Variation of Hospital Times Use the sample data given in Exerc...

Question

Variation of Hospital Times Use the sample data given in Exercise 7 “Seat Belts” and test the claim that for children hospitalized after motor vehicle crashes, the numbers of days in intensive care units for those wearing seat belts and for those not wearing seat belts have the same variation. Use a 0.05 significance level.

Accepted Answer

Hello. In this video, a researcher wants to know if the variability in test scores differs between two different teaching methods. In Method A, 13 students have a sample variance of 18.2, and in Method B, 11 students have a sample variance of 9.5. At a 0.05 significance level, we want to test the claim that the two methods have the same variances. So, let's go ahead and write down the given information in this problem. Now the claim in this problem is that the variances between the two methods are equal to each other. We'll name the variants of the first test sigma 12 and the variants of the second method sigma2 squared. And we are again, we are going to test to see if the variances are equal to each other. So we are going to suggest that the null hypothesis is going to be that the variances are equal to each other, but if the variances are not equal to each other, then that is going to be our alternate hypothesis, is that the two tests produce different vari variances between the two methods. So this is going to be our null and alternate hypothesis. Now, in the given problem, the first method has a sample size of 13 students, and the sample variance. Of the first test is going to be 18.2. And the second method, we have a sample size of 11 students and a sample variance of 9.5. So, since we are trying to test whether there's a difference between the variances of the methods, we are going to approach this problem by using the F test. Now we are going to define our F-test statistic. The F test statistic is defined as the variance of sample A. The variance of sample A divided by the variance of sample B. Now in this case here, we do have that given information that is going to be 18.2 divided by 9.5, and with the help of a calculator, this is going to give us an approximate test statistic of 1.9. Since we have our test statistic, let's go ahead and set that to the side for now. Now, the next thing we need to do is we need to determine the upper and lower critical values for the F test, but in order to do that, we need to calculate the degree of freedoms for each of the methods. Now the degree of freedom for method A is defined as the sample size of method A minus 1. That is going to be 13 minus 1, which gives us a degree of freedom of method A to be 12. The same can be said for the degree of freedom of method B. The degree of freedom of method B is going to be the sample size of method B minus 1, and that is going to be 11 minus 1, which gives us a degree of freedom of 10. So now, let's go ahead and use this to calculate the upper critical value. Now, in this case here, again, we are working with a significance level of 0.05. This means that what we need to do is we need to go ahead and divide the significance level by 2 because we are working with the upper and lower areas of the distribution curve. So the significance level divided by two is going to give us a significance level of 0.025. So, the F critical value with respect to the upper upper curve, which is the upper area of 0.025, a degree of freedom of 12. And with respect to 10 is going to be 3.621. By the F test or by the F critical chart. And we're going to repeat this process for the lower critical value. Now, the lower area with respect to our significance level is going to be an F critical value with respect to the area of 0.975, a degree of freedom of 12, and a degree of freedom of 10, and this is going to give us a lower critical value of 0.296 by the F chart. Now that we have our upper and lower critical values, we are going to compare our test statistic to the critical values. Notice that 1.9. Notice that 1.9 lies within the region between the upper critical value and the lower critical value. 0.296 is less than 1.9, which is less than 3.621. And what this tells us is that because the test statistic is between the region of the upper and lower critical values, we are going to fail to reject the null hypothesis. So what does this tell us about the original claim? Well, what this tells us is that there is insufficient evidence to say that the variances are different. And with that being said, the solution to this problem is going to be a. So I hope this video helps you in understanding how to approach this problem, and we will go ahead and see you all in the next video.

Key Concepts

Variance and Variation

F-Test for Equality of Variances

Significance Level and Hypothesis Testing

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