Question

The mean hourly wage for employees in goods-producing industries is currently $24.57 (Bureau of Labor Statistics website, April, 12, 2012). Suppose we take a sample of employees from the manufacturing industry to see if the mean hourly wage differs from the reported mean of $24.57 for the goods-producing industries.

Answer 1

Answer:

a)Null hypothesis:$\mu = 24.57$  

Alternative hypothesis:$\mu \neq 24.57$  

b) $df=n-1=30-1=29$  

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

$p_v =2*P(t_{(29)}$  

c)  If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is different from 24.57 at 5% of signficance.

d)  For this case since we have a two tailed test we need to find two critical values and we need 0.025 of the area on each tail of the t distribution with 29 degrees of freedom. The critical values are given by:

$t_{crit}=\pm 2.045$

And the rejection zone would be given by:

$(-\infty , -2.045) U(2.045,\infty)$

Since pur calculated value is not on the rejection zone we Fail to reject the null hypothesis.

Step-by-step explanation:

The mean hourly wage for employees in goods-producing industries is currently $24.57. Suppose we take a sample of employees from the manufacturing industry to see if the mean hourly wage differs from the reported mean of $24.57 for the goods-producing industries.

a. State the null and alternative hypotheses we should use to test whether the population mean hourly wage in the manufacturing industry differs from the population mean hourly wage in the goods-producing industries.

We need to conduct a hypothesis in order to check if the mean is equal to 24.57 or not, the system of hypothesis would be:  

Null hypothesis:$\mu = 24.57$  

Alternative hypothesis:$\mu \neq 24.57$  

b. Suppose a sample of 30 employees from the manufacturing industry showed a sample mean of $23.89 per hour. Assume a population standard deviation of $2.40 per hour and compute the p-value.

$\bar X=23.89$ represent the sample mean  

$s=2.4$ represent the sample standard deviation

$n=30$ sample size  

$\mu_o =24.57$ represent the value that we want to test

t would represent the statistic (variable of interest)  

$p_v$ represent the p value for the test (variable of interest)  

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:  

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$  (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:  

$t=\frac{23.89-24.57}{\frac{2.4}{\sqrt{30}}}=-1.552$    

P-value

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

$df=n-1=30-1=29$  

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

$p_v =2*P(t_{(29)}$  

c. With  α = 0.05  as the level of significance, what is your conclusion?

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is different from 24.57 at 5% of signficance.

d. Repeat the preceding hypothesis test using the critical value approach.

For this case since we have a two tailed test we need to find two critical values and we need 0.025 of the area on each tail of the t distribution with 29 degrees of freedom. The critical values are given by:

$t_{crit}=\pm 2.045$

And the rejection zone would be given by:

$(-\infty , -2.045) U(2.045,\infty)$

Since pur calculated value is not on the rejection zone we Fail to reject the null hypothesis.