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

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

Step-by-step explanation:

Information given

[tex]\bar X=10345[/tex] represent the mean height for the sample

[tex]s=1540[/tex] represent the sample standard deviation for the sample

[tex]n=900[/tex] sample size

[tex]\mu_o =10460[/tex] represent the value to verify

t would represent the statistic

[tex]p_v[/tex] 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:[tex]\mu = 10460[/tex]

Alternative hypothesis:[tex]\mu \neq 10460[/tex]

The statistic is given by:

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

Replacing the info given we got:

[tex]t=\frac{10345-10460}{\frac{1540}{\sqrt{900}}}=-2.24[/tex]

The degrees of freedom are given by:

[tex]df=n-1=900-1=899[/tex]

The p value for this case would be given by:

[tex]p_v =2*P(t_{(899)}<-2.24)=0.0127[/tex]

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