91影视

Given:

$n=23$

$b=-9.695$

$S{E}_b=1.889$

Determine the hypothesis:

$H_0:尾=0$

$H_a:尾<0$

Compute the value of the test statistic:

localid="1650543668177" $$\begin{gathered} t=\frac{b-{尾}_0}{S{E}_b} \\ =\frac{-9.695-0}{1.889} \\ 鈮�-5.132 \end{gathered}$$

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 localid="1650543673841" $$\begin{gathered} df=n-2 \\ =23-2 \\ =21: \end{gathered}$$

$P<0.0005$

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

$P<0.05鈬�\text{Reject}{H}_0$

There is sufficient evidence to support the claim that there is a negative linear relationship between Professor Moore's swim time and his pulse rate in the population of days on which he swims 2000 yards.

Given:

$n=23$

$b=-9.695$

$S{E}_b=1.889$

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

$$\begin{gathered} df=n-2 \\ =23-2 \\ =21 \end{gathered}$$

The critical t-value can be found in table B in the row of d f=21 and in the column of c=95% :

$t^{*}=2.080$

The boundaries of the confidence interval then become:

localid="1650543691786" $$\begin{gathered} b-t^{*}脳S{E}_b \\ =-9.695-2.080脳1.889 \\ =-13.6241 \end{gathered}$$

localid="1650543696751" $$\begin{gathered} b+t^{*}脳S{E}_b \\ =-9.695+2.080脳1.889 \\ =-5.7659 \end{gathered}$$