Answer:
(64.92, 65.48)
Step-by-step explanation:
The given details are;
The number of adults in the survey, n = 381 adults
The mean height of the adults in the survey, [tex]\overline x[/tex] = 65.2 inches
The standard deviation of the height of the adults in the survey, s = 2.8 inches
The confidence interval is given as follows;
[tex]CI=\bar{x}\pm t_{\alpha /2} \cdot \dfrac{s}{\sqrt{n}}[/tex]
The test statistic for a 95% confidence interval with α = 0.05, the degrees of freedom, df = n-1 = 381 - 1 = 380, [tex]t_{\alpha /2, df}[/tex] = [tex]t_{\frac{0.05}{2} , \ 381}[/tex] = 1.97
Therefore, we get;
C.I =
[tex]CI=65.2\pm 1.97 \times \dfrac{2.8}{\sqrt{381}} \approx (64.92, 65.48)[/tex]
