In: Statistics and Probability
Use MS Excel to answer this. From the data given below test a hypothesis for variances being equal. Use α = 0.05.
Section A | Section B |
15 | 16 |
26 | 52 |
52 | 55 |
53 | 57.5 |
54 | 58 |
56.5 | 60 |
61 | 61 |
61.5 | 70 |
63 | 70 |
66 | 71 |
66 | 72 |
66.5 | 73 |
69 | 74.5 |
71 | 75 |
77 | 75.5 |
77 | 76 |
78 | 77 |
79 | 81 |
81 | 85 |
86 | 85.5 |
87 | 86 |
90 | 88 |
90 | 88.5 |
91 | 91 |
91 | 93 |
94 | |
95 | |
96 | |
98 | |
98 | |
99.5 | |
100 |
as instructed,i am using excel to solve the problem .
steps:-
copy the data in excel including the labels data analysis F test two sample for variances ok in variable 1 range select the range of section A including label and in variable 2 range select the range of section B including label tick labels in alpha type 0.05 in output range select any blank cell of excel sheet ok.
your output be:-
F-Test Two-Sample for Variances | ||
Section A | Section B | |
Mean | 68.3 | 77.25 |
Variance | 365.8125 | 326.8548 |
Observations | 25 | 32 |
df | 24 | 31 |
F | 1.119189 | |
P(F<=f) one-tail | 0.379346 | |
F Critical one-tail | 1.875073 |
solution to our problem:-
hypothesis:-
as we have to test a hypothesis for variances being equal. i have used the two sided alternative hypothesis .
test statistic(F) = 1.119
p value = 2 * p value for one tail = (2*0.379346) 0.7587
decision:-
p value = 0.7587 > 0.05 (alpha)
so, we fail to reject the null hypothesis.
conclusion:-
there is sufficient evidence to claim that variances are equal at α = 0.05.
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