91Ó°ÊÓ

The following summary information is given, and the sample size(n) is 265.

Let,

The above figure shows that the area in each of the shaded tails is $\frac{\mathrm{Î\pm }}{2}=0.05$

To find $z_{\frac{\mathrm{Î\pm }}{2}}=1.645$by noting that the cumulative area to its left must be 1-0.05 or 0.95,refer to the standard normal table values to find the area of 0.95 correspondings . $z=1.645$For the 90% confidence level, the critical value is therefore $z_{\frac{\mathrm{Î\pm }}{2}}=1.645$

The margin of error is calculated using the following formula

$E=z_{\frac{\mathrm{Î\pm }}{2}}×\frac{\mathrm{σ}}{\sqrt{n}}$with$z_{\frac{\mathrm{Î\pm }}{2}}=1.645$

Therefore,

$\begin{array}{rcl}E& =& 1.645×\frac{65}{\sqrt{265}}\\ & =& 1.645×3.9929\\ & =& 6.5684\\ & & \end{array}$

Since the values of role="math" localid="1659734982378" $\stackrel{¯}{x}$and E to construct the confidence interval estimate of the mean failure time are obtained.

Now, substitute those values in the general formal for the confidence interval as follows:

$\begin{array}{rcl}\stackrel{¯}{x}-E& <& \mathrm{μ}<\stackrel{¯}{x}+E\\ 19-6.5684& <& \mathrm{μ}<19+6.5684\\ 12.432& <& \mathrm{μ}<25.568\\ & & \end{array}$

Let’s find the 90% confidence interval for all students' mean change in score of the SAT-Verbal tests.

Also, the margin of error by using the following formula:

$E=z_{\frac{\mathrm{Î\pm }}{2}}×\frac{\mathrm{σ}}{\sqrt{n}}$with $z_{\frac{\mathrm{Î\pm }}{2}}=1.645$

$\begin{array}{rcl}E& =& 1.645×\frac{49}{\sqrt{265}}\\ & =& 1.645×3.010\\ & =& 4.9515\\ & & \end{array}$

Now, the values of $x$and E to construct the confidence interval estimate of the mean failure time are obtained;substitute those values in the general formal for the confidence interval as follows:

$\begin{array}{rcl}\stackrel{¯}{x}-E& <& \mathrm{μ}<\stackrel{¯}{x}+E\\ 7-4.9515& <& \mathrm{μ}<7+4.9515\\ 2.048& <& \mathrm{μ}<11.952\\ & & \end{array}$

Given the true population Mean change in score on one of the SATs for all students who paid a private tutor is 15. The above results show that the SAT–Mat is the most likely to have mean changes.