In Exercises 14.48-14.57, we repeat the information from Exercises 14.12-14.21.

a. Decide, at the lore significance level, whether the data provide sufficient evidence to conclude that \(x\) is useful for predicting y.

b. Find a 90rk confidence interval for the slope of the population regression line.

$y=2.875-0.625x$

STEP 1: The null and alternative hypotheses must be established.

${\mathrm{H}}_0:{\beta }_1=0$($\mathrm{X}$is ineffective at forecasting $\mathrm{y}$)

$H_a:{\beta }_1\ne 0$($\mathrm{X}$ can forecast $\mathrm{y}$)

STEP 2: Determine the $\mathit{\alpha }$significance level.

The hypothesis test will be run at a significance level of $10\%$ or $\alpha =0.10$.

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

$=0-\left(15\right)\left(5\right)/5$

$=0-75/5$

$=0-15$

$=-15$

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

$=69-(15{)}^2/5$

$=69-225/5$

$=69-45$

$=24$

The entire amount of square SST is calculated as follows:

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

$=25-(5{)}^2/5$

$=25-25/5$

$=25-5$

$=20$

SSR stands for regression sum of squares.

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

$=\frac{(-15{)}^2}{24}$

$=\frac{225}{24}$

$=9.375$

SSE = SST - SSR

$=20-9.375$

$=10.625$

The slope of a regression line is calculated using the formula:

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

$=\frac{-15}{24}$

$=-0.625$

The standard error of the estimate is calculated using the formula:

$=\sqrt{\frac{10.625}{5-2}}$

$=1.881931632$E

$\approx 1.881$

STEP 3: Calculate the test statistic's value

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

$=\frac{-0.625}{1.881/\sqrt{24}}$

$=\frac{-0.625}{0.383957517}$

$=-1.627784252$

$\approx -1.63$

STEP 4: $\alpha =0.10$from Step 2. For $n=5$,

$\mathrm{df}=n-2$

$=5-2$

$=3$

The crucial values are $\pm t_{\omega 2}=\pm t_{0.05}=\pm 2.355$, as determined by technology.

Or

STEP 4: The test statistic's value is $t=-1.63$, as determined in Step 3. The P-value is the probability of seeing a value of $t$of $-1.63$or larger in magnitude if the null hypothesis is true, because the test is two-tailed. We get $P=0.2022$by using technology.

STEP 5: Because the test statistic's value is less than the critical value. We do not reject $H_0$, i.e. $t=-1.63<t_{\text{e0.05}3}=2.355$.

Or

STEP 5: Since the P-value is $=0.2022>\alpha =0.10$, this is the final step. Our null hypothesis $H_e$is not rejected.

STEP 6: The data do not give sufficient evidence to determine that the slope of the population regression line is not 0, and so the variable $x$is not useful for predicting the variable $y$at the $10\%$ significance level.

STEP 1: $\alpha =0.10$for a $90\%$confidence interval. Since $n=4$,

$\mathrm{df}=n-2$

$=5-2$

$=3$

Because of technology, $t_{\alpha 2}=t_{0.10/2}=t_{0.05}=2.355$

STEP 2: For ${\beta }_1$, the formula for finding the end points of the confidence interval is

$\Rightarrow b_1\pm t_{\omega 2}\times \frac{s_e}{\sqrt{S_{xx}}}$

We have $b_1=-0.625$,

$s_e=1.881$

$S_{xx}=24$

So, $-0.625\pm 2.355\times \frac{1.881}{\sqrt{24}}$

Or $-0.625\pm 0.904219953$, Or $-1.529$or to $0.279$.

STEP 3:

Interpretation: The slope of the population regression line is most likely $-1.529$ to $0.279$.