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)

Magnet Treatment of Pain People spend around $5 billion annually for the purchase of magnets used to treat a wide variety of pains. Researchers conducted a study to determine whether magnets are effective in treating back pain. Pain was measured using the visual analog scale, and the results given below are among the results obtained in the study (based on data from “Bipolar Permanent Magnets for the Treatment of Chronic Lower Back Pain: A Pilot Study,” by Collacott, Zimmerman, White, and Rindone, Journal of the American Medical Association, Vol. 283, No. 10). Higher scores correspond to greater pain levels.

a. Use a 0.05 significance level to test the claim that those treated with magnets have a greater mean reduction in pain than those given a sham treatment (similar to a placebo).

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Accepted Answer

Welcome back, everyone. This problem, 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, and the data are as follows. For the new drug group, the sample size was 22, the mean reduction was 15.6, and the standard deviation was 4.2. While for the placebo group, the sample size was 22, the mean reduction was 414.9, and the standard deviation was 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. At the 0.0 0.05 significance level, test the claim that the new drug leads to a greater mean reduction in cholesterol than the placebo. Now what are we doing here? Well, we are testing if the new drugs leads to a greater mean reduction, OK? So in that case, another hypothesis or default assumption is that the new drug is not more effective. In other words, the mean reduction time for the new drug is less than the mean reduction time for the placebo. While our alternative hypothesis then is that the new drug is more effective. In other words, mu1 is greater than mu 2. Now if we view the new drug as sample one and the placebo group as sample two, OK, then for sample one. OK. For a sample one. Then our. Mean reduction in cholesterols or X bar 1 would be 15.6. Our standard deviation S1 would be 4.2 and our sample size N1 is 22. On the other hand, for the placebo group, OK, then we know that X bar 2 will be 14.9. S2 is 5. And N2 is 22. Now, if we're going to test this claim, then we can use the two sample T test, OK. Where we're going to use a one-tailed test. OK, a one-tailed right or a right one tail test. In other words, like we said, since we know that the populations are normally distributed, then if we can find our critical T value here, OK, and our test statistic is greater than the critical T value, then we would reject the null hypothesis. But if it's less than the critical value, then we don't have enough evidence to support this claim. don't have enough evidence to reject that non hypothesis. So let's find the T value from the test statistic and then find a critical T value and then make a decision. So, since we're using the two sample T test, OK. We know that the test statistic T is gonna be equal to the difference between the sample means, OK. Minus the difference between the population means where it's often assumed to be zero. Divided by the square root, OK, of S1 square divided by N1 plus S2 quad divided by N2. When we substitute what we know, that's going to be 15.6 minus 14.9, OK. Minus 0 divided by the square root, OK, of 4.2 quad divided by 22 plus 5 2 divided by 22. Now let me go ahead and solve here. Then we should get our test statistic to be approximately 0.503. So now we have our test statistic. Next we need to find the critical value. So here, let me write write a critical value to the right, OK. For a one-tailed test. For a one-tailed test. OK. Add a significance level alpha of 0.05. And a degree of freedom to be the smaller of N1 minus 1 and N2 minus 1, where N1 and N2 are 22 and 22 minus 1 is 21. So that means we're working with 21 degrees of freedom. Then if we can, if we, if we look at our tea table, then the critical T value that we'd find is that it would be approximately equal to 1.71. Now that we have a T value from the test statistic and the critical T value, what do you notice? Well, since T is less than the critical value, 0.503 is less than 1.721, OK, then we fail to reject, we fail to reject the null hypothesis. So what's our conclusion? Well, if I were to write it out here, find it a little easier to write, OK. Then we should find that at the 0.05 significance level, OK, there is not enough evidence to support the claim. That the new drug leads to a greater mean reduction in cholesterol. OK, than the placebo. In other words, we don't have enough evidence to support our claim. Thanks, oops, as I wrote clam, but this should actually be clear, my apologies here, clear. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Hypothesis Testing

Independent Samples

Significance Level

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