High School

You wish to test the following claim (H) at a significance level of [tex] \alpha = 0.05 [/tex]. For the context of this problem, let [tex] \mu = \text{PostTest} - \text{PreTest} [/tex], where the first data set represents a pre-test and the second data set represents a post-test. Each row represents the pre and post-test scores for an individual.

- Be careful when you enter your data and specify what your [tex] \mu [/tex] is so that the differences are computed correctly.

You believe the population of difference scores is normally distributed, but you do not know the standard deviation. You obtain the following sample of data:

Pre-test/Post-test:

1. 33.5 / -1.2
2. 41.1 / 42.2
3. 47.9 / 68.9
4. 44.3 / 43.9
5. 54.9 / 89.1
6. 36.2 / 34.3
7. 58.4 / 69.4
8. 40.6 / 74.8
9. 64.4 / -1.3
10. 59.8 / 11.8
11. 55.1 / -40.6
12. 53.3 / 51.4
13. 52 / -12.2
14. 37.6 / 2.9
15. 59.1 / 90.2
16. 49.2 / 25.2
17. 50 / 10.6
18. 48.5 / -72.8

For the data above, the sample mean of the differences is -22.1833, and the sample standard deviation of the differences is 44.8331.

Hint: You should use a Hypothesis Test MS Excel Template or Minitab for calculations.

Answer :

The t-statistic can be compared to the critical value at a significance level of 0.05 to determine whether to reject or fail to reject the null hypothesis.

To test the claim (H) at a significance level of a = 0.05, a two-sample t-test for the mean difference between pre-test and post-test scores can be performed. As the standard deviation is not known, the t-statistic can be calculated using the sample mean and standard deviation of the differences. The null hypothesis is that the population mean difference is equal to zero, and the alternative hypothesis is that the population mean difference is not equal to zero. The t-statistic can be compared to the critical value at a significance level of 0.05 to determine whether to reject or fail to reject the null hypothesis.

To learn more about the standard deviation, visit:

brainly.com/question/23907081

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