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ElenaW [278]
3 years ago
12

The average starting salary for the 108 students was $38,584 with a sample standard deviation of $7,500. The mean starting salar

y 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?
Mathematics
1 answer:
White raven [17]3 years ago
8 0

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

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