Use the given set of points to test the null hypothesis $H _ { 0 } : \beta _ { 1 } = 0 \text { versus } H _ { 1 } : \beta _ { 1 } \neq 0 . \text { Use the } \alpha = 0.01$ level of significance.

$\begin{array} { c | c c c c c c c c } x & 14 & 7 & 15 & 10 & 9 & 6 & 14 & 15 \\\hline y & 32 & 17 & 36 & 26 & 21 & 15 & 34 & 37\end{array}$

Question

Use the given set of points to test the null hypothesis

H _ { 0 } : \beta _ { 1 } = 0 \text { versus } H _ { 1 } : \beta _ { 1 } \neq 0 . \text { Use the } \alpha = 0.01

level of significance.

\begin{array} { c | c c c c c c c c } x & 14 & 7 & 15 & 10 & 9 & 6 & 14 & 15 \\\hline y & 32 & 17 & 36 & 26 & 21 & 15 & 34 & 37\end{array}

Quizplus · Accepted Answer

We need to perform a hypothesis test to determine if there is evidence to reject the null hypothesis that the slope of the population regression line is zero (i.e., no linear relationship between x and y) at the 0.01 level of significance. We can use the simple linear regression model: $$y_i = \beta_0 + \beta_1 x_i + \varepsilon_i$$ where $y_i$ is the i-th observation of the response variable, $x_i$ is the i-th observation of the explanatory variable, $\beta_0$ is the intercept, $\beta_1$ is the slope, and $\varepsilon_i$ is the error term associated with the i-th observation. We can use software or manual calculations to obtain the estimated regression equation based on the given data: $$\hat{y} = \hat{\beta_0} + \hat{\beta_1}x$$ where $$\hat{\beta_1}=\frac{\sum_{i=1}^{n}(x_i-\bar{x})(y_i-\bar{y})}{\sum_{i=1}^{n}(x_i-\bar{x})^2}$$ and $$\hat{\beta_0}=\bar{y}-\hat{\beta_1}\bar{x}$$ Using the given data, we obtain: \begin{align*} \hat{\beta_1} &= \frac{(14-10.125)(32-26.375)+(7-10.125)(17-26.375)+(15-10.125)(36-26.375)+(10-10.125)(26-26.375)\ &\quad+(9-10.125)(21-26.375)+(6-10.125)(15-26.375)+(14-10.125)(34-26.375)+(15-10.125)(37-26.375)}{(14-10.125)^2+(7-10.125)^2+(15-10.125)^2+(10-10.125)^2\ &\quad+(9-10.125)^2+(6-10.125)^2+(14-10.125)^2+(15-10.125)^2} \ &= \frac{119.75}{104.125} \ &= 1.150 \end{align*} and \begin{align*} \hat{\beta_0} &= \bar{y}-\hat{\beta_1}\bar{x} \ &= 27 - 1.150 imes 10.875 \ &= 14.51 \end{align*} The estimated regression equation is therefore: $$\hat{y} = 14.51 + 1.150 x$$ Now we need to test whether the population slope is significantly different from zero. The null and alternative hypotheses are: $$H_0 : \beta_1 = 0$$ $$H_1 : \beta_1 e 0$$ We can calculate the t-statistic for testing the null hypothesis: $$t = \frac{\hat{\beta_1}-0}{ ext{SE}(\hat{\beta_1})}$$ where $ ext{SE}(\hat{\beta_1})$ is the standard error of the estimate of $\hat{\beta_1}$, given by: $$ ext{SE}(\hat{\beta_1}) = \frac{ ext{RSS}}{\sqrt{n-2} \sqrt{\sum_{i=1}^n (x_i-\bar{x})^2}}$$ where RSS is the residual sum of squares: $$ ext{RSS} = \sum_{i=1}^n \hat{\varepsilon_i}^2 = \sum_{i=1}^n (y_i - \hat{y_i})^2$$ Using the values obtained above, we have: \begin{align*} ext{RSS} &= (32-25.25)^2 + (17-18.06)^2 + (36-32.15)^2 + (26-23.60)^2 + (21-20.75)^2 \ &\quad+ (15-14.35)^2 + (34-25.25)^2 + (37-32.15)^2 \ &= 97.24 \ ext{SE}(\hat{\beta_1}) &= \frac{97.24}{\sqrt{8} \sqrt{\sum_{i=1}^8 (x_i-\bar{x})^2}} \ &= 0.5039 \ t &= \frac{1.150-0}{0.5039} \ &= 2.281 \end{align*} Using a t-table with $n-2=6$ degrees of freedom at the 0.01 level of significance (two-tailed test), the critical value is $t_{0.005,6}=3.707$. Since $|t|=2.281<3.707$, we do not reject the null hypothesis. However, we can also obtain the p-value using software or a t-table. The two-tailed p-value is the probability of observing a t-statistic as extreme or more extreme than the observed one (in absolute value) under the null hypothesis. For $n-2=6$ degrees of freedom and $|t|=2.281$, the p-value is approximately 0.056, which is greater than the chosen level of significance $\alpha=0.01$. Therefore, we do not reject the null hypothesis. Therefore, the correct answer is A, Reject H₀.

Use the Given Set of Points to Test the Null H0:β1=0 versus H1:β1≠0. Use the α=0.01H _ { 0 } : \beta _ { 1 } = 0 \text { versus } H _ { 1 } : \beta _ { 1 } \neq 0 . \text { Use the } \alpha = 0.01H0​:β1​=0 versus H1​:β1​=0. Use the α=0.01

Use the Given Set of Points to Test the Null $H _ { 0 } : \beta _ { 1 } = 0 \text { versus } H _ { 1 } : \beta _ { 1 } \neq 0 . \text { Use the } \alpha = 0.01$