Answer: [tex](56.03\ ,63.97)[/tex]
Step-by-step explanation:
Given : The speeds of vehicles traveling on a highway are normally distributed .
Standard deviation: [tex]\sigma: 7[/tex]
Since , sample size : [tex]n=20[/tex] , which is small sample (n<30) so the test applied here is a t-test.
Degree of freedom= [tex]n-1=20-1=19[/tex]
Sample mean : [tex]\overline{x}=60[/tex]
Level of significance : [tex]1-\alpha:0.98[/tex]
Then , significance level : [tex]\alpha: 1-0.98=0.02[/tex]
Using t-distribution table , Critical value : [tex]\text{t-score}=t_{n-1, \alpha/2}= 2.539483[/tex]
The confidence interval for population mean is given by :-
[tex]\overline{x}\pm t_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}[/tex]
[tex]60\pm( 2.539483)\dfrac{7}{\sqrt{20}}\approx60\pm3.97=(56.03\ ,63.97)[/tex]
Hence, a 98% confidence interval for the population mean=[tex](56.03\ ,63.97)[/tex]