"Performing a Two-Sample F-Test In Exercises 19–26, (a) identi...

Question

"Performing a Two-Sample F-Test In Exercises 19–26, (a) identify the claim and state H0 and Ha, (b) find the critical value and identify the rejection region, (c) find the test statistic F, (d) decide whether to reject or fail to reject the null hypothesis, and (e) interpret the decision in the context of the original claim. Assume the samples are random and independent, and the populations are normally distributed.

Carbon Monoxide Emissions An automobile manufacturer claims that the variance of the carbon monoxide emissions for a make and model of one of its vehicles is less than the variance of the carbon monoxide emissions for a top competitor’s equivalent vehicle. A sample of the carbon monoxide emissions of 19 of the manufacturer’s specified vehicles has a variance of 0.008. A sample of the carbon monoxide emissions of 21 of its competitor’s equivalent vehicles has a variance of 0.045. At α=0.10, can you support the manufacturer’s claim? (Adapted from U.S. Environmental Protection Agency)"

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 university administrator claims that the variance in student satisfaction scores for online courses is less than the variance for in-person courses. A random sample of 22 online course students has a satisfaction score variance of 1.5. A random sample of 16 in-person course students has a variance of 3.2. At alpha equals 0.01, is there enough evidence to support the administrator's claim? Awesome. So it appears for this particular problem, we're asked to take a specific level of significance alpha of 0.01, and we're asked to determine whether or not there is enough evidence to support the administrator's claim. And once again, the administrator is claiming that the variance in student satisfaction scores for online courses is less than the variance for in-person courses. So now that we know what we're trying to solve for, let's read off our multiple choice answers to see what our final answer might be. A is fail to reject the null hypothesis. There is not enough evidence to support the claim. B is reject the null hypothesis. There is not enough evidence to support the claim. C is fail to reject the null hypothesis. There is enough evidence to support the claim. And finally, D is reject the null hypothesis. There is enough evidence to support the claim. Awesome. So let's quickly note that the null hypothesis is denoted as H0. So with that in mind, our first step that we need to take to solve this problem is we need to note once again, the claim is that the variance for online courses is less than for in-person courses, so we could write this as sigma 1 squared is less than sigma 2 squad, which will mean we could write our null hypothesis as Sigma 1 squad is equal to sigma 22, whereas our alternative hypothesis, which will denote as HA will state that sigma 1 squared is less than sigma 2 squared. Fantastic. So moving right along here. Our second step now is we need to determine the degrees of freedom. And we need to determine the degrees of freedom for both. Online course students and in-person course students. So DF1 is equal to 22 minus 1, which is equal to 21, and DF2 is equal to 16 minus 1, which is equal to 15. And we need to recall that for a left tail test, which you'll call LT, so for a left tail test, At a level of significance alpha is equal to 0.01, we need to use F subscript 0.012115. So with that in mind, we need to find F subscript 0.9915,21. And when we use our F tables, we will find that F subscript 0.99.15.21 will be equal to 2.82. And once again, this is when we use our F table, and you should have an F table in your textbook. If not, you can quickly look one up online. So with that in mind, we can now solve for the critical value, which I'll denote as CV and our critical value is equal to 1 divided by 2.82 and 1 divided by 2.82 will be equal to 0.355 will be round to three decimal places. Now we need to focus our attention on solving for the rejection region, so we need to recall that we can reject the null hypothesis if, and let's write this in blue, if F is less than 0.355. So we can reject the null hypothesis if F is less than 0.355. So with that in mind, our third step is we need to note that F is equal to 1.5 divided by 3.2, which is equal to 0.469. And that will mean moving right along here, our fourth step we need to perform our comparison and since 0.469 is greater than 0.355, we do not reject the null hypothesis, so I'll take H0 and put an X through it. So once again, because 0.469 is. Greater than 0.355, we do not reject the null hypothesis, and because we do not reject the null hypothesis, that will mean as a result, there is not enough evidence at alpha equals 0.01 to support the administrator's claim. So that will mean our final answer, without a doubt, will have to be multiple choice answer letter A, which once again is fail to reject the null hypothesis, there is not enough evidence to support the claim, 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.

Key Concepts

Hypothesis Testing

F-Test for Variances

Critical Value and Rejection Region

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