Question

14. Graduation Rates. Refer to Problems $11$and $12$

a. At the $5\%$significance level, do the data provide sufficient evidence to conclude that student-to-faculty ratio is useful as a predictor of graduation rate?

b. Determine a $95\%$confidence interval for the slope, ${\beta }_1$, of the population regression line that relates graduation rate to student-to faculty ratio. Interpret your answer.

Short answer

(a) The data do not provide sufficient evidence to conclude that student-to-faculty ratio is useful as a predictor of graduation rate.

(b) The required confidence level is$0.75\text{to}4.80$.

Step-by-step solution

Part (a) Step 1: Given Information

Null and Alternate hypothesis:

${\mathrm{H}}_0:{\beta }_1=0$

$H_a:{\beta }_1\ne 0$

Perform hypothesis at $5\%$significance table gives

$x$	$y$	$xy$	$x^2$	$y^2$
$16$

Part (a) Step 2:  CalculatingSxy,Sxx

It is calculated as $S_{xy}=\sum x_i{y}_i-\left(\sum x_i\right)\left(\sum y_i\right)/n$

$=9088-\left(173\right)\left(515\right)/10$

$=9088-8909.5$

$=178.5$

and

$S_{xx}=\sum x_i^2-{\left(\sum x_i\right)}^2/n$

$=3081-(173{)}^2/10$

$=3081-2992.9=88.1$

Part (a) Step 3: Calculating Regression Sum

Sum of squares SST is

$S_{yy}=\sum y_i^2-{\left(\sum y_i\right)}^2/n$

$=27907-(515{)}^2/10=27907-26522.5=1384.5$

Regression Sum Square is

$SSR=\frac{S_{xy}^2}{S_{xx}}$

$=\frac{(178.5{)}^2}{88.1}=361.6600454$

Now,

$\mathrm{SSE}=\mathrm{SST}-\mathrm{SSR}$

$=1384.5-361.6600454=1022.839955$

Part (a) Step 4: Standard Error and Test Static

Slope of regression line is calculated as

$b_1=\frac{S_{xy}}{S_{xx}}$

$=\frac{178.5}{88.1}=2.026106697$

Standard Error of estimate is calculated as

$s_e=\sqrt{\frac{SSE}{n-2}}$

$\approx 11.3073$

Test Static is solved as

$t=\frac{b_1}{s_e/\sqrt{S_{xx}}}$

$=\frac{2.026107}{11.3073/\sqrt{88.1}}$

$\approx 1.682$

Part (a) Step 5: Critical Values

From table $\alpha =0.05$

For $n=10$

$\mathrm{df}=n-2$

$\mathrm{df}=10-2=8$

Hence, critical values$\pm t_{\alpha 2}=\pm t_{0.025}=\pm 2.306$

Part (a) Step 6: Conclusion

• Above step gives $t=1.682$
• As test is double tailed, $P$value is probability of observing values of $t=1.682$or greater if null hypothesis is true,
• From Technology, $P=0.131$
• The value of test static is less than critical value. We won't reject null hypothesis. $H_0\text{. i.e.}t=1.682<t_{0.025,8}=2.306$
• $\text{P-value}=0.131>\alpha =0.05$, we can't reject null hypothesis.
• At $5\%$significance data, data do not provide sufficient evidence to conclude that student-to-faculty ratio is useful as a predictor of graduation rate.

Part (b) Step 1: Given Information

Confidence level is$95\%$,$\alpha =0.05.n=10$

Part (b) Step 2: Simplification

Using Formula $\text{df}=n-2=10-2=8$

Also $t_{\alpha /2}=t_{0.05/2}=t_{0.025}=2.306$

End points of confidence interval for ${\beta }_1$are $b_1\pm t_{\alpha 2}\times \frac{s_e}{\sqrt{S_{xx}}}$

Now, role="math" localid="1654097940510" $b_1=2.026107,s_e=11.31,S_{xx}=88.1$

$\Rightarrow 2.026107\pm 2.306\times \frac{11.31}{\sqrt{88.1}}$

${\beta }_1=2.026107\pm 2.777987349,\text{or}0.75\text{to}4.80$

There is $95\%$confidence that relates graduation rate to student-to faculty ratio