Question

Young children in the United States are exposed to an average of 4.0 hours of backgroundtelevision per day. Having the television on in the background while children are doing other activitiesmay have adverse consequences on a child’s well-being. You have a research hypothesis that childrenfrom low-income families are exposed to more than 4.0 hours of daily background television. In orderto test this hypothesis, you have collected a random sample of 60 children from low-income familiesand found that these children were exposed to a sample mean of 4.5 hours of daily backgroundtelevision. Based on a previous study, you are willing to assume that the population standard deviationis 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 yourhypothesis 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

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

Explanation:

We are given the following in the question:

Population mean, μ = 4 hours

Sample mean,
`\bar{x}` = 4.5 hours

Sample size, n = 60

Alpha, α = 0.01

Population standard deviation, σ = 1 hour

First, we design the null and the alternate hypothesis

`H_(0): \mu = 4.0\text{ hours}\\H_A: \mu > 4.0\text{ hours}`

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

Formula:

`z_(stat) = \displaystyle\frac{\bar{x} - \mu}{(\sigma)/(√(n)) }`

Putting all the values, we have

`z_(stat) = \displaystyle(4.5 - 4.00)/((1)/(√(60)) ) = 3.872`

Now,
`z_(critical) \text{ at 0.05 level of significance } = 2.33`

Since,

`z_(stat) > z_(critical)`

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

The P-Value is 0.000054.

p-value < 0.01

On the basis of p-value we again reject the null hypothesis. Thus, children from low-income families are exposed to more than 4.0 hours of daily background television.

Answer 2

Answer with explanation:

Let
`\mu` be the population mean.

As per given , we have

a)
`\mu=4\\\\ \mu>4`

Since, the alternative hypothesis is right-tailed , so the test is a right-tailed test.

Given :

`n=60\\\\\overline{x}=4.5\\\\ \sigma=1.0`

z-score :
`z=\frac{\overline{x}-\mu}{(\sigma)/(√(n))}`

`z=(4.5-4)/((1)/(√(60)))=3.87298334621\approx3.87`

b) P-value :
`P(x>4.5)=P(z>3.87)=1-P(z<3.87)`

`\\\\=1-0.9999456=0.0000544`

Critical value for 0.01 significance level =
`z_(0.01)=2.33`

c) Test decision : The test statistic value (3.87) is greater than the critical value (2.33), so we reject the null hypothesis.

d) Conclusion : We have enough evidence to support the claim that children from low-income families are exposed to more than 4.0 hours of daily background television.