Question

Stacy randomly survey 20 people and asked them how many people live in their home
Let x=the number of people in each household
Let n=the number of people surveyed

response of the 20 people surveyed:

6	4	4	3	2	2	3	6	4	4
3	3	3	4	3	2	3	2	3	3

1.Obtain a 95% confidence interval for the mean number of people per household
2.Using a=.05,
a)If your point estimate, μ>2.8, test to see if there is enough significant evidence to support the claim that the average number people per household is greater than 2.8.
b.If your point estimate, μ<2.8, test to see if there is enough significant evidence to support the claim that the average number people per household in is less than 2.8.

Answer 1

	Values ( X )
	6	7.0225
	4	0.4225
	4	0.4225
	3	0.1225
	2	1.8225
	2	1.8225
	3	0.1225
	6	7.0225
	4	0.4225
	4	0.4225
	3	0.1225
	3	0.1225
	3	0.1
	4	0.4225
	3	0.1225
	2	1.8225
	3	0.1225
	2	1.8225
	3	0.1225
	3	0.1225
Total	67	24.55

Mean

Standard deviation

Part a)

Confidence Interval



Lower Limit =
Lower Limit = 2.818
Upper Limit =
Upper Limit = 3.882
95% Confidence interval is ( 2.818 , 3.882 )

Part 2)

To Test :-

H0 :- μ <= 2.8

H1 :-  μ>2.8

Test Statistic :-


t = 2.1639

Test Criteria :-
Reject null hypothesis if


Result :- Reject null hypothesis

Decision based on P value
P - value = P ( t > 2.1639 ) = 0.0217
Reject null hypothesis if P value < level of significance
P - value = 0.0217 < 0.05 ,hence we reject null hypothesis
Conclusion :- Reject null hypothesis

Part b)

To Test :-   

H0 :- μ >= 2.8

H1 :-  μ < 2.8

Test Statistic :-


t = 2.1639

Test Criteria :-
Reject null hypothesis if


Result :- Fail to reject null hypothesis

Decision based on P value
P - value = P ( t > 2.1639 ) = 0.9783
Reject null hypothesis if P value <    level of significance
P - value = 0.9783 > 0.05 ,hence we fail to reject null hypothesis
Conclusion :- Fail to reject null hypothesis