91Ó°ÊÓ

Given:

$n=\text{Sample size}=19$

$c=\text{Confidence level}=95\%$

The estimate of the slope is given in the row "X Variable 1 and in the column "Coefficients" of the output:

$b≈-22.969$

The estimate of the standard error of the slope is given in the row "X Variable 1 and in the column "standard error" of the output:

$S{E}_b≈3.557$

The degrees of freedom is the sample size decreased by 2 :

localid="1650543375677" $$\begin{gathered} df=n-2 \\ =19-2 \\ =17 \end{gathered}$$

The critical t-value can be found in table B in the row of d f=17 and in the column of c=95% :

$t^{*}=2.110$

The boundaries of the confidence interval then become:

localid="1650543397407" $$\begin{gathered} b-t^{*}×S{E}_b \\ =-22.969-2.110×3.557 \\ =-30.4743 \end{gathered}$$

localid="1650543408380" $$\begin{gathered} b+t^{*}×S{E}_b \\ =-22.969+2.110×3.557 \\ =-15.4637 \end{gathered}$$

Given:

$n=19$

$b=-22.969$

$S{E}_b=3.557$

Determine the hypothesis:

$\mathrm{β}=0$

$\mathrm{β}<0$

Compute the value of the test statistic:

localid="1650543432666" $$\begin{gathered} t=\frac{b-{\mathrm{β}}_0}{S{E}_b} \\ =\frac{-22.969-0}{3.557} \\ ≈-6.457 \end{gathered}$$

The P-value is the probability of obtaining the value of the test statistic, or a value more extreme. The P-value is the number (or interval) in the column title of Table B containing the t-value in the row d f=n-2=19-2=17 :

$P<0.0005$

If the P-value is less than or equal to the significance level, then the null hypothesis is rejected:

$P<0.05⇒\text{Reject}{H}_0$

There is sufficient evidence to support the claim that there is a negative linear relationship between wine consumption and heart disease deaths in the population of countries.