Answer to Question #327744 in Statistics and Probability for Zenablyn

Question

A sample of 100 Grade 9 students’ ages was obtained to estimate the mean age of all Grade 9 students. =16 years and the population variance is 16.1.) What is the point estimate for ?

2.) Find the 95% confidence interval for .

3.) Find the 99% confidence interval for .

4.) What conclusions can you make based on each estimate?

Answer 1

$n=100\\\bar{x}=16\\s^2=16\\1:\\\hat{\mu}=\bar{x}=16\\2:\\\left( \bar{x}-t_{\frac{1+\gamma}{2},n-1}\frac{s}{\sqrt{n}},\bar{x}+t_{\frac{1+\gamma}{2},n-1}\frac{s}{\sqrt{n}} \right) =\left( 16-0.4\cdot 1.9842,16+0.4\cdot 1.9842 \right) =\\=\left( 15.2063,16.7937 \right) \\3:\\\left( \bar{x}-t_{\frac{1+\gamma}{2},n-1}\frac{s}{\sqrt{n}},\bar{x}+t_{\frac{1+\gamma}{2},n-1}\frac{s}{\sqrt{n}} \right) =\left( 16-0.4\cdot 2.6264,16+0.4\cdot 2.6264 \right) =\\=\left( 14.9494,17.0506 \right) \\4:\\With\,\,95\% confidence\,\,the\,\,mean\,\,is\,\,in\,\,\left( 15.2063,16.7937 \right) \\With\,\,99\% confidence\,\,the\,\,mean\,\,is\,\,in\,\,\left( 14.9494,17.0506 \right)$