Question

Repeat the test in Exercise 10.83, but this time combine the data from Laps 1 and 5.

Short answer

(a) The null hypothesis:$H_0:{\mu }_1={\mu }_2$
(b) The alternate hypothesis: $H_a:{\mu }_1\ne {\mu }_2$
(c) The difference between the mean times for completing a lap in races and in practices is denotes by $\overline{X_1}-\overline{X_2}$.
(d) Student's t distribution.
(e) The test statistic: $-5.087$.
(f) The p-value: $0.0000084$
(g) Check student's solution.
(h) (i) $\alpha =0.05$
(ii) Decision: Null hypothesis is rejected.
(iii) $\alpha >p$-value
(iv) Conclusion: There is enough evidence to conclude that the mean time for completing a lap in races is different from that in practices, at the $5\%$ significance level.

Step-by-step solution

Given information

To repeat the test in Exercise 10.83, but this time combine the data from Laps 1 and 5.

Explanation

(a) The null hypothesis is defined as follows:
$H_0:{\mu }_1={\mu }_2$
(b) The alternate hypothesis is defined as follows:
$H_a:{\mu }_1\ne {\mu }_2$
(c) The difference between the mean times for completing a lap in races and in practices is denotes by $\overline{X_1}-\overline{X_2}$
(d) Student's t distribution.
(e) The paired test static for the mean difference is,

After ENTER all values the OUTPUT will look like this:

Hence, the test statistics $-5.087$.

Explanation

(f) The $p$-value from the output is $0.0000084$.
(g) Using the data from the last exercise, create a mental picture of the situation.
The horizontal axis should be clearly labelled and scaled, and the $p$-value-corresponding region(s) on the graph should be shaded.

(h)

(i) $\alpha =0.05$
(ii) Decision: Null hypothesis is rejected.
(iii) $\alpha >p$-value
(iv) Conclusion: There is enough evidence to conclude that the mean time for completing a lap in races is different from that in practices, at the $5\%$significance level.