Performing a One-Way ANOVA Test In Exercises ...

Question

Performing a One-Way ANOVA Test In Exercises 5–14, (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, the populations are normally distributed, and the population variances are equal.

[APPLET] Statistician Salaries The table shows the salaries of a sample of entry level statisticians from six large metropolitan areas. At α=0.05, can you conclude that the mean salary is different in at least one of the areas? (Adapted from Salary.com)

[APPLET] Well-Being Index The well-being index is a way to measure how people are faring physically, emotionally, socially, and professionally, as well as to rate the overall quality of their lives and their outlooks for the future. The table shows the well-being index scores for a sample of states from four regions of the United States. At α=0.10, can you reject the claim that the mean score is the same for all regions? (Adapted from Gallup and Healthways)

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 researcher wants to determine whether the average number of study hours per week differs based on students' primary study environment. Students report their weekly study hours while studying in one of four primary environments library, home, coffee shop, or campus lounge. The researcher performs a one-way ANOVA using the collected data and obtains the following results. Mean square between groups MSB 16.0992. Mean square within groups, MSW 0.4193 degrees of freedom between groups, 3 degrees of freedom within groups, 16. At the alpha equals 0.05 level of significance, can you conclude that the mean number of study hours differs by primary study environment? Awesome. So it appears for this particular problem, given a specific level of significance of 0.05, we're asked whether or not we can conclude that the mean number of study hours differs by primary study environment. So now that we know what we're ultimately trying to solve for, let's read off our multiple choice answers to see what our final answer might be. A is there is not enough evidence to conclude that study hours differ across environments. B is the mean study hours are equal across all four environments. C is there is strong evidence that at least one environment has a different average study time, and finally, D is the test fails because the sample size is too small. OK. So our first step that we need to take in order to solve this problem is we need to state our hypotheses. So we need to state our null and our alternative hypotheses. So let us quickly recall that we want to know the means differ between groups, and we want to know if the means differ between groups. So for our null hypothesis, which will denote as H0, we need to note that the mean of the library, which I'll call mu L, is equal to the mean of home, which I'll call muh. Which is equal to mu of the coffee shop, which I'll call M like M C. is equal to new campus lounge, which I'll call CL, so new CL. So once again, our no hypothesis states that the mean for the library is equal to the mean for the home, which is equal to the mean of the coffee shop, which is equal to mean of the of the compass lounge. So, with that in mind, all group means are equal to each other, whereas for our alternative hypothesis, HA. We need to recall a note and be able to recognize that at least one group mean will be different. So with that in mind. Our second step now is we need to compute the F statistic, which, as we should recall, the F statistic is equal to MSB divided by MSW, which is equal to when we plug in our known variables, 16.0992, divided by 0.4193, so 0.4193. And when we perform that calculation in our calculator, we will find that it's equal to 38.400. Fantastic. When we round to 3 decimal places. Fantastic. So moving right along here, our third step now is we need to compare with the critical F value. So we need to recall how to use an F table, which you should have an F table in your textbook, or you could use a calculator, but we're told that alpha, or level of significance, is equal to 0.0. 5 and we're also told that our degrees of freedom 1 is equal to 3 and our degrees of freedom 2 is equal to 16. So that will mean that our critical F value, which I'll call FC, will be equal to once again using our F table or our calculator, it will be equal to 3.2389. Fantastic. Or you can also quickly look up an F table online too. That's always an option. Our fourth step now is we need to make our final decision. So F is equal to 38.400, and this is greater than our F critical value, which is equal to 3.2389. And because 38.400 is greater than 3.2389. As a result, because once again 38.400 is greater than our F critical value, we can therefore, as a result, reject the null hypothesis. So I'll put a red X next to H0. We can reject our null hypothesis, and with that in mind, we can make the final conclusion, so we can in conclusion, state. That at the 5% significance level, we have sufficient evidence to conclude that the average study hours per week differ on or may differ, so once again the, the study hours per week differ based on the study environment. So looking at our multiple choice answers, our final answer without a doubt will have to be multiple choice answer letter C, which once again is, there is strong evidence that at least one environment has a different average study time. 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

One-Way ANOVA Test

Hypothesis Testing and Rejection Region

F-Statistic and Critical Value in ANOVA

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