Teachers’ salaries in one state are very low that the educators in that state regularly complain about their compensation. The state mean is $33,600, but teachers in one district claim that the mean their district is significantly lower. They survey a simple random sample of 22 teachers in the district and calculate a mean salary of $32,400 with a standard deviation s = $ 1520. Test the teachers’ claim at the 0.05 level of significance.

The p value is significantly lower than the significance level so then we have enough evidence to conclude that the true mean is significantly lower from 33600

Step-by-step explanation:

Information given

[tex]\bar X=33400[/tex] represent the sample mean

[tex]s=1520[/tex] represent the sample standard deviation

[tex]n=22[/tex] sample size

[tex]\mu_o =33600[/tex] represent the value that we want to analyze

[tex]\alpha=0.05[/tex] represent the significance level

t would represent the statistic

[tex]p_v[/tex] represent the p value for the test

System of hypothesis

We want to check if the true mean is lower than 33600, the system of hypothesis would be:

Null hypothesis:[tex]\mu \geq 33600[/tex]

Alternative hypothesis:[tex]\mu < 33600[/tex]

The statistic is given:

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

Replacing the data given we got:

[tex]t=\frac{32400-33600}{\frac{1520}{\sqrt{22}}}=-3.702[/tex]

The degrees of freedom are given by:

[tex]df=n-1=22-1=21[/tex]

The p value is given by:

[tex]p_v =P(t_{(21)}<-3.702)=0.00066[/tex]

The p value is significantly lower than the significance level so then we have enough evidence to conclude that the true mean is significantly lower from 33600