Question

The average starting salary for the 108 students was $38,584 with a sample standard deviation of $7,500. The mean starting salary for economics majors from other universities is $36,000. At a 5% significance level, is there statistical evidence showing that the mean for your university is different from the other universities’ mean?

Answer 1

Answer:

$t=\frac{38584-36000}{\frac{7500}{\sqrt{108}}}=3.58$    

$p_v =2*P(t_{(107)}>3.58)=0.00052$  

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis and we can conclude that the true mean is different from 36000 at 5% of significance

Step-by-step explanation:

Data given and notation  

$\bar X=38584$ represent the sample mean

$s=7500$ represent the sample standard deviation

$n=108$ sample size  

$\mu_o =36000$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

$p_v$ 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 is different from 36000, the system of hypothesis would be:  

Null hypothesis:$\mu = 36000$  

Alternative hypothesis:$\mu \neq 36000$  

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:  

$t=\frac{38584-36000}{\frac{7500}{\sqrt{108}}}=3.58$    

P-value

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

$df=n-1=108-1=107$  

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

$p_v =2*P(t_{(107)}>3.58)=0.00052$  

Conclusion  

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis and we can conclude that the true mean is different from 36000 at 5% of significance