In Exercises 1–10, based on the nature of the given data, do t...

Question

In Exercises 1–10, based on the nature of the given data, do the following:

a. Pose a key question that is relevant to the given data.

b. Identify a procedure or tool from this chapter or the preceding chapters to address the key question from part (a).

c. Analyze the data and state a conclusion.

IQ Scores of Twins Listed below are IQ scores of twins listed in Data Set 12 “IQ and Brain Size” in Appendix B. The data are pairs of IQ scores from ten different families.

Accepted Answer

Alright, hello, everyone. So this question says, listed below are the heights and centimeters of siblings from 8 different families. Part A says to pose a key question that is relevant to the given data. B says identify a statistical procedure or tool to address the key question from Part A. And she says. Analyze the data and state a conclusion. So first, I want to take a moment to actually answer part B before answering Part A and then proceeding with the calculations, right? Because notice how the data values are inherently paired, because they each represent siblings from different families. So, As for the method we can use, that would be a paired tea test because as mentioned before. The data values themselves are inherently paired because they represent siblings. So, as for the key question, what we can do is ask if there's a significant difference between the average heights of sibling A and sibling B. Which now brings us to the table that I went ahead and pre-drew on the screen. Now, for the 1st 3 columns, I went ahead and just copied the data that was already given, because we're going to need this information for the additional. Values, or in other words, the other two columns of the table that I made. So the first thing on here. Or the first blank column, I should say. Is the difference in heights between both siblings for each family. So the difference or D. Lower case d is equal to the height of sibling A subtracted by the height of sibling B. And so when you repeat this calculation for all 8 families, the differences are listed as follows in order from family 1 through 8. So that's 2, -2, -1, 2, -2. 22 and 2. So here With the differences listed in the 2nd to last column of my table here, we can now calculate the mean difference or D bar. Now, D bar is going to be equal to the sum of all of the differences you previously calculated, divided by 8, because that's the total number of. Differences in this case. Because it's an average. So when you do so, right, D bar the mean difference is equal to 0.125. Which brings us now to the last column of the table. Now, for the last column of the table, what you first need to do is subtract D from each family by D bar. That represents the deviation from the mean for each pair of siblings. And then you square the deviation from the mean. So that's D subtracted by D bar squared. And when you do this, The values for the squared deviations are as follows, once again listed from in order from family 1 through 8. So that's see 3.51. 56 4. 5156. One. 2652 3.5156 4.5156. 3.5156. 4.5156. And 3.5156. And so now that you have all of the squared deviations, what you want to do is find the sum of all of your squared deviations. And so when you calculate this, your total comes out to 28.875. And now this leads you into the sample variants. Now, the sample variance is S sub D square. And this is equal to the sum of squared differences, so that's 28. 875 divided by the sample size subtracted by 1. So that's 28.875 divided by 7. And this gives you 4.125. So recall also that the square root. Of the sample variants is equal to the sample or the standard deviation. So the standard deviation is equal to the square root of 4.125. Which equals approximately 2.031. And now you can find the tea statistic. So T for our data set. Is equal to D bar. Divided by the standard deviation, divided by the square root of the sample size. So that would be equal to. 0.125. Divided by 2.031. Divided by the square root of 8. And this equals approximately 0.1741. So now Our calculated value is 0.1741. And the degrees of freedom is equal to the sample size subtracted by 1, so that's equal to 7. Now, we can check for the critical T value for a two-tailed test. And then compare this to our calculated value. So The critical value in this case, the critical T value. For a two-tailed test with alpha equals 0.05 and 7 degrees of freedom. Is equal to positive or negative 2.365. And it's at this point that we can see that our calculated value is much smaller than our critical value. Which means that we fail to reject the null hypothesis, meaning that there is no significant difference between the average heights of sibling A and sibling B, at least based on this sample. And now we've arrived at our final answer, which reads as follows. For part A, is there a significant difference between the average heights of sibling A and sibling B? For Part B, use a paired tea test or a dependent samples tea test because the heights are naturally paired, siblings from the same family. For Part C, we fail to reject the null hypothesis. There is no significant difference between the average heights of sibling A and sibling B based on the sample. And there you have it. So with that being said, thank you so very much for watching, and I hope you found this helpful.

Key Concepts

Descriptive Statistics

Hypothesis Testing

Correlation and Regression

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