Winter 2017 ME 364 Final Exam 2. write the word or phrase that best completes ea

Question

Winter 2017 ME 364 Final Exam 2. write the word or phrase that best completes each statement to answer the question. Assume simple random sample has been selected from a normally d population and test the given claim. Identity the null and alternative hypotheses, test statistic, critical value(s) or P-value (or range of as propriate, and state the final conclusion that addresses the original claim. company gives job applicants a test of programming ability and the mean for that test has been the past. Twenty-five job applicants are randomly s and 12, 183 from one large university and they produce a mean and standard deviation of from a score respectively. level significance to test the claim that this sample comes population Use a of P-value method of testing hypotheses with a mean score greater than 160. Use the

Explanation / Answer

Answer to part I)

Claim: Mean is greater than 160

.

Hypothesis:

Null hypothesis: Mean score is 160

Alternate hypothesis: Mean score > 160 ......[thus it is a right tailed test]

.
Population Mean (M) = 160

Sample size (n) = 25

Sample mean (x bar) = 183

Standard deviation (s) = 12

alpha = 0.05

.

Test statistic used is t , because the population standard deviation is not provided, only sample standard deviation is provided

t = (183 - 160) /(12/25)

t = 9.583

df = 25-1 = 24

Thus the P value is :0.0000

.

Conclusion: Since the P value is 0.000 , which is less than the alpha value 0.05 , we reject the null and conclude that the sample is drawn from a population with mean greater than 160

Answer to part II)

Claim: The true average differs from 5.5

.

hypothesis:

Null hypothesis: The true population mean is 5.5

Alternate hypothesis: The true population mean is different from 5.5.......[it is a two tailed test]

.

Population mean (M) = 5.5

Sample Proportion xbar = 5.25

Alpha = 0.01

Sample size (n) = 16

Standard deviation (SD) = 0.3

.

Z = (5.25 - 5.5) / (0.3/16)

Z = -3.33

.

P value for Z = -3.33 ,is =2 * 0.00043 = 0.00086

.

Since the P value 0.00086 < 0.01, we reject the null and conclude that the Mean is different from 5.5

.

Conclusion:

Thus we conclude that there is sufficient evidence to support the claim that the true average percentage differs from 5.5

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

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