Use a 0.01 level of significance to test whether the difference between the mean
ID: 3440767 • Letter: U
Question
Use a 0.01 level of significance to test whether the difference between the means of these two samples is significant. Assume equal variance between populations.
The t-statistic for this test is t0 =
b)Based on the t-statistic calculated above, the null hypothesis must be rejected at a level of significance equal to 0.01. We conclude that the average heat-producing capacity of the coal from the two mines is not the same. The very small P value 0.0024 strengthens this conclusion. Enter 0 if this statement is FALSE or 1 otherwise
The following are the average weekly losses of worker-hours due to accidents in 10 industrial plants before and after a certain safety program was put into operation BEFORE Plant 1: 45 Plant 2: 73 Plant 3: 46 Plant 4:124 Plant 5: 33 Plant 6: 57 Plant 7: 83 Plant 8: 34 Plant 9: 26 Plant 10: 17 AFTER Plant 1: 36 Plant 2: 60 Plant 3: 44 Plant 4:119 Plant 5: 35 Plant 6: 51 Plant 7: 77 Plant 8: 29 Plant 9: 24 Plant 10: 11 Conduct a paired t test to analyze this problem. Use the 0.05 level of significance to test whether the safety program is effectiveExplanation / Answer
Let ud = u2 - u1.
Formulating the null and alternative hypotheses,
Ho: ud = 0
Ha: ud =/ 0
At level of significance = 0.01
As we can see, this is a two tailed test.
Calculating the standard deviation of the differences (third column):
s = 311.5875284
Thus, the standard error of the difference is sD = s/sqrt(n):
sD = 98.53262804
Calculating the mean of the differences (third column):
XD = 102.8
As t = [XD - uD]/sD, where uD = the hypothesized difference = 0 , then
t = 1.043309227 [ANSWER, T STATISTIC]
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Additional info:
As df = n - 1 = 9
Then the critical value of t is
tcrit = +/- 3.249835542
Thus, comparing t and tcrit, we WE FAIL TO REJECT THE NULL HYPOTHESIS.
Also, using p values,
p = 0.324017857
Also, comparing this p to the significance level, WE FAIL TO REJECT THE NULL HYPOTHESIS.
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