Answer:
There is not sufficient evidence to support the claim that average adult listens to the radio less than 26 hours per week at 0.01 significance level.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 26 hours per week
Sample mean, [tex]\bar{x}[/tex] = 22.4 hours per week
Sample size, n = 25
Alpha, α = 0.01
Population standard deviation, σ = 8 hours per week
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 26\text{ hours per week}\\H_A: \mu < 26\text{ hours per week}[/tex]
We use one-tailed z test to perform this hypothesis.
Formula:
[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]z_{stat} = \displaystyle\frac{22.4 - 26}{\frac{8}{\sqrt{25}} } = -2.25[/tex]
Now, we calculate the p-value from the table.
P-value = 0.0122
Since the p-value is greater than the significance level, we fail to reject the null hypothesis and accept it.
Conclusion:
Thus, there is not sufficient evidence to support the claim that average adult listens to the radio less than 26 hours per week at 0.01 significance level.