Question

That is just the chart. Disregard question 6. 7. 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.O (64.92, 65.48)© (64.83, 65.57)O (65.19, 65.21)© (64.96, 65.44)

Answer 1

Step 1

Given;

`\begin{gathered} n=381\text{ adults} \\ Mean\text{ \lparen}\mu)=65.2 \\ \sigma=2.8 \end{gathered}`

Step 2

The confidence interval is given as follows;

`CI=\mu\pm t_{(\alpha)/(2)}((s)/(√(n)))`

The test statistic for a 95% confidence interval with α = 0.05, the degrees of freedom, df= n-1 = 381-1=380

`t_{(\alpha)/(2),df}=t_{(0.05)/(2),380}`
`t_{(\alpha)/(2),df}=t_{(0.05)/(2),380}=1.97`
`\begin{gathered} C.I=65.2\pm1.97((2.8)/(√(381)))=65.2\pm0.2825932406 \\ C.I=(64.92,65.48) \end{gathered}`

Answer;

`C.I=(64.92,65.48)`