In a survey of 2566 adults in a recent year, 1420 say they have made a New Year's resolution. Construct 90% and 95% confidence intervals for the popul

In a survey of 2566 adults in a recent​ year, 1420 say they have made a New​ Years
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

SchepperJ

Step 1 Given : $n=2566$ $x=1420$ Sample proportion:
$\stackrel{^}{p}=\frac{1420}{2566}$
a) The​ 90% confidence interval for the population proportion p $\alpha =0.10$
${z}_{c}={z}_{\frac{\alpha }{2}}$
${z}_{c}=0.05$
${z}_{c}=1.645$ Confidence interval: $CI=\stackrel{^}{p}±{z}_{c}×\sqrt{\frac{\stackrel{^}{p}×\left(1-\stackrel{^}{p}\right)}{n}}$
$CI=0.553±1.645×\sqrt{\frac{0.553×\left(1-0.553\right)}{2566}}$

Step 2 $CI=0.553±0.0161$
$CI=\left(0.5369,0.5691\right)$ 90% confidence interval for the population proportion p is lies between ( 0.5369, 0.5691 ) b) The​ 95% confidence interval for the population proportion p $\alpha =0.05$
${z}_{c}={z}_{\frac{\alpha }{2}}$
${z}_{c}=0.025$
${z}_{c}=1.96$ Confidence interval: $CI=\stackrel{^}{p}±{z}_{c}×\sqrt{\frac{\stackrel{^}{p}×\left(1-\stackrel{^}{p}\right)}{n}}$
$CI=0.553±1.96×\sqrt{\frac{0.553×\left(1-0.553\right)}{2566}}$
$CI=0.553±0.0192$
$CI=\left(0.5338,0.5722\right)$ 95% confidence interval for the population proportion p is lies between ( 0.5338, 0.5722 ) Hence the 95% confidence interval is wider.