Question

The Rocky Mountain district sales manager of Rath Publishing Inc., a college textbook publishing company, claims that the sales representatives make an average of 42 sales calls per week on professors. Several reps say that this estimate is too low. To investigate, a random sample of 40 sales representatives reveals that the mean number of calls made last week was 44. The standard deviation of the sample is 1.9 calls. Using the .01 significance level, can we conclude that the mean number of calls per salesperson per week is more than 42.

Answer 1

Answer:

$t=\frac{44-42}{\frac{1.9}{\sqrt{40}}}=6.657$    

$p_v =P(t_{(39)}>6.657)=3.17x10^{-8}$  

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 42 at 1% of signficance.  

Step-by-step explanation:

Data given and notation  

$\bar X=44$ represent the sample mean

$s=1.9$ represent the sample standard deviation

$n=40$ sample size  

$\mu_o =42$ represent the value that we want to test

$\alpha=0.01$ 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 higher than 42, the system of hypothesis would be:  

Null hypothesis:$\mu \leq 42$  

Alternative hypothesis:$\mu > 42$  

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:  

$t=\frac{44-42}{\frac{1.9}{\sqrt{40}}}=6.657$    

P-value

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

$df=n-1=40-1=39$  

Since is a one side test the p value would be:  

$p_v =P(t_{(39)}>6.657)=3.17x10^{-8}$  

Conclusion  

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 42 at 1% of signficance.