On a certain day, a large number of fuses were manufactured, each rated at 15 A. A sample of 75 fuses is drawn from the day’s production, and 17 of them were found to have burnout amperages greater than 15 A. Find a 95\% confidence interval for the proportion of fuses manufactured that day whose burnout amperage is greater than 15 A.

abondantQ 2021-08-04
On a certain day, a large number of fuses were manufactured, each rated at 15 A. A sample of 75 fuses is drawn from the day’s production, and 17 of them were found to have burnout amperages greater than 15 A.
a) Find a \(\displaystyle{95}\%\) confidence interval for the proportion of fuses manufactured that day whose burnout amperage is greater than 15 A.
b) Find a \(\displaystyle{98}\%\) confidence interval for the proportion of fuses manufactured that day whose burnout amperage is greater than 15 A.
c) Find the sample size needed for a \(\displaystyle{95}\%\) confidence interval to specify the proportion to within \(\displaystyle\pm{0.05}.\)

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Answered 2021-08-17 Author has 28674 answers

a) Given \(\displaystyle{x}={17}\) and \(\displaystyle{n}={75}\)
\(\displaystyle\hat{{{p}}}={\frac{{{x}}}{{{n}}}}={\frac{{{17}}}{{{75}}}}={0.2267}\)
The critical value corresponding with \(\displaystyle{95}\%\) confidence level is 1.96
\(\displaystyle{95}\%{C}{I}=\hat{{{p}}}\pm\ {z}_{{{c}{r}{i}{t}{i}{c}{a}{l}}}\sqrt{{{\frac{{\hat{{{p}}}{\left({1}-\hat{{{p}}}\right)}}}{{{n}}}}}}\)
\(\displaystyle={0.2267}\pm{1.96}\sqrt{{{\frac{{{0.2267}{\left({1}-{0.2267}\right)}}}{{{75}}}}}}\)
\(\displaystyle={\left({0.1319},\ {0.3215}\right)}\)
b) The critical value corresponding with \(\displaystyle{98}\%\) confidence level is 2.33
\(\displaystyle{95}\%{C}{I}=\hat{{{p}}}\pm\ {z}_{{{c}{r}{i}{t}{i}{c}{a}{l}}}\sqrt{{{\frac{{\hat{{{p}}}{\left({1}-\hat{{{p}}}\right)}}}{{{n}}}}}}\)
\(\displaystyle={0.2267}\pm{2.33}\sqrt{{{\frac{{{0.2267}{\left({1}-{0.2267}\right)}}}{{{75}}}}}}\)
\(\displaystyle={\left({0.1141},\ {0.3393}\right)}\)
c) The critical value corresponding with \(\displaystyle{95}\%\) confidence level is 1.96
\(\displaystyle{M}{E}={0.05}\)
sample size \(\displaystyle={\left({\frac{{{z}_{{{c}{r}{i}{t}{i}{c}{a}{l}}}}}{{{M}{E}}}}\right)}^{{{2}}}{p}{\left({1}-{p}\right)}\)
\(\displaystyle={\left({\frac{{{1.96}}}{{{0.05}}}}\right)}^{{{2}}}{0.5}{\left({1}-{0.5}\right)}\) Let \(\displaystyle{p}={0.5}\)
\(\displaystyle={348.16}\)
\(\displaystyle\approx{384}\)

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