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).

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Accepted Answer

Welcome back, everyone. A pharmaceutical company wants to determine if their new drug reduces cholesterol more effectively than a placebo. The reduction in cholesterol in milligrams per deciliter was measured after 3 months. The data are as follows for the new drug group, the sample size is 22. The mean reduction in cholesterol is 15.6, and the standard deviation is 4.2. While for the placebo group, the sample size is 22, the mean reduction in cholesterol is 14.9, and the standard deviation is 5. 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 95% confidence interval for the difference in mean reduction in cholesterol, where the difference is defined as the mean for the new drug group minus the mean for the placebo group. Now, how can we construct this 95% confidence interval for or different? Well, notice here that we have independent samples and unequal variances. So the formula for the confidence interval that we will use is the formula that says the difference between the sample means X 1 minus X2. Is going to add or subtract T multiplied by the square root of S1 S1 square minus divided by N1 plus S2 2 divided by N2 to get the confidence interval, OK? So this is how we can find it. Now what do we already know that can help us? Well, if we consider group one the new drug and group 2 the placebo, then X1 X bar 1 will be equal to 15.6. S1 will be equal to 4.2 and N1 will be equal to 22, while for the placebo group, X bar 2 will be 14.9. S2 is 5 and N2 is 22. So we have almost all the information that we need to construct our confidence interval. However, we need the value of T. That is, we need our critical T value for a 95% confidence interval. Let's try to get that from a tea table. Now 4, OK. For a 95%. Confidence interval. OK, in a two-tailed test. Yeah, cuz notice we're testing if there's a difference, so either there is a difference or there is no difference. Then the critical T value for half of that significance level, OK? Because it's a two-tail test and for the degrees of freedom, which is the smaller of N11 and N2 minus 1, both N1 and N2 are 22, so we're going to use 21 degrees of freedom. Then. We're going to be finding the criticality value for 0.025 and 21, which is equal to 2.080. So now that we have our T value, then we can substitute it into our confidence interval formula. No, that means then we're gonna have X X 1 15.6 minus X2, 14.9. Plus R minus T 2.080 multiplied by the square root, OK. Of 4.2 squared divided by 22 plus 5 square divided by 22. OK? So when we simplify that, what will it be? What do we uh end up with? Well, here 15.6 minus 14.9 is 0.7. And if we work out what's under our square root, OK, then here we'll get the square root of 42.64 divided by 22, which is approximately 1.392. OK? And now, if we expand this here, then our lower limit, OK? The lower limit for our interval is going to be 0.7. Minus uh our product here, which I should have, I should have probably written here that this would be 0.7 plus or minus 2.9. So it's 0.7 minus 2.9 for a lower, lower limit which is negative 2.2. And for our upper limit. OK, it's gonna be 0.7 plus 2.9, which is 3.6. So in context, what does this mean? Well, this implies then that the 95% confidence interval and the 95% confidence interval. For the difference in mean reduction in cholesterol is. From -2.2. milligrams per deciliter to 3.6 mg per deciliter. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Confidence Interval

Independent Samples

Hypothesis Testing

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