It was observed that 30 of the 100 randomly selected students smoked. Find the confidence interval estimate of 95\% confidence level for \pi ratio of smokers in the population.(0.45<p<0.78)=0.95P(0.11<p<0.28)=0.95

UkusakazaL 2021-08-06 Answered

It was observed that 30 of the 100 randomly selected students smoked. Find the confidence interval estimate of \(\displaystyle{95}\%\) confidence level for \(\displaystyle\pi\) ratio of smokers in the population.
a) \(\displaystyle{P}{\left({0.45}{<}{p}{<}{0.78}\right)}={0.95}\)
b) \(\displaystyle{P}{\left({0.11}{<}{p}{<}{0.28}\right)}={0.95}\)
c) \(\displaystyle{P}{\left({0.51}{<}{p}{<}{0.78}\right)}={0.95}\)
d) \(\displaystyle{P}{\left({0.21}{<}{p}{<}{0.59}\right)}={0.95}\)
e) \(\displaystyle{P}{\left({0.21}{<}{p}{<}{0.39}\right)}={0.95}\)

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Expert Answer

geduiwelh
Answered 2021-08-09 Author has 16800 answers

The \(\displaystyle{95}\%\) CI is given by
\(\displaystyle{C}{I}={p}\pm{z}_{{{0.05}}}\sqrt{{{\frac{{{p}{q}}}{{{n}}}}}}\)
\(\displaystyle{C}{I}={0.3}\pm{1.96}\sqrt{{{\frac{{{0.3}\times{0.7}}}{{{100}}}}}}\)
\(\displaystyle{C}{I}={\left[{0.21},\ {0.39}\right]}\)
Option (e) is correct.
\(\displaystyle{P}{\left({0.21}{<}{p}{<}{0.39}\right)}={0.95}\)

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