In Exercises 9–12, refer to the sample data from the given exe...

Question

In Exercises 9–12, refer to the sample data from the given exercises in Section 13-2. Use the Wilcoxon signed-ranks test for the claim about the median of a population.

Exercise 15 “Cotinine in Smokers”

Accepted Answer

All right, hello, everyone. So this question says, a study records the hemoglobin levels in grams per deciliter of 10 athletes. Use the Wilcoxen sign ranks test at a 0.01 significance level to test the claim that athletes have a median hemoglobin level of 13.5 g per deciliter. Assume that the population of differences has an approximately symmetric distribution. So, the last sentence, if you recall, is important for checking the requirements for the Wilcox and sign ranks test, because one of the criteria is that the population of differences be approximately normal in their distribution or symmetric. Now the other requirement. is to assume that the data is in a simple random sample. So that's something to keep in mind as we go through these calculations. So in this context, we're testing the null hypothesis. Which would state That the median hemoglobin level of athletes is 13.5 g per deciliter. So here On the screen, I went ahead and made a table using the values that were already given to us, and also additional information that we need for the Wilcox and sign ranks test. So first, I already have the second row, or excuse me, the first row filled out with the given data values. Now, the second row, which is the median, in this case would be 13.5, because that's the median that we are testing. So now this leads us to the third row, which corresponds to D. Now D is the difference between each sample value and the claimed median of 13.5. So, you would take the value itself and subtract it by the median in each case. All right, so, after repeating the calculation for each data value given, the values for D in their corresponding order are as follows. We have 0.6, 0.2, 0.7, 1.2, 1.1, 0.4, 1, 0.9, 0.5, and negative 0.8. So it's at this point that you now want to rank the absolute values of the differences, the absolute value of D. So Ignoring whether or not it has a positive or negative sign, what you want to do is rank each value, with 1 being the smallest and 10 being the largest. So when you do so, the ranks of the absolute values of D. Are as follows. We have 415, 10, 9. 2 873, and 6. So now that you've ranked the absolute values of all of the differences, what you now want to do is attach the sign of the difference. To each rank. So when you do so, you get -4, -1, 5, 10, -9. Positive 2 Positive 8 7 3 and -6. So now we scroll down to open up some space here. You want to calculate the sum of positive ranks and the absolute value of the sum of negative ranks. So for the positive ranks. That would be 5 added to 10. Added to 2 8 and 3. With that sum being 28. And now you want to do the same thing, but this time you're finding the sum of the absolute values of all the negative ranks. So that's The absolute value Of -4 added to -1. Added to -9 7 and -6. And this gives you 27. So, the two rank sums are 28 and 27. For the next step, we want to select the smaller of the two and allow it to be capital T. So capital T in this case is equal to 27. Now Now that we have tea and. Corresponds to the number of pairs of data for which the difference d is not zero. So, N, in this case is equal to all 10, because none of the pairs in our data set actually had a difference of zero. All right, so now we can proceed with the test statistic and also the critical value. What's worth noting in this case is that our sample size, that is N, is less than or equal to 30. And when that's the case, the test statistic happens to be T. So our test value is T equals 27. Now The critical value can be found using a table. So, at alpha equals 0.01, with a two-tailed test and N equal to 10, the critical value. is equal to 3. So now we make a decision based on the comparison between the test value and the critical value. So, notably, our test value T is greater than our critical value, which means that we fail to reject the null hypothesis. And now this leads us to our final answer, which reads as follows. There is insufficient evidence to warrant rejection of the claim that athletes have a median hemoglobin level of 13.5 g per deciliter. And there you have it. So with that being said, thank you so very much for watching, and I hope you found this helpful.

In Exercises 9–12, refer to the sample data from the given exercises in Section 13-2. Use the Wilcoxon signed-ranks test for the claim about the median of a population.

Exercise 15 “Cotinine in Smokers”

Key Concepts

Wilcoxon Signed-Ranks Test

Median

Non-parametric Tests

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