In Exercises 5–20, assume that the two samples are independent...

Question

In Exercises 5–20, assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. (Note: Answers in Appendix D include technology answers based on Formula 9-1 along with “Table” answers based on Table A-3 with df equal to the smaller of n1-1 and n2-1)

Queues Listed on the next page are waiting times (seconds) of observed cars at a Delaware inspection station. The data from two waiting lines are real observations, and the data from the single waiting line are modeled from those real observations. These data are from Data Set 30 “Queues” in Appendix B. The data were collected by the author.

a. Use a 0.01 significance level to test the claim that cars in two queues have a mean waiting time equal to that of cars in a single queue.

Accepted Answer

Hi, everyone. Let's take a look at this practice problem. This problem says a psychologist is studying the effectiveness of two different stress reduction programs. Group X using program one has the following post-program stress scores 32, 28, 35, 30, 27, 29, 31, 33, and 25. Group Y using program 2 has the test scores of 25, 22, 28, 24, 27, 23, 22, 26, 29, and 24. Assume independent samples from normal populations with unequal variances. Use the degrees of freedom equal to the smaller of N X minus 1 and N Y minus 1. At the 0.05 significance level, test the claim that the mean stress scores are equal for both programs. So the first thing we're gonna do is write down our hypotheses. So, our null hypothesis, H knot is going to be that U X is equal to U Y. Where muse of X is going to be the mean for Group X and muse of Y is going to be the mean for Group Y. And our alternative hypothesis, H1, is going to be that muse of X is not equal to muse of Y. So, we're going to be using a two sample T test to test our hypothesis here. And for that, we're going to need the mean and variants for both Group X and Group Y's data. So we're going to calculate those first, and that is actually best done in a table. And on the screen, we have a table for both Group X and Group Y. So, for Group X in our table, in the first column, we have just numbered our points, going from 1 to 9, and we leveled that as I. And that means that Nex is going to be equal to 9, since we have 9 data points, and later on, we're going to need the quantity of Nex minus 1, which is going to be 9 minus 1, which is equal to 8. The next column is X a buy, and so these are just our data points. And so we have the values of 32, 28, 35, 30, 27, 29, 31, 33, and 25. And so we'll need to calculate our average. So the first thing we'll need to do is add up all of our data points, so the sum of X of I is going to be equal to. 270. And so our mean X bar is going to be equal to the sum of X by divided by n x, which is going to be equal to 270 divided by 9, which is equal to 30. In the next column, we have the quantity of X minus X bar. And so calculating that for each of our data points, we get 2, -2, 50, -3, 113, and minus 5. And in the final column, we have the quantity of X of I minus X bar in quantity squared. So we're just squaring the previous column and doing so, we get for our values of 44 250911, 9 and 25. So the first step in calculating our variance is to sum the quantity of X by minus X in quantity squared, and when we add up all those values in the final column there, we get 78. And so our variances of x 2 is going to be equal to the sum of the quantity of X of minus X in quantity squad divided by the quantity of n x minus 1 in quantity, which is equal to 78 divided by 8, which when we calculate that is going to be equal to 9.75. Now we're going to have to do the same thing with our data for Group Y. So again, we'll use a table, and in the first column we'll just label our points with I and that value goes from 1 to 10. So that means that N Y is equal to 10 and N Y minus 1 is going to be equal to 10 minus 1, which is equal to 9. In our next column, we have Y sub I, so these are just our data points for Group Y, and those values are 25, 22, 28, 24, 27, 23, 22, 26, 29, and 24. And so again, we need to calculate the mean. And so the first step is to sum Y sub I, and when we add up all those quantities, that's going to be equal to 250. And so Y bar, which is our mean, is going to be equal to the sum of Y sub I divided by N Y, which is equal to 250 divided by 10, which is equal to 25. Now the next column is going to be Y I minus Y bar, and so calculate that quantity for each data point, we will get the values of 0, minus 33122314, and minus 1. And the final column we'll have the quantity of Yi minus Y bar in quantity squared. So we're just squaring the previous column and we get the values of 09914491, 16, and 1. And so now we need to calculate our variants. So the first step is to sum the quantity of Y minus Y in quantity squad. In doing so, we'll just add up all the values in that final column, and this is going to be equal to 54. And so our variance, sorry S Y squared, that's going to be equal to the sum of the quantity of Y minus Y bar in quantity squad, divided by the quantity of N Y minus 1 in quantity, and that's going to be equal to 54 divided by 9, which is equal to 6. So now we actually need to calculate our test statistic. And for this, we're again using a 2 sample T test. And so recall your formula 4 or 2 sample T test, that's gonna be T, it's going to be equal to the quantity of X minus Y bar. In quantity minus the quantity of use of X minus use of Y. In quantity, and all of that is divided by the square root of the quantity of S X squared, divided by N X. Plussy squared divided by n y. And so substituting in our values for each of those quantities that we just calculated, this is going to be equal to. The quantity of for X bar we have 30. We have minus Y bar, which is 25 in quantity minus and here we're testing the hypothesis that muse of X is equal to mu of Y, so the quantity of mu x minus muse of Y is going to be equal to 0. And then all that is divided by the square root. Of the quantity of S X squared, which is 9.75, that is divided by NX which is 9, then we'll have plus S Y 2, which is 6, and that is divided by Nuby which is 10. And so, evaluating this expression, this is going to be approximately equal to. 3.854 So next we need to calculate our critical value. And for that, we'll first need to determine the degrees of freedom. So we'll label that as DF and this is going to be equal to the smaller of the two, so we'll have the minimum of N X minus 1 or N Y minus 1. And this is going to be equal to the minimum. Of 8 or 9, which the smaller those two values is 8, so that means there are degrees of freedom is going to be equal to 8. So, if we look up in our table for a two-tailed test with alpha equal to 0.05, since that is the significance level that we're given to the problem, and with degrees of freedom of 8. We will see that our critical value. is equal to. 2. 306. So now when we compare our critical value to our T value that we calculated, we see that the 3.8 by 4 that we calculated is greater than our critical value of 2.306. So that means that we are in the critical area, and so that means that we need to reject our null hypothesis. So we will reject. Each night And what that means is, That at a 0.05 significance level, there is sufficient evidence to conclude that the mean stress scores are different for the two programs. So that's going to be the answer for this problem. So I hope that this explanation has been useful, and I'll see you in the next video.

Key Concepts

Independent Samples

Hypothesis Testing

Significance Level

Watch next