Question

Provide the null and alternative hypotheses in formal and plain language for appropriate two-tailed test (viz., dependent or independent) at the 0.05 significance level

 Do the math and reject/accept null at a=.05. State your critical t value.

 Explain the results in plain language.

 Calculate the 95% confidence interval for the difference in means and state both formally and in plain language if appropriate.|

A local Alcoholics Anonymous chapter normally holds meeting at 5:45. Some similarly sized chapters think 6:45 will increase attendance. They take a random of ten chapters with bimonthly meeting starting times of 5:45 and ten with 6:45 bimonthly meetings from the Fall of last year. Is there a difference in attendance?

5:45 : 90, 80, 70, 80, 70, 70, 70, 70, 80, 70

6:45 : 80, 60, 70, 90, 60, 70, 60, 50, 50, 70

Answer 1

Rank	5:45 (X)	6:45 (Y)
1	90	80	225	196
2	80	60	25	36
3	70	70	25	16
4	80	90	25	576
5	70	60	25	36
6	70	70	25	16
7	70	60	25	36
8	70	50	25	256
9	80	50	25	256
10	70	70	25	16
Total	750	660	450	1440
Average	75	66
Standard Deviation	7.071	12.649

Mean = 750 / 10 = 75

Standard Deviation S =

S_x = = = 7.071

Mean = 660 / 10 = 66

Standard Deviation S_y =

S_y = =   = 12.649

To Test :-

H₀ :-

There is no difference in the attendence size even if there is change in the meeting time

H₁ :-  

There is a difference in the attendence size if we change the meeting time

Test Statistic :-

t = ( 75 - 66) / ( (7.071² / 10) + ( 12.649² / 10 ) )

t = 1.94

Test Criteria :-

Reject null hypothesis if | t | > t_{/2 , df}

df =

df = 14

t_{/2 , df} =  t_{0.025 , 14} = 2.145

| t | > t_{/2 , df} = 1.94 < 2.145, hence fail to reject null hypothesis

Conclusion :- Accept Null Hypothesis

, There is no difference in the attendence size even if we change the timing of the meeting.

Confidence Interval

Lower limit = - 0.830

Upper limit = 18.830