Question

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.

Answer 1

Answer:

$t=\frac{32400-33600}{\frac{1520}{\sqrt{22}}}=-3.702$    

The degrees of freedom are given by:

$df=n-1=22-1=21$  

The p value is given by:

$p_v =P(t_{(21)}$  

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

$\bar X=33400$ represent the sample mean

$s=1520$ represent the sample standard deviation

$n=22$ sample size  

$\mu_o =33600$ represent the value that we want to analyze

$\alpha=0.05$ represent the significance level

t would represent the statistic

$p_v$ 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:$\mu \geq 33600$  

Alternative hypothesis:$\mu < 33600$  

The statistic is given:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$  (1)  

Replacing the data given we got:

$t=\frac{32400-33600}{\frac{1520}{\sqrt{22}}}=-3.702$    

The degrees of freedom are given by:

$df=n-1=22-1=21$  

The p value is given by:

$p_v =P(t_{(21)}$  

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