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A certain market research institute is carrying out a study on employees’ salary level in company G. Before the study, researchers believe that those employees earn an average of $50,000 per year. Researchers are not allowed to investigate salary data of all employees in company G but only salary data of 150 employees at random. They find that those 150 employees on average have an annual income of $53,000, and the sample standard deviation is $20,000. Suppose the researchers want to test if the data are consistent with the belief that the true average salary is $50,000 per year. What would the p-value be for this test?Select one:a. Below 0.01b. Between 0.05 and 0.1c. Between 0.025 and 0.05d. Between 0.01 and 0.025

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Answer:

[tex]p_v =2*P(t_{149}>1.837)=0.0682[/tex]  

b. Between 0.05 and 0.1

Step-by-step explanation:

Data given and notation

[tex]\bar X=53000[/tex] represent the sample mean  

[tex]s=20000[/tex] represent the standard deviation for the sample

[tex]n=150[/tex] sample size  

[tex]\mu_o =50000[/tex] represent the value that we want to test  

[tex]\alpha[/tex] represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

[tex]p_v[/tex] represent the p value for the test (variable of interest)  

State the null and alternative hypotheses to be tested  

We need to conduct a hypothesis in order to determine if the true mean is 50000, the system of hypothesis would be:  

Null hypothesis:[tex]\mu = 50000[/tex]  

Alternative hypothesis:[tex]\mu \neq 50000[/tex]  

Compute the test statistic  

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

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

We can replace in formula (1) the info given like this:  

[tex]t=\frac{53000-50000}{\frac{20000}{\sqrt{150}}}=1.837[/tex]  

Now we need to find the degrees of freedom for the t distirbution given by:

[tex]df=n-1=150-1=149[/tex]

What do you conclude?   Use the p-value approach

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

[tex]p_v =2*P(t_{149}>1.837)=0.0682[/tex]  

So on this case the best answer would be:

b. Between 0.05 and 0.1

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