## A researcher claims that the mean annual cost of raising a child (age 2 and under) by husband-wife families in the U.S. is $13,960. In a ran

Question

A researcher claims that the mean annual cost of raising a child (age 2 and under) by husband-wife families in the U.S. is $13,960. In a random sample of husband-wife families in the U.S. the mean annual cost of raising a child (age 2 and under) is $13,725. The sample consists of 500 children and the population standard deviation is $2,345. At the α = 0.10, is there enough evidence to reject the claim? Use the p-value approach.

in progress
0

Math
3 months
2022-02-02T09:54:36+00:00
2022-02-02T09:54:36+00:00 1 Answer
0 views
0
## Answers ( )

Answer:[tex]z=\frac{13725-13960}{\frac{2345}{\sqrt{500}}}=-2.24[/tex]

[tex]p_v =2*P(z<-2.24)=0.0251[/tex]

If we compare the p value and the significance level given [tex]\alpha=0.1[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from 13960 at 10% of signficance.

Step-by-step explanation:Data given and notation[tex]\bar X=13275[/tex] represent the sample mean

[tex]\sigma=2345[/tex] represent the sample standard deviation

[tex]n=500[/tex] sample size

[tex]\mu_o =68[/tex] represent the value that we want to test

[tex]\alpha=0.1[/tex] represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value for the test (variable of interest)

State the null and alternative hypotheses.We need to conduct a hypothesis in order to check if the true mean is 13960, the system of hypothesis would be:

Null hypothesis:[tex]\mu = 13690[/tex]

Alternative hypothesis:[tex]\mu \neq 13690[/tex]

If we analyze the size for the sample is > 30 and we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (1)

z-test:“Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value”.Calculate the statistic

We can replace in formula (1) the info given like this:

[tex]z=\frac{13725-13960}{\frac{2345}{\sqrt{500}}}=-2.24[/tex]

P-value

Since is a two sided test the p value would be:

[tex]p_v =2*P(z<-2.24)=0.0251[/tex]

ConclusionIf we compare the p value and the significance level given [tex]\alpha=0.1[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from 13960 at 10% of signficance.