Question

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

a. Decide, at the $10\%$significance level, whether the data provide sufficient evidence lo conclude that a is useful for predicting $y$:

b. Find a $90\%$confidences interval for the slope of the population regression line.

$y=1.75+0.25x$

Short answer

The slope of the population regression line is most likely $-0.782$to$1.282$.

Step-by-step solution

Step 1:Given Information (part a)

STEP 1: Determine the null and alternative hypotheses

${\mathrm{H}}_e:{\beta }_1=0(\mathrm{X}$is not useful for predicting $\mathrm{y})$

$H_a:{\beta }_1\ne 0$(X is useful for predicting $\left.y\right)$

STEP 2: Determine the $\alpha$significance level.

The hypothesis test should be run at a significance threshold of $10\%$, or $\alpha =0.10$.

Step 2:Explanation (part a)

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

$=34-\left(12\right)\left(10\right)/4$

$=34-120/4$

$=34-30$

$=4$

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

$=52-(12{)}^2/4$

$=52-144/4$

$=52-36$

$=16$

The entire amount of square SST is calculated as follows:

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

$=30-(10{)}^2/4$

$=30-100/4$

$=30-25$

$=5$

Step 3:Explanation (part a)

SSR stands for the regression sum of squares.

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

$=\frac{(4{)}^2}{16}$

$=\frac{16}{16}$

$=1$

SSE $=$SST $-$SSR

$=5-1$

$=4$

The slope of the regression line is calculated using the formula,

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

$=\frac{4}{16}$

$=0.25$

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

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

$=\sqrt{\frac{4}{4-2}}$

$=1.414213562$

$\approx 1.414$

Step 4:Explanation (part a)

STEP 3: Calculate the test statistic's value

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

$=\frac{0.25}{1.414/\sqrt{16}}$

$=\frac{0.25}{0.3535}$

$=0.709219858$

$\approx 0.71$

STEP 4: Repeat Step 2 with $\alpha =0.10$. For $n=4$,

$df=n-2$

$=4-2$

$=2$

Step 5:Conclution (part a)

We discovered the crucial values by using technology $\pm t_{\omega 2}=\pm t_{\cos }=\pm 2.920$.

Or

STEP 4: The test statistic's value is $t=0.71$as determined in Step 3. The P-value is the probability of seeing a value of t of $0.71$or larger in magnitude if the null hypothesis is true, because the test is two-tailed. We get the $P=0.5527$using technology.

STEP 5: The test statistic's value is lower than the crucial value. $H_0$i.e. $t=0.71<t_{eos,2}=2.920$is not rejected.

Or

STEP 5: Since the P-value is $=0.5527>\alpha =0.10$, the next step is to calculate the alpha. The null hypothesis, $H_0$, is not rejected.

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

Given Information (part b)

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

$\mathrm{df}=n-2$

$=4-2$

$=2$

$t_{a/2}=t_{0.10/2}=t_{\text{e.05}}=2.920$comes from technology.

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

$\Rightarrow b_1\pm t_{a2}\times \frac{s_e}{\sqrt{S_{xx}}}$

We have $b_1=0.25$,

$s_e=1.414$

$S_{xx}=16$

So, $0.25\pm 2.920\times \frac{1.414}{\sqrt{16}}$

Or $0.25\pm 1.03222,-0.7822$or to $1.2822$

STEP 3:

Interpretation: The slope of the population regression line is most likely between$-0.782$ and $1.282$.