A sample selected from a population gave a sample proportion equal to 0.59 .

A sample selected from a population gave a sample proportion equal to 0.59 .
a) Make a $$\displaystyle{95}\%$$ confidence interval for p assuming $$\displaystyle{n}={150}$$
( Enter your answer; confidence interval assuming $$\displaystyle{n}={150}$$, lower bound ,Enter your answer; confidence interval assuming $$\displaystyle{n}={150}$$, upper bound )
b) Construct a $$\displaystyle{95}\%$$ confidence interval for p assuming $$\displaystyle{n}={600}$$. ( Enter your answer; confidence interval assuming $$\displaystyle{n}={600}$$, lower bound ,Enter your answer; confidence interval assuming $$\displaystyle{n}={600}$$, upper bound )
c) Make a $$\displaystyle{95}\%$$ confidence interval for p assuming $$\displaystyle{n}={1275}$$. ( Enter your answer; confidence interval assuming $$\displaystyle{n}={1275}$$, lower bound ,Enter your answer; confidence interval assuming $$\displaystyle{n}={1275}$$, upper bound )

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unett
Step 1
a) Given,
$$\displaystyle\hat{{{p}}}={0.59}$$
$$\displaystyle{n}={150}$$
t-table at $$\displaystyle{95}\%$$ confidence is
$$\displaystyle{t}_{{\frac{\alpha}{{2}}}}={1.96}$$
Confidence interval $$\displaystyle=\hat{{{p}}}\pm{t}_{{\frac{\alpha}{{2}}}}\sqrt{{{\frac{{\hat{{{p}}}{\left({1}-\hat{{{p}}}\right)}}}{{{n}}}}}}$$
$$\displaystyle={0.54}\pm{1.96}\sqrt{{{\frac{{{0.59}{\left({1}-{0.59}\right)}}}{{{150}}}}}}$$
$$\displaystyle={0.59}\pm{1.96}\sqrt{{{0}.\infty{16126666}}}$$
$$\displaystyle={0.59}\pm{0.0402}$$
$$\displaystyle={\left({0.550},\ {0.630}\right)}$$
$$\displaystyle\therefore{L}{o}{w}{e}{r}\ {B}{o}{u}{n}{d}={0.550}$$
Upper Bound $$\displaystyle={0.630}$$
Step 2
b) $$\displaystyle{n}={600}$$
$$\displaystyle\hat{{{p}}}={0.59}$$
$$\displaystyle{t}_{{\frac{\alpha}{{2}}}}={1.96}$$
$$\displaystyle{C}{I}=\hat{{{p}}}\pm{t}_{{\frac{\alpha}{{2}}}}\sqrt{{\hat{{{p}}}{\left({1}-\hat{{{p}}}\right)}}}{\left\lbrace{n}\right\rbrace}\rbrace$$
$$\displaystyle={0.59}\pm{1.96}\sqrt{{{0.59}{\left({1}-{0.59}\right)}}}{\left\lbrace{600}\right\rbrace}\rbrace$$
$$\displaystyle={0.59}\pm{1.96}\sqrt{{{0.00040316666}}}$$
$$\displaystyle={0.59}\pm{0.0201}$$
$$\displaystyle={\left({0.570},\ {0.610}\right)}$$
Lower Bound $$\displaystyle={0.570}$$
Upper Bound $$\displaystyle={0.610}$$
Step 3
c) $$\displaystyle{n}={1275}$$
$$\displaystyle\hat{{{p}}}={0.59}$$
$$\displaystyle{t}_{{\frac{\alpha}{{2}}}}={1.96}$$
$$\displaystyle{C}{I}=\hat{{{p}}}\pm{t}_{{\frac{\alpha}{{2}}}}\sqrt{{{\frac{{\hat{{{p}}}{\left({1}-\hat{{{p}}}\right)}}}{{{n}}}}}}$$
$$\displaystyle={0.59}\pm{1.96}\sqrt{{{\frac{{{0.59}{\left({1}-{0.59}\right)}}}{{{1275}}}}}}$$
$$\displaystyle={0.59}\pm{1.96}\sqrt{{{0.00018972549}}}$$
$$\displaystyle={0.59}\pm{0.0138}$$
$$\displaystyle={\left({0.576},\ {0.604}\right)}$$
Lower bound $$\displaystyle={0.576}$$
Upper bound $$\displaystyle={0.604}$$