Question

I need to have this done within 1 hour and a half. Thank YOU!:)

17. Minnesota Department of Health wants to test the claim that variability in COVID-19 within Minnesota is greater than 12.75. For this, they took a sample of 25 days data and found that the mean is 215 and standard deviation was 14.17. At alpha=5% what will be the decision?  

Question 17 options:

	Reject H0 at alpha =5%
	Do not reject H0 at alpha=5%
	We do not have enough information to make decision

21. To test the claim that mean weight is not equal to 156.3 lbs., a sample of 14 people showed that the sample mean was 172.55 and the sample standard deviation (s) was found to be 26.33 lbs. Find the p value for this test (note that population standard deviation sigma is unknown so that this is t-test).

Question 21 options:

	6.2%
	1.9%
	3.8%
	4.4%
	2.6%

Answer 1

Solution:

17) Given

n= 25 sample size

population variance

. Sample mean

s = 14.17 sample standard deviations

. Level of significance.

To test the hypothesis

.

Vs.

  

Test statistic

Test statistic

P Value = 0.196832 from online P value calculator

P Value = 0.1968

Decision : P Value

Fail to reject Ho.

Do not reject Ho at

Conclusion : Fail to reject Ho ,. there is insufficient evidence to conclude that the variability in covid 19 within mineseta is greater than 12.75

21) Given that

n= 14 sample size

lbs population mean.

.lbs Sample mean .

s = 26.33 lbs Sample Standard deviations.

To test the hypothesis

Test statistic

t = 2.3092264

Test statistic t = 2.3092

P Value = 0.038004 from online P value calculator

P Value = 3.8%

3.8 % is the correct option