Use a 001 level of significance to test whether the differen

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 t_⁰ =

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 effective

Solution

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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The part B seems to be a different problem, and has insufficient given.

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