In Exercises 5–20, conduct the hypothesis test and provide the...

Question

In Exercises 5–20, conduct the hypothesis test and provide the test statistic and the P-value and/or critical value, and state the conclusion.

Flat Tire and Missed Class A classic story involves four carpooling students who missed a test and gave as an excuse a flat tire. On the makeup test, the instructor asked the students to identify the particular tire that went flat. If they really didn’t have a flat tire, would they be able to identify the same tire? The author asked 41 other students to identify the tire they would select. The results are listed in the following table (except for one student who selected the spare). Use a 0.05 significance level to test the author’s claim that the results fit a uniform distribution. What does the result suggest about the likelihood of four students identifying the same tire when they really didn’t have a flat?

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Accepted Answer

All right, hello everyone. So this question says, a teacher suspects that students randomly guess when asked to select one of 4 doors as the exit in a psychology experiment. Out of 44 students, the choices were as follows. 10 selected door A, 12 selected door B, 9 selected door C, and 13 selected door D. At the 0.05 significance level, test whether the selections are uniformly distributed. What is the test statistic and what conclusion should be drawn? All right, so first, let's go ahead and distinguish between our two different hypotheses, right? We have the null hypothesis and the alternative. Now the null hypothesis would state that the selections here are normally distributed. Which means that each door is equally likely to be selected. So here the alternative hypothesis is that that is not the case and that the selections are not normally distributed instead. So first and foremost, what is the expected frequency if. This is indeed normally distributed. Well, here we have a total of 44 students selecting one of 4 doors. So the expected frequency is equal to 44 divided by 4, which gives you 11. And we can compare this with the observed frequencies that were already given. And it's from here that we can recall the formula for the chi square test statistic. So here chi squared. Is equal to the sum of. The observed frequency. Subtracted by the expected frequency. Divided by the expected frequency. Now, the importance of the sum here states that for each data point for each. Of the doors listed, you're going to subtract the observed frequency from the expected frequency, and then divide that by the expected frequency. Then you add up those numbers together to include in the chi square formula. But actually, I need to correct myself very quickly because it's not just the sum of. The differences between observed and expected. It's the sum of the difference between the observed and expected frequencies squared, divided by the expected frequency. And so once you evaluate the expression, once you determine. The difference between observed and expected for each value, square it and then divided by. The expected frequency, the total, which equals your chi squared comes out to 0.91. And from here, you have to determine the critical value. So, for the critical value, recall that you need the degrees of freedom. In this case, the degrees of freedom. Would be equal to. 4, which is the number of doors in the experiment subtracted by 1, which gives you 3. So, at alpha equals 0.05, the critical value from the chi square table is approximately 7.815. And now, the final step is to compare the test statistic to the critical value. And notice how the test statistic that's 0.91 is less than our critical value of 7.815. Because of that, we failed to reject the null hypothesis. So now this brings us to our final answer, which reads as follows. First, chi squared is equal to 0.910. And because of that, there is insufficient evidence to support the claim. That the selections are not uniformly distributed. 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

Hypothesis Testing

Uniform Distribution

P-value

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