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.

The Freshman 15 The “Freshman 15” refers to the belief that college students gain 15 lb (or 6.8 kg) during their freshman year. Listed below are weights (kg) of randomly selected male college freshmen (from Data Set 13 “Freshman 15” in Appendix B). The weights were measured in September and later in April.

b. Construct the confidence interval that could be used for the hypothesis test described in part (a). What feature of the confidence interval leads to the same conclusion reached in part (a)?

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says a fitness program is evaluated by measuring the number of push-ups completed by 12 participants before and after a 6-week training program. The results are, and here we're given a table where the first row is labeled 4, and it has the values of 1518, 2022, 1719, 1621, 1820, 17, and 19. And the second row is labeled after, and it has the values of 1821, 23, 24, 1921, 1823, 2022, 19, and 21. Construct the 95% confidence interval for the mean difference in pushup counts. What does the interval suggest about the program's effectiveness? So the first part of this problem, we need to construct the 95% confidence for the mean difference in push up counts. So the first thing we need to do is actually calculate the difference for each of our participants. And that is actually best done in a table. And so, on the screen here, we have a table, where the first column is labeled participants, and it just has the values of 123456789, 1011, and 12. So, we just numbered our participants as going from 1 to 12. The next column is. Labeled before, and so these are the push-ups that they did before the training, and this has the value of 1518, 2022, 1719, 1621, 1820, 17, and 19. And the next column is labeled as after, and here we have the values of 1821, 23, 24, 1921, 1823, 2022, 19, and 21. And the final column, we have D, which is the difference, and this is after minus before to calculate our difference. And here we do this for each of our pairs. We get the values of 333. 22222222 and 2. So, the first thing we need to do is to calculate our sample mean difference, as well as the standard deviation. And so, for the sample mean difference, that's gonna be D bar. And that's just going to be equal to the sum of all of our D values, divided by the number of participants. And here we can take a little bit of a shortcut since we really only have two values of D, 3 and 2. So, for the value of 3, it appears 3 times. So that means that D is going to be equal to the quantity of 3, multiplied by 3. Plus 2 multiplied by the number of times that 2 appears, which in this case is 9, so we'll have 2 multiplied by 9. And then that whole quantity is divided by the number of participants, which is in this case is 12. And so when we evaluate this expression, this is going to be equal to 2.25. Next, we need to calculate our standard deviation. And in order to calculate our standard deviation, we'll actually need to sum up the squared deviation. So, what we need to do is calculate the deviation for each of our points that it is away from the mean. And since we only have really two values of D again, 3 and 2, we really only need to calculate 2 of these squared deviations. So first, we'll calculate it for D equal to 3, so we'll have the quantity of 3 minus. 2.25, since we're looking at the deviation from the mean, and then we're going to square that quantity. In doing so, this is going to be equal to. 0.5625. And the next one that we need to calculate was is when D is equal 2, so we'll have the quantity of 2 minus 2.25 in quantity squared, and this is going to be equal to when we evaluate that. 0.0625. And so that means that the sum. Of the quantity of DI minus D bar in quantity squared is going to be equal to. 0.5625. Multiplied by 3, since we have a value of D equal to 33 times, plus 0.0625. Multiplied by 9. Which again, um we we're multiplying by 9 since we have 9 values of 2. And when we evaluate this expression, this is going to be equal to 2.25. So, now we can calculate our standard deviation, S sub D. And recall that that is going to be equal to the square root of the quantity of the sum. Of the quantity of D I minus D bar in quantity squared. Divided by the quantity of n minus 1. So this is going to be equal to the square root of the quantity of 2.25. Divided by the quantity of 12 minus 1. And when we evaluate this expression, this is going to be approximately equal to. 0.45. 2. And now we have all the information that we need to calculate our confidence interval. So, recall your formula for the paired sample confidence interval. Which is D bar plus or minus. The The alpha divided by 2. Multiplied by the quantity of S sub D. Divided by the square root of N in quantity. So, before we can use this formula, we need to figure out our T value. And since we're looking for the 95% confidence interval, that means that alpha is going to be equal to 0.05. And so then alpha divided by 2 is going to be equal to 0.025. And in order to figure out RT value, we'll also need the degrees of freedom. So, label that SDF and recall that the degrees of freedom is going to be equal to N minus 1, which is going to be equal to, in this case, 12 minus 1, which is equal to 11. That means that our T value, with an alpha divided by 2 of 0.025 and 11 degrees of freedom, If you look this up in our table. Then we will get a value of 2.201. So substituting in all of our values into our formula here, we will have 2.25. Plus or minus? 2.201, multiplied by the quantity of 0.452. Divided by the square roots of 12 in quantity. Which when simplified in this expression will get 2.25. Plus or minus. 0. 287. And so if we want to write this in interval notation, we'll need to determine our lower bound and upper bound. So, for a lower bound, we'll have 2.25 minus 0.287. Which gives us a value of 1.96. And for upper bound, we'll have the 2.25 plus the 0.287. Which will give us a value of 2.54. We could also rewrite this as an inequality. So we will have 1.96. is less than new sub D. Which is less than 2.54. And here, this is the confidence interval that we were actually looking for for this problem. So now we need to use this confidence interval to answer the question, what does the interval suggest about the program's effectiveness. Now, since the entire interval is positive, it does not include 0. We can say that there is strong evidence that the fitness program led to an increase in the number of push-ups that the participants can do. So that's gonna be the answer to the second half of this problem. So I hope that this explanation has been useful, and I'll see you in the next video.

Key Concepts

Paired Sample Data

Confidence Interval

Hypothesis Testing

Watch next