Question

Question 1 5 pts A laboratory tested n=121 chicken eggs and found that the mean amount of cholesterol was a = 90 milligrams with o - 10 milligrams. Find the margin of error E corresponding to a 95% confidence interval for the true mean cholesterol content, , of all such eggs. Time Atten 1 Hc octor fied Question 2 5 pts es Use the confidence level and sample data to find a confidence interval for estimating the population. Find the margin of error E and round your answer to the nearest tenth. n-51, -88, 5 - 14; 90% confidence Question 3 5 pts Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. A survey of 865 voters in one state reveals that 408 favor approval of an issue before the legislature. Construct the 95% confidence interval for the true proportion of all voters in the state who favor approval

Answer 1

1).z critical value for 95% confidence level, both tailed test be:-

$z^*=1.96$ (as the population sd is known, we have used z distribution )

the margin of error be:-

$=\left ( z^**\frac{\sigma}{\sqrt{n}} \right )=\left ( 1.96*\frac{10}{\sqrt{121}} \right )\approx \mathbf{1.782}$

2). df = (n-1) = (51-1) = 50

t critical value for alpha=0.10, df=50, both tailed test be:-

$t^*=1.676$ (as the sample sd is known, we have used t distribution )

the margin of error be:-

$=\left ( t^**\frac{s}{\sqrt{n}} \right )=\left ( 1.676*\frac{14}{\sqrt{51}} \right )\approx \mathbf{3.3}$

3). given data are:-

sample size (n) = 865

sample proportion ($\hat{p}$) = 408/865 = 0.471676

z critical value for 95% confidence level, both tailed test be:-

$z^*=1.96$

the 95% confidence interval for the true population proportion be:-

$=\hat{p}\pm z^**\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

$=0.471676\pm 1.96*\sqrt{\frac{0.471676(1-0.471676)}{865}}$

$\approx (\mathbf{0.438,0.505)}$

*** if you have any doubt regarding the problem ,please write it in the comment box...if satisfied,please UPVOTE.