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)

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b. Construct the confidence interval appropriate for the hypothesis test in part (a). What is it about the confidence interval that causes us to reach the same conclusion from part (a)?

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Accepted Answer

Hi everyone. Let's take a look at this practice problem. This problem says researchers are studying the effects of background music on concentration. One group of students studied with classical music while the others studied in silence. Their concentration scores were recorded and summarized as follows. And we're given a table that has 4 columns in it. The first column is the group. The second column is the sample size N. The third column is the mean XY, and the fourth column is the standard deviation S. So, our first row is for classical music. It has a sample size of 35, a mean of 7.2, and standard deviation of 1.1. And the next row is for silence, and it has a sample size of 38, a mean of 6.5, and a standard deviation of 1.4. 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. Use degrees of freedom equal to the smaller of N1 minus 1 and N2 minus 1. Construct a 99% confidence interval for the difference in mean concentration scores, where the difference is defined as the mean for the classical group minus the mean for the silence group. So for this problem, we're asked to construct a 99% confidence interval for the difference in mean concentration scores. So, we're looking for the quantity of X 1 minus X 2 in quantity where group 1 is going to be our classical music group and group 2 is going to be our silence group. And so we need to construct a confidence interval about this difference in means. So, we're gonna have that difference in means, plus or minus. Tea store. Multiplied by the square root of the quantity of S1 2 divided by N1 plus S2 2 divided by N2. Where T star here is going to be our critical value corresponding to the 99% confidence interval. Now, we were given the mean for each of our groups as well as their standard deviations and sample sizes. So, we actually have all the information to use this formula except for T. So we're actually going to have to calculate that. And the first thing we need to do is determine the number of degrees of freedom that we have. So we'll label that as DF. And we're told the problem to use the smaller of the two, which is going to be the minimum of either N1 minus 1, or N 2 minus 1. And substituting the values for N1 and in 2, we'll see that this is going to be the minimum of 35 minus 1. Or 38 minus 1. Which is going to be equal to the minimum. Of 34 or 37, and the smaller of those two is gonna be 34, so this means that our degrees of freedom is going to be equal to 34. So now we need to use that information to find our critical value, and here we need to determine our value of alpha divided by 2. Now alpha in this case is going to be equal to 0.01. So since we're looking for the 99% confidence interval, that means alpha is going to be equal to 0.01. What we're actually interested in is alpha divided by 2, which is going to be equal to 0.01 divided by 2, which is equal to 0.005. So that means that our critical value that we're looking for, which is going to be T star. is going to be equal to the T value. When alpha divided by 2 is equal to 0.005, and the degrees of freedom is equal to 34. And so we can look this up in a table. In doing so, we get a value of 2.728. So now we have all the information that we need to calculate our confidence interval. So substituting in the values into our confidence interval formula, we'll have the quantity of X1, which is going to be 7.2, minus X2, which is 6.5, in quantity, then we'll have plus or minus. For T, we'll have 2.7 to 8, and that gets multiplied by the square root of the quantity. For S1, we have a standard deviation of 1.1, and then that is going to be squared. And that's going to be divided by N1, which is 35. Then we'll have plus for S2, we have a value of 1.4, and that quantity is going to be squared, and that gets divided by N2, which is 38 in quantity. So, evaluating this expression, for our difference in our means, we have 0.7, then we'll have plus or minus, and when we evaluate the next expression, we get approximately 0.8. And here we've round it to just one decibel place, since that's the same number of decimal places that we have for our difference in means. So now we can use this to construct our confidence interval. So for our lower limit, we're going to have 0.7 minus 0.8, which gives me 0, 0.1. And then for our upper limit, we'll have 0.7 plus 0.8, and we'll get 1.5. So our 99% confidence interval is from -0.1 to 1.5, and so that's going to be the answer to this question. So, I hope that this explanation has been useful, and I'll see you in the next video.

Key Concepts

Confidence Interval

Hypothesis Testing

Independent Samples

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