Question

The average amount of lunch bills per person reimbursed in an international company has been u=$25.5. To test if this average has increased recently, a random sample n=64 of bills is selected. The sample mean and standard deviation are x=$26 and s=$6.25, respectively. Conduct the hypothesis test at a significance level alpha=0.05.

Answer 1

Answer:

$t=\frac{26-25.5}{\frac{6.25}{\sqrt{64}}}=0.64$      

$p_v =P(t_{63}>0.64)=0.262$  

If we compare the p value and the significance level given for example $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis, and the the actual true mean is not significantly higher than 25.5.      

Step-by-step explanation:

Data given and notation      

$\bar X=26$ represent the mean completion time for ABC -5500 form for the sample      

$s=6.25$ represent the standard deviation for the sample      

$n=64$ sample size      

$\mu_o =25.5$ 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 has increased recently, the system of hypothesis would be:      

Null hypothesis:$\mu \leq 25.5$      

Alternative hypothesis:$\mu > 25.5$      

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:      

$t=\frac{26-25.5}{\frac{6.25}{\sqrt{64}}}=0.64$      

Calculate the P-value      

First we need to calculate the degrees of freedom given by:  

$df=n-1=64-1=63$  

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

$p_v =P(t_{63}>0.64)=0.262$  

Conclusion      

If we compare the p value and the significance level given for example $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis, and the the actual true mean is not significantly higher than 25.5.