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.

a. Use a 0.01 significance level to test the claim that for the population of freshman male college students, the weights in September are less than the weights in the following April.

Accepted Answer

Hi, everyone. Let's take a look at this practice problem. This problem says researchers want to determine if the average weight of sophomore male students increases from September to March. The following are weights in kilograms from seven randomly selected students. And we're given the data that for September, we have the values of 75, 80, 68, 72, 77, 69, 74. And for March, we have the values of 76, 81, 69, 74, 78, 70, and 76. At the 0.01 significance level, test to claim that the mean weight in September is less than the mean weight in March for sophomore male students. So the first thing we need to do is write down our hypotheses, and so, for our null hypothesis, H knot, we're going to have mu sub D equal to 0. Where mu sub D is the mean difference in weight, so that is the mean weight in September minus the mean weight in March. So, with muse of D equal to 0, that means that there is no weight change. And for our alternative hypothesis, it's A, we're gonna have muse of D. Less than 0, and this would correspond to the mean weight in September, being less than the mean weight in March. Next, we need to actually calculate the weight difference for each of the 7 students, and this is best done in a table. So, on the screen, we actually have a table that has our students in the first column, and we've just labeled them with a number, starting from 1 and going to 7. In the next column, we have the weight in September, which has the values of 75, 80, 68, 72, 77, 69, and 74. The next column has the weights in March, which is 76, 81, 69, 74, 78, 70, and 76. And in the last column, we have D, which is equal to the weight of September minus the weight in March, and we have for its values of -1, -1, -1, -2, -1, -1, and -2. Now, in order to calculate our test statistic, we're gonna need our sample mean and the sample standard deviation. So, we're going to begin by looking at the sample mean, which we'll label as D bar. And we call for the mean, that is going to be equal to. The sum of D I divided by N. Where DI here is just one of our data points, and N is the total number of data points that we have, which in this case is 7. So that means that the bar is going to be equal to the quantity of -1 + -1 + -1. Plus -2 + -1 + -1. Plus Minus 2 in quantity divided by 7. And when we evaluate this expression, this is going to be approximately equal to -1.29. So next we'll calculate our standard deviation, and to calculate standard deviation, we actually need to actually calculate our squared deviations. And we really only need to calculate two of these, since we really only have two values of D minus 1 and -2. So, calculating that 4D equal to -1 will have the quantity of -1, minus -1.29. In quantity squared. And this is going to be equal to 0.0841. And then for our other value with the equal to -2, we'll have the quantity of -2 minus -1.29. In quantity squared and this is going to be equal to. 0.5041. And so now we actually need to sum up. These Squared deviations. So we'll have the sum. Of the quantity of DI minus D bar in quantity squared. And this is going to be equal to. For the first value, which is going to be the 0.0841. That actually, the value of D equal to minus 1 appears 5 times. So we'll multiply that by 5. And then we'll have plus, our second value, which is the 0.5041. And since D D equal to minus 2 appears twice, we'll multiply that by 2. And when we evaluate this expression, we get this that this is approximately equal to. 1.4287. So now we can actually calculate our variant, which is S D squared, and recall that this's going to be equal to the sum. Of the quantity of DI minus D bar in quantity squared, divided by the quantity of N minus 1. And so this is going to be equal to 1.4287 divided by the quantity of 7 minus 1. Which when you evaluate this expression, this is going to be approximately equal to. 0.2381. So, therefore, our standard deviation as sub D is going to be equal to the square root of 0.2381. Which is approximately equal to. 0.4880. So now we have all the information that we need in order to calculate our test stat. So, here we're gonna be using a paired T test. And so recall that T is going to be equal to. D divided by the quantity of S sub D. Divided by the square root of N in quantity, and we have values for all those quantities, so that means that T is going to be equal to. For D bar, we have minus 1.29. That's going to be divided by S of D, which is 0.4880, and that's going to be divided by the square root of 7 since N is equal to 7 in this case. So that means that T is going to be approximately equal to. -7.00. So, next we need to find our critical value, and for that, we'll need first to determine the number of degrees of freedom that we have. So we'll label that as DF and recall that the degrees of freedom is going to be equal to N minus 1, which in this case is going to be equal to 7 minus 1, which is equal to 6. So we can look up our critical value from a table. With a significance level of alpha equal to 0.01, and we have a one-tailed left test with 6 degrees of freedom. So doing so, we will see that our critical value, T critical, is going to be equal to. -3. 143. So, here we see that our calculated T value of minus 7.00 is actually less than our critical value of minus 3.143. So, that means that we will actually need to reject. Oral hypothesis, so we will reject. H knot. And by rejecting each knot, that means that there is sufficient evidence to conclude that the average weight. Average weight increases from September to March. So that is going to be the answer for this problem. So, I hope that this explanation has been useful, and I'll see you in the next video.

Key Concepts

Paired Sample T-Test

Significance Level (α)

Normal Distribution

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