In Exercises 5–16, use the listed paired sample data, and assu...

Question

In Exercises 5–16, use the listed paired sample data, and assume that the samples are simple random samples and that the differences have a distribution that is approximately normal.

Heights of Presidents A popular theory is that presidential candidates have an advantage if they are taller than their main opponents. Listed are heights (cm) of presidents along with the heights of their main opponents (from Data Set 22 “Presidents” in Appendix B).

a. Use the sample data with a 0.05 significance level to test the claim that for the population of heights of presidents and their main opponents, the differences have a mean greater than 0 cm.

Accepted Answer

All right, hello everyone. So this question says, a study compares the average number of hours slept per night by college students before and after final exams. The following data and hours were collected from 6 students. At the 0.05 significance level, test the claim that students sleep more hours after finals than before finals. So here, what I did on the screen was go ahead and copy the values that were given into a table that I made myself, because there's more information that we have to calculate to test the hypothesis given, or the claim rather. So first and foremost, The first blank column that I have on the table represents the differences in the number of hours slept before and after finals. So D, in this case is equal to the number before, subtracted by the number after for each student. Or actually I had that backwards. The difference D is actually the number of sleep, the number of hours of sleep after subtracted by the number before. So when calculating the difference for each student, the values for D are as follows. So we have 0.7. 0.7 0.6. 0.7, 0.4, and 0.6. And so now that we have calculated all of the differences, right, all of the values for D for each student, we can now find the sum. And that sum comes out to 3.7. With the sum we can now find D bar, which represents the mean of all of the differences. So D bar is equal to 3.7 divided by 6 because there were 6 calculated values for D. And so that gives you 0.6167. No For the final column of the table, the calculation per each student is as follows, right? You take the difference D. Subtract it by the bar, which is the mean, and then square the difference. And when you do so, the values for D subtracted by D bar squared are as follows, again for each student in order from 1 through 6. So that's going to be 0.00. 694 0.00694 again. 0.000. 278 0.00694 0.0469 And 0.000278. So now what you want to do is find the sum of all of the values calculated or des subtracted by D bar squared. And that sum comes out to approximately 0.0683 after rounding. So now with the sum of. All values calculated for D subtracted by D bar squared, you then calculate the standard deviation or S sub D. Because that's equal to the square root of that sum, so that's the square root of 0.0683. Divided by the sample size subtracted by 1, so that's equal to 5. And that gives you a standard deviation of approximately 0.1168. Which you then use to calculate the test statistic. So in this case, the test statistic. Is equal to. Dear Divided by the standard deviation. Multiplied by the square root of N. Actually, let me correct that. So T is equal to D bar divided by. S sub d divided by the square root of N. So that equals 0.6167. Divided by 0.1168 divided by the square root of 6. And this equals approximately 12.93. So now, before we can find our critical value, recall that our degrees of freedom in this case would be the sample size subtracted by 1, so that equals 5. Now The critical value for a right tailed tea test at 0.05 significance level and 5. 5 degrees of freedom. That critical value. Is equal to 2.015. So here we can see that our calculated value, our 12.93, is greater than the critical value, which means that we can reject the null hypothesis. So now, we've arrived at our final answer, which reads as follows. At the 0.05 significance level, there is strong evidence to support the claim that students sleep more hours after finals than before. And there you have it. So with that being said, thank you so very much for watching, and I hope you found this helpful.

Key Concepts

Paired Sample Data

Hypothesis Testing

Normal Distribution

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