Improving SAT scores. Refer to the Chance (Winter 2001) and National Education Longitudinal Survey (NELS) study of 265 students who paid a private tutor to help them improve their SAT scores, Exercise 2.88 (p. 113). The changes in both the SAT–Mathematics and SAT–Verbal scores for these students are reproduced in the table. Suppose the true population meansa change in score on one of the SAT tests for all students who paid a private tutor is 15. Which of the two tests, SAT–Mathematics or SAT–Verbal, is most likely to have this mean change? Explain.

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 $\frac{\alpha }{2}=0.05$

To find $z_{\frac{\alpha }{2}}=1.645$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 $z_{\frac{\alpha }{2}}=1.645$

The margin of error is calculated using the following formula

$E=z_{\frac{\alpha }{2}}\times \frac{\sigma }{\sqrt{n}}$with$z_{\frac{\alpha }{2}}=1.645$

Therefore,

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

Since the values of role="math" localid="1659734982378" $\overline{x}$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:

$\begin{array}{rcl}\overline{x}-E& <& \mu <\overline{x}+E\\ 19-6.5684& <& \mu <19+6.5684\\ 12.432& <& \mu <25.568\\ & & \end{array}$

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:

$E=z_{\frac{\alpha }{2}}\times \frac{\sigma }{\sqrt{n}}$with $z_{\frac{\alpha }{2}}=1.645$

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

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:

$\begin{array}{rcl}\overline{x}-E& <& \mu <\overline{x}+E\\ 7-4.9515& <& \mu <7+4.9515\\ 2.048& <& \mu <11.952\\ & & \end{array}$

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.