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.

Measured and Reported Weights Listed below are measured and reported weights (lb) of random female subjects (from Data Set 4 “Measured and Reported” in Appendix B).

a. Use a 0.05 significance level to test the claim that for females, the measured weights tend to be higher than the reported weights.

Accepted Answer

All right. Hello, everyone. So, this question says, listed below are the actual study hours logged by students and the study hours those same students reported in a survey. Assume the samples are simple random samples and the differences have an approximately normal distribution. Use a 0.05 significance level to test the claim that the actual study hours logged by students are higher than their reported study hours. And here we have 4 different entry choices labeled A through D. So first, let's take a moment to identify the hypotheses. Now, for this problem, let's allow for capital, excuse me, lowercase d to represent the difference between logged and reported hours. So it's the logged hours subtracted by the reported hours. In this case we're testing to see if the logged hours are higher than the reported ones, so The null hypothesis as a result or not. Would state that mu of D is equal to 0, which means that there is no difference between the logged and reported hours. The alternative hypothesis on the other hand, or a sub a. would state that muse of D is greater than 0, which means that the logged hours are greater than reported hours. So, let's go back to the data table that we are given, because here on the right side of the screen, I added some columns of information that we're going to need for this question. So The first column I've added is 4D, which is again the difference between logged and reported hours. So, first, let's take a moment to do that, right? Let's find the difference between the logged and reported hours for each student. So when you go through and you subtract these, you get 2. 2 One 2122 and 1. Now Of the differences of the values for D that you've just calculated, let's take a moment to find D bar, which is the mean of the differences. So the mean of the differences is equal to the sum of all D values. Divided by 8, which is the number of students in this study. And so D bar is ultimately equal to 1.625. And it's this number that you're going to need for the next column that I've filled out. Because the next column that I put in is des subtracted by D bar. So it's the difference calculated for each student, subtracted by the mean of the differences. So now, these are the values that you obtain after calculating Ds subtracted by D bar for each student. Now, if the difference was 2, then Ds subtracted by D bars equal to 3, 0.375. If the difference was 1, then Ds subtracted by D bar is -0.625. So now this leads you to the final column that I've added on the screen here, which is desubtracted by D bar squared. So you take the values of the previous column. And squirt So if des subtracted by D bar was equal to 0.375, then that value squared is 0.1406. If des subtracted by D bar was. -0.625, then that value squared is 0.3906. All right, so for the calculation of the standard deviation, we have to know the sum of all values for des subtracted by D bar squared. So when you add together all of the values in the very last column of this table, that sum is equal to, oops. 1.875. So now I need to scroll down to open up some space for our calculations. As mentioned before, we need to find S sub D or the standard deviation. Which if you recall. Is equal to the square root of. The sum Of all values for Ds subtracted by D bar squared. Divided by and subtracted by 1. So plugging in the information you have. S sub D is equal to the square root of. 1.875 Divided by 8 subtracted by 1, which gives you 7. And so S sub D is equal to approximately 0.5175. So now, recall that the standard deviation is necessary to find the test statistic, and in this case, we're using paired data. So the test statistic d. For paired data. Is T equals? Dear Divided by S sub D divided by the square root of N. So plugging in the information you have, that's. 1.625. Divided by 0. 5175 divided by the square root of 8. And so T is approximately equal to 8.88. And so here, the test statistic that we just calculated is compared to the critical value. Keep in mind that the degrees of freedom is equal to N subtracted by 1, so 8 subtracted by 1 is equal to 7. Using a T distribution table with 7 degrees of freedom, and alpha equals 0.05 for a one-tailed test, The critical T value. Is approximately 1.895. And now we compare. Right. Notice how our calculated T value, that's 8.88, is greater than our critical T value. Because it's greater, we can reject the null hypothesis. So we can reject each knot. So what does that mean? Well, because we reject the null hypothesis, we're saying that there's sufficient evidence at the 0.05 significance level to support the claim that students' logged study hours are higher than their reported study hours. And so there you have it. Going back now to our inter choices. Our correct answer is going to be option C, which reads there is sufficient evidence that logged study hours are higher than reported hours. 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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