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.

b. Construct the confidence interval suitable for testing the claim in part (a).

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 1 has the following post-program stress scores 32, 28, 35, 30, 27, 29, 31, 33, and 25. Group Y using program 2 has the scores 25, 22, 28, 24, 27, 23, 22, 26, 29, and 24. Assume independent samples from normal populations with unequal variances. Use degrees of freedom equal to the smaller of n x minus 1 and ends of Y minus 1. Construct a 95% confidence interval for the difference in mean post-program stress scores, where the difference is defined as the mean for Group X minus the mean for group Y. We're also getting four possible choices as our answers. For choice A, we have the interval from 2.0 to 7.5. For choice B, we have the interval from 1.5 to 7.5. For choice C, we have the interval from 1.5 to 8.0. And for choice D, we have the interval of 2.0 to 8.0. So we're asked to construct a 95% confidence and for the difference in the mean post-program stress scores. So we need to recall our formula for the confidence interval for the difference in means using independent samples and unequal variances. So, for our formula here, we have the quantity of X bar minus Y bar. Plus or minus T multiplied by the square root. Of the quantity of S X squared divided by N X plus S Y 2 divided by NY in quantity. Where X and Y are the sample means for groups X and Y respectively, T is our critical value. S X2 and SY2d are the variants for Group X and Group Y, respectively, and NX and Ny are the number of data points again in Group X and Y, respectively. So, in order to use this formula, we'll have to calculate the mean and variance for each of our groups. And so, this is best done in a table. And so On the screen here, we have created 2 tables. The first table is for Group X. And in the first column in this table, we have just labeled our data points. Um, with the values of 1 through 9. And so that's gonna be labeled as I. And so, when we look at the number of data points that we have, we have N X is equal to 9, and so N X minus 1, which we'll need later, it's just gonna be 9 minus 1, which is going to be equal to 8. The next column is our actual data points. And that is labeled X of I. And for X of I, we have the values of 32, 28, 35, 30, 27, 29, 31, 33, and 25. And so we need to use these values to calculate our mean X bar. So the first step is to sum all of our values Xi, and in doing so, we get that that is equal to 270. And so that means that X bar is going to be equal to the sum of X of by divided by NX, which is going to be equal to 270 divided by 9, which is equal to 30. In the next column, we have X by minus X bar. And if we calculate this for each of our data points, we get the values of 2 minus 250, minus 3, -1, 13, and minus 5. And in our final column, we calculate the quantity of X of I minus X bar in quantity squared. And so we're just squaring the previous column, and we get the values of 44, 25, 09, 11, 9, and 25. And so we need to calculate our variants. And so the first step is to sum the quantity of X minus X bar in quantity squared. So we're just summing that last column, and in doing so, we get that this is going to be equal to 78. And so our variance S X 2 is equal to the sum of the quantity of X I minus X in quantity squad, divided by the quantity of NX minus 1 in quantity, and this is going to be equal to 78 divided by 8. And when we evaluate the expression, that's going to be equal to 9.75. We're gonna do the same thing with group Y. And so in our table in our first column is labeled as I. So here we've just numbered our data points from 1 to 10, since we have 10 data points. So that means that N Y is equal to 10, and NY minus 1 is going to be 10 minus 1, which is equal to 9. In the next column, we just have our data points Y sub I, and we have the values of 25, 22, 28, 24, 27, 23, 22, 26, 29, and 24. And so calculator mean, the first step is to sum all of our data points. We'll have the sum of Y subby. In doing so, that's going to be equal to 250. And so that means that Y bar, our mean, is going to be equal to the sum of Y sub I divided by NY, and that's going to be equal to 250 divided by 10, which is equal to 25. The next column, we have the quantity of Yvi minus Y bar. And if we calculate this for each data point, we get the values of 0, minus 33122314, and minus 1. And in our final column, we have the quantity of Y is the body minus Y bar in quantity squared. So we're just squaring our previous column and we get the values of 0991449, 1, 16, and 1. So again, we need to calculate our variants. So the first step is to sum the quantity of Y minus Y in quantity squad. So summing that last column, and this is going to be equal to 54. So our variance Sysquad. is going to be equal to the sum of the quantity of Yi minus Y in quantity square divided by the quantity of N Y minus 1 in quantity, and this is going to be equal to 54 divided by 9, which is equal to 6. So our next step is to find our critical value T star. And for this, we'll need to determine the number of degrees of freedom. So we'll label that as DF. And this is going to be equal to the minimum. Of N X minus 1, or N Y minus 1. And so this is going to be equal to the minimum. Of 8 or 9, and the smaller of those two values is 8, so that means that the degrees of freedom is equal to 8. Next, we're gonna need our quantity of alpha divided by 2. And this is going to be equal to, and recall that alpha here. It's just gonna be 1 minus our confidence interval percentage. So we'll have 1 minus. 0.95, here we just converted the 95% into a decimal, and then that quantity is going to be divided by 2. So that means that alpha divided by 2 is going to be equal to 0.025. And so with our degrees of freedom of 8 and alpha divided by 2, equal to 0.025, we can look up in our tables and see that our critical value T is going to be equal to 2.306. So now we have all the information that we need to calculate our confidence interval at the 95% confidence level. And so, coming back to our formula, we will have the quantity of X bar, which is 30, minus Y bar, which is 25 in quantity, then plus or minus T, which is 2.306. And that gets multiplied by the square root of the quantity of S X squad, which is 9.75, and that's divided by n X, which is 9. Then we'll have plus S Y quad, which is 6 divided by N Y, which is 10 in quantity. And so evaluating this expression. For the first part, we'll have 30 minus 25, which is 5, then we'll have plus or minus, and when we evaluate the expression for the second part, this is going to be approximately equal to 3.0. And so if we want to write our our confidence info here in info notation, we will have for the lower bound, 5 minus 3, which is going to be 2.0, and for the upper bound, we'll have 5 + 3, which is going to be 8.0. So this is the confidence and that we were looking for for this problem. And when we compare that to our answer choices, we see that the correct answer choice corresponds to answer choice D. So I hope that this explanation has been useful and I'll see you in the next video.

Key Concepts

Confidence Interval

Independent Samples

Normal Distribution

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