Following are the data on total hours studied over 2 weeks and test score at the end of the 2 weeks, use$\alpha =0.01$presuming that the assumption for regression inference are met, decide at the specified significance level whether the data provide sufficient evidence to conclude that the predictor variable is useful for providing the response variable.

Given table is

$\alpha =0.01$we have to find out whether the data provide sufficient evidence to conclude that the predictor variable is useful for providing the response variable.

STEP-1 The null and alternative hypothesis are:

$H_0:{\beta }_1=0\&H_{\alpha }:{\beta }_1\ne 0$

STEP-2 determine significance level

The hypothesis test should be run at a significance level of $1$percent, or $\alpha =0.01$

Table of computation:

$$\begin{gathered} S_{xy}=\sum x_i{y}_i-\left(\sum x_i\right)\left(\sum y_i\right)/n \\ =-133.5 \end{gathered}$$

$$\begin{gathered} S_{xx}=\sum x_i^2-{\left(\sum x_i\right)}^2/n \\ =157.875 \end{gathered}$$

SST is given by

$$\begin{gathered} S_{yy}=\sum y_i^2-{\left(\sum y_i\right)}^2/n \\ =188 \end{gathered}$$

SSR is given by

$$\begin{gathered} SSR=\frac{S_{xy}^2}{S_{xx}} \\ =112.88361 \end{gathered}$$

$$\begin{gathered} SSE=SST-SSR \\ =188-112.88361 \\ =75.11163895 \end{gathered}$$

$b_1=-0.8454065701$

$s_e\approx 3.54$

STEP-3 The value of test statistic can be calculated as

$$\begin{gathered} t=\frac{b_1}{s_e/\sqrt{S_{xx}}} \\ \approx -3.00 \end{gathered}$$

STEP-4

$$\begin{gathered} df=n-1 \\ =8-2 \\ =6 \end{gathered}$$

We discovered that critical values are $\pm t_{\alpha /2}=\pm t_{0.005}=\pm 3.707$using technology.

STEP 4: The test statistic's value is$t=-3.00$, as determined in Step 3. The P-value is the chance of seeing a value of $tof-3.00$or larger in magnitude if the null hypothesis is true, because the test is two-tailed. We obtain the$p=0.023917$using technology.

STEP 5: Because the test statistic's value exceeds the critical value. Our null hypothesis is not rejected. $H_0$i.e. $t=-3.00<t_{0.005,6}=3.707$

STEP 5: Because of the$P=0.023917>\alpha =0.01$. Our null hypothesis $H_0$is not rejected.

STEP 6: The data do not provide adequate evidence at the $1\%$significance level to establish that the variable $\left(x\right)$is useful as a predictor of test result for students in beginning calculus courses.