The confidence interval for population mean (when population standard deviation is unknown) is given by :-
[tex]\overline{x}-t^*\dfrac{s}{\sqrt{n}}< \mu<\overline{x}+z^*\dfrac{s}{\sqrt{n}}[/tex]
, where n= sample size
[tex]\overline{x}[/tex] = Sample mean
s= sample size
t* = Critical value.
Given : n= 25
Degree of freedom : [tex]df=n-1=24[/tex]
[tex]\overline{x}= \$93.36[/tex]
[tex]s=\ $19.95[/tex]
Significance level for 98% confidence interval : [tex]\alpha=1-0.98=0.02[/tex]
Using t-distribution table ,
Two-tailed critical value for 98% confidence interval :
[tex]t^*=t_{\alpha/2,\ df}=t_{0.01,\ 24}=2.4922[/tex]
⇒ The critical value that should be used in constructing the confidence interval = 2.4922
Then, the 95% confidence interval would be :-
[tex]93.36-(2.4922)\dfrac{19.95}{\sqrt{25}}< \mu<93.36+(2.4922)\dfrac{19.95}{\sqrt{25}}[/tex]
[tex]=93.36-9.943878< \mu<93.36+9.943878[/tex]
[tex]=93.36-9.943878< \mu<93.36+9.943878[/tex]
[tex]=83.416122< \mu<103.303878\approx83.4161<\mu<103.3039[/tex]
Hence, the 98% confidence interval for the mean repair cost for the dryers. = [tex]83.4161<\mu<103.3039[/tex]
