# discussion question and peer reviews

ng>DISCUSSION POST MUST BE 10+ SENTENCES AND THE PEER REVIEWS MUST BE 7+ SENTENCES PLEASE USE YOUR OWN WORDS AND DO NOT COPY FROM OTHER SITES STAY ON TOPIC ,BE POSITIVE , AND DO NOT COMMENT ON GRAMMAR ERRORS. TALK DIRECTLY TO CLASSMATES IN PEER REVIEWS!!! YOU MUST COMPLETE BOTH PEER REVIEWS AND DISCUSSION QUESTION!! THANK YOU.

1) DISCUSSION QUESTION

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Employees from Company A and Company B both receive annual bonuses. What information would you need to test the claim that the difference in annual bonuses is greater than \$100 at the 0.05 level of significance? Write out the hypothesis and explain the testing procedure.

2) PEER REVIEW #1 (TAMMY)

Employees from Company A and Company B both receive annual bonuses. What information would you need to test the claim that the difference in annual bonuses is greater than \$100 at the 0.05 level of significance? Write out the hypothesis and explain the testing procedure.

Explanation:

First, a sample of employees in each company, with their bonuses

nA ,nB

second, compute the average and standard deviation of the bonds in each company

third, hypothesis

H0:?A??B=100

H1:?A??B>100

fourth, value of the test statistic

ttest=spn11+n21(x1??x2?)?(?1??2)

fifth, value of p

P=P(t>ttest )

sixth, if p is less than 0.05, H0 is rejected….

From my understanding of hypothesis and what I recently learned from this course is that I know it is an educated guess. But in this scenario, the average between the two companies must be found weather the testing of the hypothesis is rejected or not. I feel that there could have been more information provided for this question but overall, the null hypothesis will be rejected if the maximum percent is 0.05. Also, if the p-value is less than then a two tailed test must be conducted and the null hypothesis should be rejected.

I really don’t know quite if I’m on the right track in regards to this discussion but I need feedback please.

3)PEER REVIEW #2 (MONICA

For this problem we have two independent variable, company A and company B, each having a bonus Mu 1 and Mu 2.  The claim is the difference between these variables is greater than \$100. Mu 1 – Mu 2 > 100.  This is the alternative hypothesis. The opposite is the null hypothesis Mu 1 – Mu 2 <= 100.  From here we must determine which test statistic to use t or z. The t test is used because we don’t have a large enough sample size and standard deviation is unknown.Then calculate the p value. If the p value is less than 0.05 we will reject the null hypothesis. If it’s greater then we fail to reject it.

I am not sure how to notate the equation using my keyboard so I hope this explanation will suffice.

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