91Ó°ÊÓ

Given:

\(\begin{align}\text x̄ &=0 \text{.9255} \\ \text s&=0 \text{.0809} \\ \text n&=20 c=95 \end{align}\)

(a)

Determine the t-value by looking in the row starting with degrees of freedom \({\rm{df = n - 1 = 20 - 1 = 19}}\) and in the column with \({\rm{\alpha = (1 - c)/2 = (1 - 0}}{\rm{.95)/2 = 0}}{\rm{.025}}\) in table A. 5 :

\({{\rm{t}}_{{\rm{0}}{\rm{.025,19}}}}{\rm{ = 2}}{\rm{.093}}\)

The margin of error is then:

\({\rm{E = }}{{\rm{t}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{s}}}{{\sqrt {\rm{n}} }}{\rm{ = 2}}{\rm{.093 \times }}\frac{{{\rm{0}}{\rm{.0809}}}}{{\sqrt {{\rm{20}}} }}{\rm{\gg 0}}{\rm{.0379}}\)

Hence The boundaries of the confidence interval then become:

\(\begin{aligned}{\rm{\bar x - E = 0}}{\rm{.9255 - 0}}{\rm{.0379 = 0}}{\rm{.8876}}\\{\rm{\bar x + E = 0}}{\rm{.9255 + 0}}{\rm{.0379 = 0}}{\rm{.9634}}\end{aligned}\)

(b)

Determine the t-value by looking in the row starting with degrees of freedom \({\rm{df = n - 1 = 20 - 1 = 19}}\)nd in the column with \({\rm{\alpha = (1 - c)/2 = (1 - 0}}{\rm{.95)/2 = 0}}{\rm{.025}}\)in tableA. 5 :

\({{\rm{t}}_{{\rm{0}}{\rm{.02x,19}}}}{\rm{ = 2}}{\rm{.093}}\)

The margin of error is then:

\({\rm{E = }}{{\rm{t}}_{{\rm{\alpha /2}}}}{\rm{ \times s}}\sqrt {{\rm{1 + }}\frac{{\rm{1}}}{{\rm{n}}}} {\rm{ = 2}}{\rm{.093 \times 0}}{\rm{.0809}}\sqrt {{\rm{1 + }}\frac{{\rm{1}}}{{{\rm{20}}}}} {\rm{\gg 0}}{\rm{.1735}}\)

Hence The boundaries of the confidence interval then become:

\(\begin{aligned}{\rm{\bar x - E = 0}}{\rm{.9255 - 0}}{\rm{.1735 = 0}}{\rm{.7520}}\\{\rm{\bar x + E = 0}}{\rm{.9255 + 0}}{\rm{.1735 = 1}}{\rm{.0990}}\end{aligned}\)

(c)

Given:

\({\rm{k = 99\% = 0}}{\rm{.99}}\)

The tolerance critical value is given in the row with $n=20$ and in the column with confidence level \(95\% \)and\(\% \)of population captured \( \ge 99\% \)

\({\rm{tol = 3}}{\rm{.615}}\)

The margin of error is then the product of the tolerance critical value and the standard deviation:

\({\rm{E = tol \times s = 3}}{\rm{.615 \times 0}}{\rm{.0809\gg 0}}{\rm{.2925}}\)

Hence The boundaries of the confidence interval then become:

\(\begin{aligned}{\rm{\bar x - E = 0}}{\rm{.9255 - 0}}{\rm{.2925 = 0}}{\rm{.6330}}\\{\rm{\bar x + E = 0}}{\rm{.9255 + 0}}{\rm{.2925 = 1}}{\rm{.2180}}\end{aligned}\)