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

Short Answer

Expert verified

(a) The null hypothesis:H0:μ1=μ2
(b) The alternate hypothesis: Ha:μ1≠μ2
(c) The difference between the mean times for completing a lap in races and in practices is denotes by X1¯-X2¯.
(d) Student's t distribution.
(e) The test statistic: -5.087.
(f) The p-value: 0.0000084
(g) Check student's solution.
(h) (i) α=0.05
(ii) Decision: Null hypothesis is rejected.
(iii) α>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

01

Given information

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

02

Explanation

(a) The null hypothesis is defined as follows:
H0:μ1=μ2
(b) The alternate hypothesis is defined as follows:
Ha:μ1≠μ2
(c) The difference between the mean times for completing a lap in races and in practices is denotes by X1¯-X2¯
(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.

03

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) α=0.05
(ii) Decision: Null hypothesis is rejected.
(iii) α>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.

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Table.localid="1653641957518" 10.28


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