Answer:
b) 95 percent confidence interval for this single-sample t test
[11.64, 16.36]
Step-by-step explanation:
Explanation:-
Given data a study of 62 college students finds that their average interest rate is 14 percent with a standard deviation of 9.3 percent.
Sample size 'n' =62
sample mean x⁻ = 14
sample standard deviation 'S' = 9.3
95 percent confidence interval for this single-sample t test
The values are [tex](x^{-} - t_{0.05} \frac{S}{\sqrt{n} } ,x^{-}+t_{0.05}\frac{S}{\sqrt{n} })[/tex] the 95 percent confidence interval for the population mean 'μ'
Degrees of freedom γ=n-1=62-1=61
t₀.₀₅ = 1.9996 at 61 degrees of freedom
[tex](14 - 1.9996\frac{9.3}{\sqrt{62} } ,14+1.9996\frac{9.3}{\sqrt{62} })[/tex]
(14-2.361 , 14 + 2.361)
[(11.64 , 16.36]
Conclusion:-
95 percent confidence interval for this single-sample t test
[11.64, 16.36]
