Clancy, Rowling, and Tolstoy Ease of Reading Pages were random...

Question

Clancy, Rowling, and Tolstoy Ease of Reading Pages were randomly selected from three books: The Bear and the Dragon by Tom Clancy, Harry Potter and the Sorcerer’s Stone by J. K. Rowling, and War and Peace by Leo Tolstoy. Listed below are Flesch Reading Ease Scores for those pages. Higher scores correspond to pages that are easier to read. Use a 0.01 significance level to test the claim that pages from books by those three authors have the same median Flesch Reading Ease score.

Table showing Flesch Reading Ease scores for pages from Clancy, Rowling, and Tolstoy books.

Accepted Answer

All right, hello, everyone. So this question says, a study compared the reaction times in seconds of drivers using 3 different types of driving simulators, basic, intermediate, and advanced. Each driver completed a reaction test under similar conditions, and the reaction times are as follows. At the 0.01 significance level, use the Crus call Wallace test to test the claim that the median driver reaction time is the same across all three simulators. So first, let's take a moment to state the hypothesis for this problem, right? The no hypothesis, if you recall, would state that all groups, that is all of the simulators, would have the same. Median reaction time. The alternative hypothesis, therefore. I would say that at least one of the groups is different. And what I mean by that is that at least one of the groups has a different median reaction time compared to the others. So here I went ahead and made a table. Of the values of our data set, because our task right now is to rank them in order from smallest to largest, or smallest to greatest, and then calculate the average rank as needed. So for example, the smallest value in the data set, 2.1. Would take rank number one, not only because it's the smallest value, but because it happens to appear only once in the data set. So because it appears only once, the average rank would also be one. Now, 2.2, on the other hand, actually appears 3 times in the data set, which means that it takes ranks 23, and 4. And so the average rank would be the sum of 23, and 4 divided by 3, right? That's the average. And that gives you 3 as the average rate. 2.3 would then take ranks 5 and 6, giving you an average of 5.5. 2.4 would take ranks 7 and 8 with an average of 7.5. And 2.5 would take ranks 9 and 10. With an average rank of 9.5. Now 2.6 would take ranks 11 and 12. Giving you an average rank of 11.5. And 2.7, 2.8, and 2.9 all appear only once in the data set. So they take ranks 1314, and 15 respectively. Which would also be equal to their average rank. So from here Our task is to group the ranks by the. Group that they correspond to, and then find the sum of all of the ranks for each category. So for the basic simulator, for example, the corresponding ranks of its data values would be 5.5, 11.5, 7.5. 9.5 and 3. With the sum being 37. For intermediate, that would be 1 5.5. 33 again. And 7.5. With a sum of 20. And then for advanced, that would be 13. 9.5. 11.5 14 and 15. With a sum of 63. And so with this information in mind, right, we can now move on to using the Crusco Wallace formula. Which states that H is equal to 12 divided by N. Multiplied by capital and added to 1 multiplied by the sum of all rank sums squared. And divided by lower case of one. Subtracted by 3 multiplied by capital N added to 1. So, for this context, capital N is equal to 15, and N 1 is equal to 5 for each group, because that's the number of ranks that correspond to each category. So plugging in the information given, H is equal to 12. Multiplied by 15, multiplied by 16. Multiplied by the sum of 37 squared. Divided by 5 20 square divided by 5. And 63 square divided by 5. Subtracted by 3. Multiplied by 16. And so evaluating this expression. Equals approximately 9.38. Which you would then compare to the critical value. Now to find the critical value, you would need to find the degrees of freedom. And the degrees of freedom in this case. Which, if you recall equals K subtracted by 1, would equal 2. And it was already given to us that the chi squared value, or rather that alpha, is equal to 0.01. And so with this information in mind. The critical chi squared value at alpha equals 0.01 is 9.210. So now you compare. Apologies for that. I think I, I think I hovered over one of my desktop icons for too long. But in any case, right, at this point we're going to compare our calculated value to our critical value. And because the calculated value of 9.38 is greater than the critical value of 9.210, we can reject the null hypothesis. So now this leads us to our final answer, which reads as follows. We reject the null hypothesis. There is a significant difference in median reaction times among the simulators. 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

Flesch Reading Ease Score

Median

Hypothesis Testing

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