A survey of 381 adult females produced a mean height of 65.2 inches with a standard deviation of 2.8 inches. Construct a confidence interval based on a 95% confidence level.
(64.83, 65.57)
(65.19, 65.21)
(64.96, 65.44)
(64.92, 65.48)

Respuesta :

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]

ACCESS MORE
EDU ACCESS