Testing the Difference Between Two Means In Exercises 15–24, (...

Question

Testing the Difference Between Two Means In Exercises 15–24, (a) identify the claim and state Ho and Ha, (b) find the critical value(s) and identify the rejection region(s), (c) find the standardized test statistic z, (d) decide whether to reject or fail to reject the null hypothesis, and (e) interpret the decision in the context of the original claim. Assume the samples are random and independent, and the populations are normally distributed.

Repair Costs: Washing Machines You want to buy a washing machine, and a salesperson tells you that the mean repair costs for Model A and Model B are equal. You research the repair costs. The mean repair cost of 24 Model A washing machines is $208. Assume the population standard deviation is $18. The mean repair cost of 26 Model B washing machines is $221. Assume the population standard deviation is $22. At α=0.01, can you reject the salesperson’s claim?

Accepted Answer

Welcome back, everyone. In this problem, a researcher claims that the average test scores for students taught with method X and Y are the same. A sample of 30 students taught with method X has a mean score of 78, with a population standard deviation of 9, and a sample of 28 students taught with method Y has a mean score of 82 with a population standard deviation of 11. At a significant level alpha of 0.05, can you reject the researcher's claim? A says yes and B says no. Now how can we figure out if we can reject this or reject, sorry, this researcher's claim. Well, the claim is that the average test scores are the same, OK? So. What we're trying to figure here is that if we let in. M X Be the mean score for Method X. OK. And me why. Be the mean score for method Y. OK. Then Or another hypothesis is that. Mu X is equal to muy, that is both scores are the same and that matches the claim. while the alternative hypothesis, OK, let me write that properly here, alternative hypothesis HA is that mu x is not equal to muy, OK. This means that we would use a two-tailed Z test for the difference between means and we're sure of that because the population standard deviations are known. So if we think about it here. Given, let's define our variables. Given we know that for our for method X. The sample mean is 78, the sample size is 30. And our standard deviation is 9. OK. Well, for method Y, the samples mean is 82. The sample size is 28 and the population standard deviation is 11, then we can use this to figure out our Z value for our test, for our hypothesis test. But let's just make a note of the decision rule so we know exactly what we need. So based on the information we have using this two-tailed uh Z test, our decision rule is basically going to be then. That if the calculated Z value, or if the absolute value of our calculated Z value is greater than the critical Z value, OK, then we can reject the null hypothesis. But if it is not greater that it is less or that is it's less than or equal to, then we would fail to reject the null hypothesis. So, this tells us then that we will need our calculated Z value and our critical Z value for our two-tail test at a 0.05 significance level so that we can do the comparison. So let's see if we can do that. First, let's calculate the test statistic or the Z score. Recall that the Z square is going to be equal to the difference between our sample means, OK. Minus the difference between the population means. Divided by the square root, OK of Sigma X squared. OK, divided by N X. Plus sigma y2 divided by NY. And now we can substitute those values. Like we said, the difference between our sample means will be 78 minus 82. Uh, for a population means under the null hypothesis since mu xeequals muy, then their difference would be zero. That's their hypothesized difference. I know we'll be dividing that by the square root of 9 square divided by 30 plus 11 2 divided by 28. And when we calculate here, we should get this to be approximately -1.509. So this is our Z test statistic. Now what about our critical value? Well, for a two-tailed test. A two-tailed Z test. With our significance level alpha equals 0.05, then the critical Z value, given that we split alpha into two tails, is going to be equal to plus R minus 1.96. So, in this case, now we can make our decision. Our test statistic is -1.509. Our critical value is critical values are plus or minus 1.96. So, by comparison, since our critical Z value, OK, remember our decision rule, or sorry, since our calculated Z value, OK. Which is equal to 1.509. That is the absolute value of our calculated Z value is less than the absolute value of our critical Z value, which is 1.96. OK. Then, That means it does not fall into the rejection region and thus we fail to reject the nut hypothesis. What does that tell us? Well, at the 0.05 level of significance, then there would be insufficient evidence to reject the researcher's claim that the average test scores for students taught with method X and the method Y are the same. Therefore, B is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

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