Question

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

television per day. Having the television on in the background while children are doing other activities

may have adverse consequences on a child’s well-being. You have a research hypothesis that children

from low-income families are exposed to more than 4.0 hours of daily background television. In order

to test this hypothesis, you have collected a random sample of 60 children from low-income families

and found that these children were exposed to a sample mean of 4.5 hours of daily background

television. Based on a previous study, you are willing to assume that the population standard deviation

is 1.0 hours. Use .01 as the level of significance.

a) Develop hypotheses that can be used to test your research hypothesis.

b) What is the p-value based on your sample of 60 children from low-income families? What is your

hypothesis test decision based on p-value approach?

c) Repeat the test using the critical value approach. What is your hypothesis test decision?

d) State your conclusion for the study.

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.