Quality ControlSuppose the mean wait-time for a telephone rese...

Question

Quality ControlSuppose the mean wait-time for a telephone reservation agent at a large airline is 43 seconds. A manager with the airline is concerned that business may be lost due to customers having to wait too long for an agent. To address this concern, the manager develops new airline reservation policies that are intended to reduce the amount of time an agent needs to spend with each customer. A random sample of 250 customers results in a sample mean wait-time of 42.3 seconds with a standard deviation of 4.2 seconds. Using an α = 0.05 level of significance, do you believe the new policies were effective? Do you think the results have any practical significance?

Accepted Answer

Welcome back, everyone. In this problem, a fitness center claims that its new exercise class leads to weight loss. In a study of 40 participants, the mean weight loss after 6 months was 2.2 kg with a standard deviation of 4.2 kg. At the 0.01 significance level, test the claim that the mean weight loss is greater than zero. Is the result statistically significant and is it practically significant? Now, what are we trying to figure out here? Well, we're testing a claim, OK, that the mean weight loss from this exercise class is greater than 0, OK? And here, we know that we are working with a simple random sample with a sample size of 40, OK? Now, this is a claim about a population mean with a population standard deviation not known. And since the sample size is greater than 30 and it's a simple random sample, then we cannot assume it follows a normal distribution. Thus, that means we can use the T test since only the standard sample standard deviation is known and the population standard deviation is unknown. So first, let's say our state our hypothesis. Remember we're testing if the mean weight loss is greater than 0. So the null hypothesis, the default assumption is that the mean weight loss is less than or equal to 0, OK? And our alternative hypothesis is going to be that the mean is greater than 0. So this is going to be a right tailed test. If we quickly draw a normal distribution here, we're basically saying is that if we find a critical value given these uh conditions, OK, and we calculate the test statistic and it is greater than the critical value, OK, then that means there is enough evidence for us to reject the null hypothesis. But if it is not greater than the critical value, in other words, it falls on the other side of the critical value to the left, then that means there is not enough evidence for us to reject that hypothesis. So you'll tell you, you, you realize then that we need the test statistic and the critical value. What information do we have? Well, so far, we're told that we are trying to test this claim at the 0.01 significant level. So alpha equals 0.01. Or sample mean X bar is equal to 2.2 or population mean equals 0, or sample standard deviation S is 4.2 and our sample size N is 40, OK? So using the formula for the test statistic, that means T is going to be equal to. Well, we know it's X bar minus mu divided by S divided by the square root of N. So that's gonna be 2.2 minus 0 divided by 4.2. Divided by the square root of 40. And when we calculate here, we get this test statistic to be approximately equal to 3.313. Know that we have the test statistic, we need to find the critical value for us to compare. Now, for, as we said, for a right tailed test. OK, for a right tailed test. OK. With Our significance level alpha equal to 0.01 and the degrees of freedom being 1 less than the sample size, in this case 40 minus 1 or 39. For a right tail test with alpha equals 0.01 and 39 degrees of freedom, then the critical value, OK, T 0.01, 39 is gonna be approximately equal to, let me write properly here. I don't know where I got that, approximately equal to 2.426. So what does this tell us? Well, if we compare it to our test statistic. Then we notice that our test statistic, 3.313. Is greater than our critical value, OK. Of 2.426. So since our test statistic is greater than our critical value, that means it will fall in the region which would reject or which would allow us to reject the null hypothesis. So what can we conclude? Well, statistically. OK. If we think about it, based on the information we have, let me write this here. No, we have enough information. Let me make this font a bit smaller. No, we have enough information. And there is there is sufficient evidence. To reject another hypothesis and thus support the claim that the mean weight loss, OK, the mean weight loss is greater than 0. That's the conclusion we can come to given these parameters. But now, what does this mean for us? Is it statistically significant and is it practically significant? Well, the result is statistically significant at the 0.01 confidence level. There is strong evidence that the mean weight loss is greater than zero, but when we think about practical significance, will that make a difference for us? If we look back at the problem statement, we were told that this mean weight loss was after 6 months. And if the mean weight loss is only 2.2 kg over 6 months, that's, that's relatively smart. So while it's statistically significant, it does not appear to have practical significance because a 2.2 kg loss may not be very meaningful and may not have a large practical impact for most individuals. Thanks for watching everyone. I hope this video helped.

Key Concepts

Hypothesis Testing

Significance Level and p-value

Practical vs. Statistical Significance

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