You wish to test the following claim (HaHa) at a significance
level of α=0.02α=0.02. For the context of this problem,
μd=PostTest−PreTestμd=PostTest-PreTest where the first data set
represents a pre-test and the second data set represents a
post-test. (Each row represents the pre and post test scores for an
individual. Be careful when you enter your data and specify what
your μ1μ1 and μ2μ2 are so that the differences are computed
correctly.)
Ho:μd=0
Ha:μd≠0
You believe the population of difference scores is normally
distributed, but you do not know the standard deviation. You obtain
the following sample of data:
pre-test | post-test |
---|---|
67.2 | 38.2 |
51.8 | 31 |
61.4 | 78.1 |
51.3 | 108.9 |
60.9 | 66.9 |
50.9 | -2.2 |
51.1 | 69.1 |
65.8 | 31.3 |
54 | 60 |
50.1 | 75.6 |
48.7 | 45.1 |
70 | 56.2 |
42 | 47 |
65.4 | 48.5 |
67.8 | 50.9 |
59.3 | 50.1 |
What is the test statistic for this sample?
test statistic = (Report answer accurate to 4 decimal
places.)
What is the p-value for this sample?
p-value = (Report answer accurate to 4 decimal
places.)
The p-value is...
This test statistic leads to a decision to...
As such, the final conclusion is that...
Hypothesis :
Two tailed test.
We have to perform paired t test.
Test statistic :
Where , is sample mean of difference ( difference = d = PostTest−PreTest )
is standard deviation of differences
n is sample size.
= -3.9375 , = 26.53626 , n = 16
So test statistic is ,
P-value :
p-value for this two tailed test is ,
P-value =2*P( t < test statistic ) = 2*P( t < -0.5935 )
Using Excel function , =T.DIST( t , df , 1 )
df = n - 1 = 16 - 1 = 15
P( t < -0.5935 ) = T.DIST( -0.5935 , 15 , 1 ) = 0.280845
So, p-value = 2*0.280845 = 0.5617
P-value = 0.5617
P-value is greater than = 0.02
This test statistic leads to a decision to fail to reject null hypothesis .
As such, the final conclusion is that ' There is not sufficient sample evidence to support the claim that the mean difference of post-test from pre-test is not equal to 0 '
You wish to test the following claim (HaHa) at a significance level of α=0.02α=0.02. For the...
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