Waiting in LineOne aspect of queuing theory is to consider wai...

Question

Waiting in LineOne aspect of queuing theory is to consider waiting time in lines. A fast-food chain is trying to determine whether it should switch from having four cash registers with four separate lines to four cash registers with a single line. It has been determined that the mean wait-time in both lines is equal. However, the chain is uncertain about which line has less variability in wait time. From experience, the chain knows that the wait times in the four separate lines are normally distributed with σ = 1.2 minutes. In a study, the chain reconfigured five restaurants to have a single line and measured the wait times for 50 randomly selected customers. The sample standard deviation was determined to be s = 0.84 minute. Is the variability in wait time less for a single line than for multiple lines at the α = 0.05 level of significance?

Accepted Answer

Hello, I'm glad to have you back. Let's take a look at our next problem. An electronics manufacturer claims that the population variance of the lifetimes of its LED bulbs is less than 1.5 years squared. A random sample of 25 bulbs yields a sample variance of 1.2 years squared. Assume lifetimes are normally distributed. At alpha equals 0.05, is there sufficient evidence to support the manufacturer's claim? There is insufficient evidence to conclude that the true variance is less than 1.5 years squared. There is sufficient evidence to conclude that the true variance is less than 1.5 years squared. The test is inconclusive, or D, more data is needed. So, step one, always think about our null and and alternative hypotheses. So in this case, are no hypothesis. Which would be that statement of equality. Will be I'll say no hypothesis, is that the population variance, so sigma squared of the light bulbs, equals 1.5. And then the claim, and then alternative hypothesis is that the true population variance is less than 1.5. So that's the claim, also a statement of inequality, so that's our alternative hypothesis. Now, since we have this left, or excuse me, since we have this less than, we're going to look at a left tailed. Chi squared test, even though we've got 25, a sample of 25, we're told to assume a normal distribution. Now when we talk about chi squared statistics, recall that we need our degrees of freedom. And our degrees of freedom are equal to N minus 1. In this case, we have a sample of 25, so that's going to equal 24. As are degrees of freedom Now, a left tailed chi square test is always, you always have to think about it a little bit more, because the way the table is written, it's always giving us a right tailed assumption. So, how we always want to think about it, of course, drawing that graph, we have a chi squared distribution. And recall that the number from the graph will give us A critical value based on the probability on the underneath the right side of the curve. So, if we want a left tailed test, so, I'm just going to draw a line towards the left, and we're looking for a rejection region that I'll highlight in blue. In the area under the curve to the left of our critical value. So I'm going to call this chi squared L, the left tailed critical value. That means what we actually need to look for is based on the pro is the probability. On the right side when using our chart. So we know that alpha is equal to 0.05. So our rejection region. That area in blue. Has that area of 0.05. So that yellow region on the right, the probability we need to use on our table. Is 1 minus 0.05, which equals 0.95. So, you always need that extra little step when you're thinking about a left tailed chi squared volume. So, now let's go back to calculate our test statistic. So that's going to be chi squared. And we use the formula. And the numerator in parentheses n minus 1. Multiplied by Squared, so our sample standard deviation. Divided by Sigma squared, so our population variants, or sorry, on, on the numerator, let's repeat that, in parentheses and minus 1 multiplied by Squared are sample variants, not standard deviation. Divided by in the denominator, sigma squared, so our population variants. So, plug in the numbers we have. So, the numerator N minus 1, 25 minus 1. And then our Squad, our sample variants, we are told, was 1.2. So it's multiplied by 1.2. And then in the denominator, we have a sigma squared or population variance, which is 1.5, according to our null hypothesis. So go ahead and put those numbers into your calculator. And you will get an answer of 19.20. So you can see that to test whether we fall in the rejection region. To the left. Of that left tailed critical chi squared value. We need to compare this calculated chi squared value with that critical value. So we're going to look up. Our chi squared value, this left value in the table, based on our probability, or we're going to treat it as our alpha, it's our probability. Which will set as alpha, will be 0.95 to get this result. And then, of course, our degrees of freedom, as we said, are 24. So when we look at the chi square table, that's going to give us a value of approximately. 13.848. So now we need to compare that so that critical value. To our calculated test value, so that 19.20. Chi squared value. Is greater than. are 13.848 or are left. Tailed critical chi square value. And again, that means that our chi squared value is going to fall somewhere. Draw that there in that yellow region, so we fail to reject. Are no hypothesis. And that means that at that 5% significance level, Choice A, there is insufficient evidence to conclude that the true variance is less than 1.5 years squared. We did have enough data and our test is conclusive. But there is not sufficient evidence or chi squared value would have had to fall within that left-sided rejection region, to conclude that the manufacturer's claim is correct. So, again, is there sufficient evidence to support the manufacturer's claim at al equals 0.05? And that answer is A, there is insufficient evidence to conclude that the true variance is less than 1.5 years squared. See you in the next video.

Key Concepts

Hypothesis Testing for Variance

Chi-Square Distribution

Significance Level and Decision Rule

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