In Exercises 1–10, use a 0.05 significance level with the indi...

Question

In Exercises 1–10, use a 0.05 significance level with the indicated test. If no particular test is specified, use the appropriate nonparametric test from this chapter.

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. Use the sign test to test the claim that for the population of freshman male college students there is not a significant difference between the weights in September and the weights in the following April. What do you conclude about the Freshman 15 belief?

Accepted Answer

All right, hello, everyone. So this question says, a study investigates whether there is a significant difference in cholesterol levels of patients before and after a new diet program. The cholesterol levels in milligrams per deciliter of 8 patients are measured at the start and after 6 months. Data pairs are as follows. Use the sign test at the 0.05 significance level to test the claim that the median cholesterol level does not change after the diet. And here we have Two different answer choices labeled A and B stating whether or not there is sufficient or insufficient evidence to support the claim that the median cholesterol level changed after the diet. All right, so first, it's important to state both the null and alternative hypotheses. Right, the null hypothesis states that the median difference. Median difference is 0, which means that there would be no change in cholesterol levels. On the other hand, right, the alternative hypothesis would state that the median difference. Is simply not 0, which means that there, which would mean that there is a change after the diet. So here you can see that the data is already listed in pairs for each patient. So first, let's find. The difference and what I mean by that is let's first calculate. The value before, subtracted by the value after, for each of the 8 data pairs that we have. So in order, the differences with their respective sign are as follows. We have 2. 0 Positive 55 again. 2 02 and 2 again. So now, let's count the number of positive and negative signs, excluding zeros. And by doing so we can see that we have 5 positive signs. And Only one negative sign. Which means that the total of. Non-zero pairs is equal to 6. Now we move on to the binomial distribution. So, under the null hypothesis, the probability of a positive or negative sign, that's P is equal to 0.5. This means that the test, or rather the test statistic is the smaller of. The number of positive or negative signs. So it's the minimum, which would mean that our minimum, that's capital X here, the test statistic, would be equal to 1. And now we calculate the P value. Now this is a two-tailed test, which means that we're going to double the number. Of the smaller tail probability. All right, so here we have 3 different probabilities to compare, the probability of X being equal to 0, that of X equal to 1, and that of X being less than or equal to 1. So, for the binomial distribution, recall that N is equal to 6, because that's the number of non-zero pairs. And, as mentioned before, According to the null hypothesis, P is equal to 0.5, so. Starting off with the probability of X is equal to 0. That would be 6 and 0. Multiplied by 0.5. To the power of 6. So then this would be 1 multiplied by 0.015, 625. And this just gives you 0.015625. Now for the probability of X being equal to 1, our binomial would be 6 and 1. Multiplied by 0.5 to the power of 6. So this time it would be equal to 6, multiplied by that 0.015625. Which would then give you 0.09375. Now for the probability of X being less than or equal to 1. Now, this, if you recall, is equal to the probability of X being equal to 0. Added to the probability of X being equal to 1. So that would be 0.015625 added to 0.09375. And this equals 0.109. 375. So now the two-tailed p-value is going to be. The probability of X being less than or equal to 1 multiplied by 2. Now I just want to acknowledge very quickly that I had misspoken earlier when I said that there are 3 values to compare. Rather, in order to calculate the value of the probability of X being less than or equal to 1, we first needed to find the probabilities of X being 0 and 1 respectively. So not really comparing, I just wanted to correct that. But in any case, right the p value. Of the two-tailed test is 2 multiplied by 0.109375. And this equals 0.21875. And at this point, we can compare now to the significance level. So because 0.21875 is greater than 0.05, we failed to reject the null hypothesis. Which means if I scroll up here very quickly, our correct answer is going to be option B in the multiple choice. Because at the 0.05 significance level, there is insufficient evidence to support the claim that the median cholesterol level changed after the diet. And there you have it. So with that being said, thank you so very much for watching and I hope you found this helpful.

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