Mr. Graham was fired from LA Pierce College because the word got out that he was

Question

Mr. Graham was fired from LA Pierce College because the word got out that he was spending too much time with the Coyotes over by the Duck Pond. So, Mr. Graham went to work for a cereal company in the quality control department where he was in charge, of making sure that the not too much or too little cereal was put into each of the boxes. Specifically, the cereal company wanted Mr. Graham to check if they were putting too much or too little cereal in the boxes. The population, mean () weight, for a box a cereal was 15 ounces with a population standard deviation () of .5 ounces. The cereal company suspected that too much cereal was being put into the cereal boxes. Mr. Graham took a SRS of 25 boxes of cereal to check on the cereal company's concern. He found that the average (sample mean) weight for a box of cereal was M = 15.18 ounces in this sample (Use for questions 1 - 11).

1. What are the conditions and assumptions necessary to apply the ‘One Sample Mean Z Statistic for the above study?

2. State the null and alternative (research) hypotheses for the above research scenario.

3. Given the above null & alternative hypotheses and with alpha = .05, what is Zcrit equal to?

4. What are the values for µM and M?

5. What is the value of Zobt?

6. What is the p-value for the outcome (what is the probability of getting an outcome this extreme (one or two tail test?) in Mr. Graham’s cereal study?

7. With alpha = .05, what is the correct decision given the outcome of Mr. Graham’s cereal study?

8. What type of error might Mr. Graham be making?

9. Given the results, of the study, is it appropriate to conduct a confidence interval? If so conduct a 95% CI.

10. Given the results, of the study, is it appropriate to evaluate practical significance? If so calculate Cohen’s d.

11. What is the probability of that error Mr. Graham might be making?

Explanation / Answer

1) One Sample Mean Z Statistic: Known population standard deviation

2) H0: Mean= 15

Ha: Mean =/ 15

3) Z critical: Z(0.05/2) = + / - 1.96

4) Population mean(µM) = 15

Population standard deviaiton(M)= 0.5

5) Z test statistic:

Z= (X_bar- µ) / (/sqrt(n)) = (15.18-15) / (0.5/sqrt25)) = 1.8

6) P-value: (Two tail) = 0.072

7) The test statistic is not significant and failed to reject H0.

8) Type-1 error

Related Questions

Navigate

• Browse (All)
• Browse M
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.