Answer to Question #322507 in Statistics and Probability for jempoy

Question

Professor Z conducted a periodic examination on all his students enrolled in Math 11. The scores are normally distributed with a population standard deviation of 10. He randomly selected 125 students and obtained a mean score of 60.

 a, What is the population of interest?

 b. What is the point estimate of the population mean?

 c. Compute the standard error of the mean.

 d. At 95% confidence level, what is the margin of error?

 e. Find the 90% confidence interval for the population mean.

 f. Find the 95% confidence interval for the population mean.

 g. Find the 98% confidence interval for the population mean.

Answer 1

$a:\\All\,\,students\\b:\\\hat{\mu}=\bar{x}=60\\c:\\s=\frac{\sigma}{\sqrt{n}}=\frac{10}{\sqrt{125}}=0.894427\\d:\\E=\frac{\sigma}{\sqrt{n}}z_{\frac{1+\gamma}{2}}=0.894427\cdot 1.960=1.75308\\e:\\0.9:\left( \bar{x}-\frac{\sigma}{\sqrt{n}}z_{\frac{1+\gamma}{2}},\bar{x}+\frac{\sigma}{\sqrt{n}}z_{\frac{1+\gamma}{2}} \right) =\left( 60-\frac{10}{\sqrt{125}}\cdot 1.6449,60+\frac{10}{\sqrt{125}}\cdot 1.6449 \right) =\\=\left( 58.5288,61.4712 \right) \\0.95:\left( \bar{x}-\frac{\sigma}{\sqrt{n}}z_{\frac{1+\gamma}{2}},\bar{x}+\frac{\sigma}{\sqrt{n}}z_{\frac{1+\gamma}{2}} \right) =\left( 60-\frac{10}{\sqrt{125}}\cdot 1.9600,60+\frac{10}{\sqrt{125}}\cdot 1.9600 \right) =\\=\left( 58.2469,61.7531 \right) \\0.98:\left( \bar{x}-\frac{\sigma}{\sqrt{n}}z_{\frac{1+\gamma}{2}},\bar{x}+\frac{\sigma}{\sqrt{n}}z_{\frac{1+\gamma}{2}} \right) =\left( 60-\frac{10}{\sqrt{125}}\cdot 2.3263,60+\frac{10}{\sqrt{125}}\cdot 2.3263 \right) =\\=\left( 57.9193,62.0807 \right) \\\\$