Question

Use the t-distribution to find a confidence interval for a mean u given the relevant sample results. Give the best point estimate for u, the margin of error, and the confidence interval. Assume the results come from a random sample from a population that is approximately normally distributed. A 95% confidence interval for u using the sample results I = 10.5,5 = 6.9, and n = 30 Round your answer for the point estimate to one decimal place, and your answers for the margin of error and the confidence interval to two decimal places. point estimate = i margin of error = i The 95% confidence interval is to i

Answer 1

Solution

$\mathbf{given:}\;\bar x=10.5,\;s=6.9,\;n=30$

degrees of freedom (df) = n-1= 30 -1 = 29

$Confidence\;level\;(C)=0.95$

a=1-C=1-0.95 = 0.05

Refer t-table or use excel function "=T.INV.2T(0.05,29)" to find the critical value of t.

$\mathbf{\therefore t_{\frac{\alpha }{2}}=2.045}$

$\mathbf{point\;estimate}=\bar x=\mathbf{{\color{Red} 10.5}}$

$\mathbf{margin\;of\;error}=t_{\frac{\alpha }{2}}*\left \{ \frac{s}{\sqrt{n}} \right \}=2.045*\left \{ \frac{6.9}{\sqrt{30}} \right \}=\mathbf{{\color{Red} 2.58}}$

$\mathbf{The \;95\%\; confidence \;interval\; is\; {\color{Red} 7.92}\;to\; {\color{Red} 13.08} }$

$\mathbf{Lower\;limit}=point\;estimate-margin\;of\;error=10.5-2.58=\mathbf{7.92}$

$\mathbf{Upper\;limit}=point\;estimate+margin\;of\;error=10.5+2.58=\mathbf{13.08}$