Using = 0.05, test if the mean is different from 8.2 if a sample of size 36 yielded a mean of 7.4 and a sample std dev )s) od 4.
Solution:
Given the hypothesis
H0: = 8.2 ....null hypothesis
H1: 8.2 ....alternative hypothesis
n = 36 ....sample size
d.f. = n -1 = 36 -1 = 35
= 7.4 ...sample mean
s = 4 ....Sample standard deviation
= 0.05 .... level of significance
/2 = 0.025
Since population SD is unknown,we use t test.
The test statistics t is given by ..
t =
= (7.4 - 8.2)/(4/36)
= -1.20
| t | = 1.20
Now, sign in H1 indicates that the two tailed test.
So, the critical values are i.e.
The critical region : t < or t >
= 0.025,35 = 2.030 (using t table)
| t | <
We fail to reject the null hypothesis H0 at 0.05 level.
Not sufficient evidence to support the claim that the mean is different from 8.2.
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