Answer:
A) Fail to Reject the Null Hypothesis
Step-by-step 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 [tex]\bar x[/tex] = 6
standard deviation [tex]\sigma[/tex] = 1.5
level of significance ∝ = 0.05
The null hypothesis and the alternative hypothesis can be computed as:
[tex]\mathbf{ H_o: \mu \leq 7}[/tex]
[tex]\mathbf{ H_i: \mu \geq 7}[/tex]
The test statistic can be computed as:
[tex]z = \dfrac{\bar x - \mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \dfrac{6 -7} {\dfrac{1.5}{\sqrt {30}}}[/tex]
[tex]z = \dfrac{-1} {\dfrac{1.5}{5.477}}}[/tex]
[tex]z = \dfrac{-5.477} {1.5}[/tex]
z = -3.65
Given that ;
level of significance of 0.05;
z = -3.65
degree of freedom = 30 - 1 = 29
The p-value = P([tex]t_{29}[/tex] > - 3.65)
= 0.9998
Decision Rule: Reject [tex]H_o[/tex] 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
