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.

Wilcoxon Signed-Ranks Test Repeat the preceding exercise using the Wilcoxon signed-ranks test.

Accepted Answer

Hello there, today we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. A study investigates whether there is a significant difference in cholesterol levels of patients before and after starting a new diet program. The cholesterol levels in units of milligrams per deciliter of 8 patients are measured at the start and after 6 months. Data pairs are as follows. We have two rows. Our first row is for before, and we have the values 210, 195, 220, 205, 198, 215, 202, 210. And then we have our second row for after 208, 195, 215, 200, 200, 215, 200, and 208. Use the Wilcoxson sign rank test at the 0.05 significance level to test the claim that the median cholesterol level does not change after the diet. Awesome. So it appears for this particular problem, we're asked to use the Wilcoxson signed rank test at our significance level of 0.05. To test the claim that the median cholesterol level does not change after implementing the new diet program. So now that we know what we're trying to solve for, our first step that we need to take is we need to identify our two different hypothesis values. Starting with our null hypothesis, which we denote as H0, we need to note that for the null hypothesis, the median difference is 0, meaning there is no change in cholesterol levels, and for our alternative hypothesis, which will denote as H1, the median difference is not zero, meaning there is a change. So now, with that said, we now need to calculate the differences and their absolute values. So with that said, to make it simple, I've gone ahead and created a table for us to refer to. And once again, what we're doing with this table to help us organize our data is we're calculating the differences and their absolute values. So, starting with our column on the far right, we have our absolute Value which going down the column we have values of 20552022. And our next column we have sign which is positive, 0, positive, positive, negative, 0, positive, positive. Then we have our column for difference which is + 20 + 5 + 5 minus 20 + 2 + 2. Then we have our after, which is 208, 195, 215, 200, 200, 215, 200, 208. Then we have our before column which is 210, 195, 220, 205. 198215 202 and 210. Awesome. So now, considering our table, we need to exclude pairs with zero differences. So that means we will have the absolute values, which you'll call AV for short, so we'll have the absolute values of 2. 552-22. And now we need to rank the absolute differences from the smallest value to the largest value. So using our absolute differences, AD for short, we will have 222. 2, so I have 123, and 4 2s, 55. So I have 2.2, 22, 5.5. So 2 5s and 4 2s. So now we need to rank the 2's average of ranks, which is 1. 234, which is equal to 1 + 2. Plus 3. Plus 4, and then we need to divide everything by 4, which will give us an answer of 10 divided by 4, which is equal to 2.5. Then we need to rank the 5's average of rank, which is 5.6 is equal to 5 plus 6 divided by 2, which is equal to 11 divided by 2, which means that it's going to be simply equal to 5.5. So now we need to assign our ranks. So 2 is assigned to 2.5, and this is for 4 times. And 5 is assigned to 5.5, which is for 2 times. So now, With that in mind, we need to assign, we need to assign signs to the ranks based on the original differences. So we have our differences, which I'll call D or 2. 55, 2 or 2, 222. And we need to note that we have the sign ranks says 2 is assigned to 2.5. Fantastic, and that will mean, without a doubt, that 5 is assigned to 5.5. And moving right along here. We'll have a value of 5 is assigned to 5.5, and then we'll have a value of -2 is assigned to -2.5, and 2 is assigned to 2.5, and 2 is assigned to 2.5. Fantastic. So now our next step. We need to calculate the sum of positive and negative ranks. Now that is our next step. So, once again, we're calculating the sum of positive and negative ranks. So starting out with positive ranks first, so we'll call this value. PR for short, so PR for positive ranks. And we'll have + 2.5. plus or positive. 5.5. And then we need to do comma positive. 5.5, 2.5. 2.5. Fantastic. So, moving right along now, so now we have the sum of positive ranks, which is 2.5 plus 2. So it's gonna be 2.5 plus. 5.5 plus. So once again, the sum is when we add all our values together, so 2.5 plus 5.5 plus 5.5 plus 2.5 plus 2.5, which when we plug that into our calculator, we should get the falling value of 18.5. So that will mean that our negative rank, which I'll call NR for short, our negative rank will be equal to -2.5. So the test statistic, which will denote as W, is smaller, will be the smaller of the sum of positive or negative ranks, so we can ignore the sign in this case. So W is equal to the min of 18.5. 2.5, which is equal to 2.5. So, now, to find the critical value for the data sample, or the sample is equal to 6, so N is equal to 6, that's our sample size. So that's the number of non-zero differences. At our significance of alpha, which alpha denotes significance of 0.05, because we're dealing with a two-tailed situation. So, from the Wilcoxson signed rank tables, as we should recall, the critical value for N equals 6 at Alpha equals 0.05, 2 tailed will be 0. So if W is less than or equal to 0, we can reject the null hypothesis. So I'll put an X next to H0, noting that we can. Reject the null hypothesis. So now we can state our final conclusion. So since W is equal to 2.5 and 2.5 is greater than 0, we do not need to reject the null hypothesis. So therefore, in conclusion, at the 0.05 significance level, there is insufficient evidence to support the claim that the median cholesterol level changed after implementing the diet. So, going back to look at our problem itself to summarize what we've learned, that will mean, without a doubt that our final answer once again has to be the following. So, At the 0.05 significance level, there is insufficient evidence to support the claim that the median cholesterol level changed after the diet, and that's it. That is our final answer. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

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.

Wilcoxon Signed-Ranks Test Repeat the preceding exercise using the Wilcoxon signed-ranks test.

Key Concepts

Wilcoxon Signed-Ranks Test

Significance Level

Nonparametric Tests

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