Question

The US Department of Agriculture reports that the mean cost of raising a child from birth to age 2 in a rural area is $10,460. A team of economists believes this value is incorrect, so they select a random sample of 900 children (age 2) and find that the sample mean cost is $10,345 with a sample standard deviation of $1540. Is this significantly different than the US Department of Agriculture reports

Answer 1

`t=(10345-10460)/((1540)/(√(900)))=-2.24`

The degrees of freedom are given by:

`df=n-1=900-1=899`

The p value for this case would be given by:

`p_v =2*P(t_((899))<-2.24)=0.0127`

The p value is low and if we use a significance level of 0.05 we can reject the null hypothesis and we can conclude tha the true mean is different from 10460

Explanation:

Information given

`\bar X=10345` represent the mean height for the sample

`s=1540` represent the sample standard deviation for the sample

`n=900` sample size

`\mu_o =10460` represent the value to verify

t would represent the statistic

`p_v` represent the p value for the test

Hypothesis to test

We want to verify if the true mean is equal to 10460, the system of hypothesis would be:

Null hypothesis:
`\mu = 10460`

Alternative hypothesis:
`\mu \\eq 10460`

The statistic is given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

Replacing the info given we got:

`t=(10345-10460)/((1540)/(√(900)))=-2.24`

The degrees of freedom are given by:

`df=n-1=900-1=899`

The p value for this case would be given by:

`p_v =2*P(t_((899))<-2.24)=0.0127`

The p value is low and if we use a significance level of 0.05 we can reject the null hypothesis and we can conclude tha the true mean is different from 10460