Question

Young children in the United States are exposed to an average of 4.0 hours of background

Answer 1

Answer:

The critical value approach and the p-value approach gave the same result.              

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 4 hours

Sample mean, $\bar{x}$ = 4.5

Sample size, n = 460

Alpha, α = 0.051

Population standard deviation, σ = 1.00 hour

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 4.00\text{ hours}\\H_A: \mu > 4.00\text{ hours}$

We use One-tailed z test to perform this hypothesis.

Formula:

$z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Putting all the values, we have

$z_{stat} = \displaystyle\frac{4.5 - 4}{\frac{1}{\sqrt{60}} } = 3.87$

Now, $z_{critical} \text{ at 0.01 level of significance } = 2.326$

Since,  

$z_{stat} > z_{critical}$

We reject the null hypothesis and accept the alternate hypothesis. Thus, low-income families are exposed to more than 4.0 hours of daily background television.

p value at z-stat 3.87 is 0.000054. Since p-value is less than the significance level. We reject the null hypothesis.

Hence, the critical value approach and the p-value approach gave the same result.