Question

A car dealer who sells only late-model luxury cars recently hired a new salesman and believes that this salesman is selling at lower markups. He knows that the long-run average markup in his lot is $5,600. He takes a random sample of 16 of the new salesman's sales and finds an average markup of $5,000 and a standard deviation of $800. Assume the markups are normally distributed. What is the value of an appropriate test statistic for the car dealer to use to test his claim?

Answer 1

Answer:

$t=\frac{5000-5600}{\frac{800}{\sqrt{16}}}=-3$      

Explanation:

Data given and notation      

$\bar X=5000$ represent the sample mean      

$s=800$ represent the standard deviation for the sample      

$n=16$ sample size      

$\mu_o =5600$ represent the value that we want to test    

$\alpha$ 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 determine if the mean is lower than 5600, the system of hypothesis would be:      

Null hypothesis:$\mu \geq 5600$      

Alternative hypothesis:$\mu < 5600$      

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:      

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

$t=\frac{5000-5600}{\frac{800}{\sqrt{16}}}=-3$