The solution to this question is given below:
From the information given:
n=15
x bar=25.0
s=3.5
(a) I'll suppose that calculating a 95% confidence interval is necessary; other confidence intervals can be calculated in a similar way.
c=0.95 or 95%
In the table with the critical values for tt distributions in the appendix, locate the row beginning with degrees of freedom.
The maximum number of logically independent values—that is, values with the freedom to change—in the data sample is referred to as the degree of freedom.
df=n-1
=15-1
=14
and the column with alpha=(1-c)/2=0.025 =(1c)/2 =0.025 to find the t-value:
t α/2 =2.145
Therefore, the error margin is:
E=t α/2 × s/√n
=2.145× 3.5/√15
≈1.9384
The confidence interval's outer limits are thus:
xbar-E=25.0-1.9384 =23.0616
xbar+E=25.0+1.9384 = 26.9384
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