A temp agency feels that one of its recruiters has unorthodox methods and claims that the wages the recruiter secures for its temps differs from the agency's average wage of $12 per hour. The agency takes a random sample of 49 recent contracts the recruiter secured and finds an average wage of $11.25 per hour and a standard deviation of $2.25 per hour. Does the evidence support the agency's claim at the 5% significance level?

If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, we can conclude that the true mean is different from 12 at at 5% of signficance.

Step-by-step explanation:

Data given and notation

[tex]\bar X=11.25[/tex] represent the sample mean

[tex]s=2.25[/tex] represent the sample standard deviation

[tex]n=49[/tex] sample size

[tex]\mu_o =12[/tex] represent the value that we want to test

[tex]\alpha=0.05[/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 mean differes from 12, the system of hypothesis would be:

Null hypothesis:[tex]\mu = 12[/tex]

Alternative hypothesis:[tex]\mu \neq 12[/tex]

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

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

t-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]t=\frac{11.25-12}{\frac{2.25}{\sqrt{49}}}=-2.33[/tex]

P-value

The first step is calculate the degrees of freedom, on this case:

[tex]df=n-1=49-1=48[/tex]

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

[tex]p_v =2*P(t_{(48)}<-2.33)=0.024[/tex]

Conclusion

If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, we can conclude that the true mean is different from 12 at at 5% of signficance.