Question

The Mean Corporation would like to invest in the booming health food industry. It is considering the creation of a health-drink franchise called Goose Juice. The investment department of the Mean Corporation wants to investigate the feasibility of this venture by examining the profits of similar franchises. It believes that the venture will be feasible if an average annual profit of more than $89,000 can be expected from each Goose Juice that is opened. It is known that the annual profits earned by health-drink franchises has a population standard deviation of $7,800. The Mean Corporation's statisticians would like to construct a hypothesis test for the mean annual profit (μ) earned by health-drink franchises. A random sample of 60 franchises were chosen and their annual profit for the previous financial year was recorded. The mean annual profit for the sample was calculated as $90,300. The hypotheses that will be used by the statisticians are H0: μ = 89,000 and Ha: μ > 89,000.

Answer 1

Answer:

$z=\frac{90300-89000}{\frac{7800}{\sqrt{60}}}=1.29$      

$p_v =P(z>1.29)=0.0986$  

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 mean is not significantly higher than 89000 with 5% of significance.

Step-by-step explanation:

Data given and notation      

$\bar X=90300$ represent the sample mean

$\sigma=7800$ represent the standard deviation for the population    

$n=60$ sample size      

$\mu_o =89000$ 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 true mean is higher than 89000, the system of hypothesis would be:      

Null hypothesis:$\mu \leq 89000$      

Alternative hypothesis:$\mu > 89000$      

We know the population deviation, so for this case is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:      

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)      

z-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:      

$z=\frac{90300-89000}{\frac{7800}{\sqrt{60}}}=1.29$      

Calculate the P-value      

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

$p_v =P(z>1.29)=0.0986$  

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 mean is not significantly higher than 89000 with 5% of significance.