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 a 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

Welcome back, everyone. In this problem, a nutritionist wants to compare the effects of two different diets and weight loss. After 8 weeks, the following statistics are recorded the sample size, the mean weight loss, and the standard deviation for group with diet X and the other with diet Y. 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 weight loss where the difference is defined as the mean for the diet X group minus the mean for the diet Y group. Now what do we know that will help us to construct our 95% confidence interval for this difference? Well, recall that for the confidence interval for the difference in means for independent samples with unequal variances, then we can use the formula that says the difference in means X1 minus X2 X 1 minus X2 is plus R minus the criticalt value T multiplied by the standard error, OK, where the standard error. Of the difference is equal to the square root of S1 2 divided by N1 plus S2 2 divided by N2, OK? So now, in this case, as you can notice, first we'll need to find our difference between our means, or critical test value, T or critical T value, and the standard error of the difference. Now, for starters, OK. We know, given from our information, let's just make a note of that. We know that from diet X, if we consider it the first group, X1 equals 8.2. This S1 equals 2.4. Where S represents the standard deviation of the sample and N1 equals 28. On the other hand, from our table, OK, we know that ford Y, the second group, X2 equals 6.5, S2 equals 2.9, and N2 equals 30. So now we can use that to figure out our values because no this implies that the difference in sample means is going to be equal to 8.2 minus 6.5, which equals 1.7. On the other hand, for our standard error of the difference, or sorry, for our uh T critical T value, OK. For our criticalte value, we can find it using our tea tables, OK? So now remember that here we're talking about a two-tailed test, OK. We're talking about a two-tailed test where. Our confidence level, since this is a 95% confidence interval, then our confidence level is 5% or 0.5, OK? And our degrees of freedom. Is equal to this, let me write that on the next line. The degrees of freedom is equal to the smaller of the sample size is -1, so it's the smaller of 28 minus 1 and 30 minus 1, which equals the smaller value from 27 and 29, which is 27. OK. Then. The critical T value, which here is gonna be alpha divided by 20.025 since it's a two-tailed test, at 27 degrees of freedom is going to be equal to 2.052. So now we have our difference and we have our criticality value. The only thing we need to find no is our standard error of the difference. Now remember here in our formula if you were to plug in the values, then that means it's gonna be the square root of S1 2.4 squared divided by 28 plus S2 2.9 squared divided by 30, and in this case we should get it to be approximately equal to 0.697. Know that we have the three parts to our formula for the confidence interval, we can substitute and solve because now by constructing the confidence interval, OK, then it means that it will be 1.7 plus R minus T, which is 2.052 multiplied by a standard error which is 0.697. It's going to be 1.7 plus or minus 1.4. In that case, OK, this means then that the lower limit. It's going to be 1.7 minus 1.4, which equals 0.3. And upper limit. Is going to be 1.7 plus 1.4, which equals 3.1. Thus that tells us that the 95% confidence interval for the difference in mean weight loss is 0.3 kg to 3.1 kg. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Confidence Interval

Independent Samples

Hypothesis Testing

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