours. To test this claim, the Internet provider randomly samples 30 households and finds that these families' mean number of hours spent on the Internet on a Saturday was 6 hours with a standard deviation of 1.5 hours. At a level of significance of 0.05, can the Internet provider's claim be supported?

A) Fail to Reject the Null Hypothesis
B) Reject the Null Hypothesis
C) Reject The Alternative Hypothesis
D) Fail to Reject the Alternative Hypothesis
E) Accept the Null Hypothesis
F) Accept the Alternative Hypothesis

Answer 1

A) Fail to Reject the Null Hypothesis

Explanation:

Given that:

A local Internet provider wants to test the claim that the average time a family spends online on a Saturday is at least 7 hours.

sample size = 30

sample mean
`\bar x` = 6

standard deviation
`\sigma` = 1.5

level of significance ∝ = 0.05

The null hypothesis and the alternative hypothesis can be computed as:

`\mathbf{ H_o: \mu \leq 7}`

`\mathbf{ H_i: \mu \geq 7}`

The test statistic can be computed as:

`z = (\bar x - \mu)/((\sigma)/(√(n)))`

`z = \frac{6 -7} {\frac{1.5}{\sqrt {30}}}`

`z = \frac{-1} {(1.5)/(5.477)}}`

`z = \frac{-5.477} {1.5}`

z = -3.65

Given that ;

level of significance of 0.05;

z = -3.65

degree of freedom = 30 - 1 = 29

The p-value = P(
`t_(29)` > - 3.65)

= 0.9998

Decision Rule: Reject
`H_o` if p-value is less than the level of significance

But since the p -value is greater than the level of significance, we conclude that There is no enough evidence to support the Internet provider claim, Therefore;

Fail to Reject the Null Hypothesis