Answer:
The p-value of the test is 0.0119 < 0.06, which means that the data provides statistically significant evidence at the 0.06 level that the average teaching stay in the metro Atlanta area is greater than the state average.
Explanation:
The average teaching salary in Georgia is $48,553 per year. The school districts in the metro Atlanta area boast that they pay more.
At the null hypothesis, we test if they pay the average, that is:
![H_0: \mu = 48553](https://img.qammunity.org/2022/formulas/mathematics/college/objxa3pi8eicg4wed4w7uvvifwxjo34wg9.png)
At the alternate hypothesis, we test if they pay more, that is:
![H_1: \mu > 48553](https://img.qammunity.org/2022/formulas/mathematics/college/b35a552hhur4j1gwqlx6zr07maezbhsycy.png)
The test statistic is:
![z = (X - \mu)/((\sigma)/(√(n)))](https://img.qammunity.org/2022/formulas/mathematics/college/59im90558cjdobm60unnw2lrn6ewzh3ena.png)
In which X is the sample mean,
is the value tested at the null hypothesis,
is the standard deviation and n is the size of the sample.
48553 is tested at the null hypothesis:
This means that
![\mu = 48553](https://img.qammunity.org/2022/formulas/mathematics/college/5f9zhywgc1x9xb4kiki3oo4p0ekaw1wsod.png)
Sample of 228 teachers from the metro Atlanta area and record their salaries. The sample mean is $49,021, with a standard deviation of $3,127.
This means that
![n = 228, X = 49021, \sigma = 3127](https://img.qammunity.org/2022/formulas/mathematics/college/x0awp6vfh6btxglhr1lykhogvl00kkvpf5.png)
Value of the test statistic:
![z = (X - \mu)/((\sigma)/(√(n)))](https://img.qammunity.org/2022/formulas/mathematics/college/59im90558cjdobm60unnw2lrn6ewzh3ena.png)
![z = (49021 - 48553)/((3127)/(√(228)))](https://img.qammunity.org/2022/formulas/mathematics/college/gal8dtx5hdxtizo6s1zihmwd6cs7zxyicg.png)
![z = 2.26](https://img.qammunity.org/2022/formulas/mathematics/college/52vaad1720x6bdxg1u5pt9smmmk7z2qls1.png)
P-value of the test and decision:
The p-value of the test is the probability of finding a sample mean above 49021, which is 1 subtracted by the p-value of z = 2.26.
Looking at the z-table, z = 2.26 has a p-value of 0.9881
1 - 0.9881 = 0.0119
The p-value of the test is 0.0119 < 0.06, which means that the data provides statistically significant evidence at the 0.06 level that the average teaching stay in the metro Atlanta area is greater than the state average.