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Questiou 8 (Total:4 marks) A Superstore is interested in estimating the average annual expenditure on extended health for employees in the industry. In a pilot stu...

Question

Questiou 8 (Total:4 marks) A Superstore is interested in estimating the average annual expenditure on extended health for employees in the industry. In a pilot study, a random sample of 15 stores was chosen_ The mean annual expenditure was found to be $251.50 with a sample standard deviation of $9.49. Assume that the distribution across different stores in the industry is normally distributed.A) On the basis of the pilot study, determine a 99% confidence interval for |4, the actual mean annual

Questiou 8 (Total:4 marks) A Superstore is interested in estimating the average annual expenditure on extended health for employees in the industry. In a pilot study, a random sample of 15 stores was chosen_ The mean annual expenditure was found to be $251.50 with a sample standard deviation of $9.49. Assume that the distribution across different stores in the industry is normally distributed. A) On the basis of the pilot study, determine a 99% confidence interval for |4, the actual mean annual expenditure 0n benefits. marks) B) The firm wants its results to be accurate within $1.50 with 95% confidence. How large a sample size does the firm need to take? (2 marks) (Must use Z, not t_ This will be fine as long as n is greater than 30.)



Answers

(a) identify the claim and state $H_{0}$, and $H_{a},(b)$ 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. If convenient, use technology. To compare the amounts spent in the first three months by clients of two meal-replacement diets, a researcher randomly selects 20 clients of each diet. The mean amount spent for Diet $A$ is $\$ 643$ Assume the population standard deviation is $\$ 89 .$ The mean amount spent for Diet B is $\$ 588 .$ Assume the population standard deviation is $\$ 75$ At $\alpha=0.01,$ can the researcher support the claim that the mean amount spent in the first three months by clients of Diet $A$ is greater than the mean amount spent in the first three months by clients of Diet B?

Following is a solution video to number 26 and this looks at the flight times from Albuquerque to I think like Denver or something. The average flight times and minutes. And the first part is to ask for the point estimate. The point estimate to find the mean is the X bar the sample mean. And you can do that just using the formula or if you want to you can use technology and I'm gonna use technology. If you go to Staten edit this on a T. I 84 Here are the data values 117 minutes, 95 minutes, 109 minutes etc. And if you go to stat and then air over to Calcutta and then it's one of our stats List is L one and then we calculate and then this X bar is about one of 3.44 So that's what we're going to put here. So one oh 3.44 minutes is the point estimate. Then we're giving a box plot and a normal probability plot. And it says because the sample size is pretty small. If you look back at that, In fact, n equals nine. There are only nine native values. So in order to use the Z interval, um we need to see if the population is normally distributed and so we can look at a box plot in the normal probability plot to determine that. And by looking at those, the short answer here is yes, the data appears normal, so the box plot looks fairly symmetric so the data appears normal. And also the normal probability plot looks about linear, kind of has heavy tails, but that's fine. So it looks about linear. All those data values are kind of within range free from any skin this or outliers. So that's Looking at the box pot normal probability plot. We can go ahead and use those conditions for inference or we basically verify the conditions of inference. Yes, the sample size is small, but the original data appears to be approximately normal. So now we're going to find the 95% confidence interval. So again, you can use the formula if you so wish or you know, it's probably a little easier if you have a stat dishonesty I 84 tests and it's the seventh option here. The Z interval. Alright, so since we know the sigma, we know the population standard deviation, we're going to go to the Z interval and then make sure the data is highlighted. So it's not stats. We usually we have summary stats at this time, we actually have data sets of data and then sigma is eight. That was given to you in the problem. The list in my case was L one, so if you have a different column then there was changes to L. Two or three or whatever you have. And then the sea level is 95 95% of 950.95 And then calculate. And then this top line here that gives you the Confidence intervals about between 98.218 and 108.67. So the 95% confidence that it will is between 98 .218 And. 108 0.67 And then we also need to interpret that. So the interpretation here follows the same kind of pattern. We say we can be 95% confident that the main flight time mm between in Dallas and Alburquerque. I should say for all american airlines flights for all american airline flights, mm is between 98.2 and 108.7 minutes. Now, I've rounded there, you can be as accurate as you want, but Somewhere between those two numbers, we can be 95% confident notice it's for all American airline flights between Albuquerque and Dallas. So that's the 95% confidence interval. Now we're gonna do the same thing, but this time we're gonna change it to the 90% confidence interval. So we go to the Z. interval and all this stays the same except I'm gonna say .9 this time And it's between 99.058 and 107.83. So let's go and write that down. So the 90% confidence interval is 99 point 058 And 107.83. And we interpret it the same way really. The only difference here are the numbers. So we could say, you know, we can be, I'm not gonna write everything down Just, you know, the things that change. So we can be 90% confident that the mean flight time between Dallas and Albuquerque for all american airline flights is between, so is between so dot dot, dot is between 99 058 And one of 7.83 minutes. So it's between, you know, basically an hour and I don't know, maybe an hour, 40 minutes or so. Um, so let's look at these a little bit more closely. So the 95 compared to tonight. So everything stays the same except for the confidence interval or the confidence level, I should say. So, notice that the interval gets a little bit narrower. So here we've got uh, oh, maybe about a 7.5 minute difference, but here we've got almost actually more than a 10 minute difference between the low end and the high end. So whenever we get less confident we can widen up that, I'm sorry, narrow that interval down. So that's kind of the trade off. The more confident you are, the wider your interval has to be the less confident you are, then you can be a little bit narrower on that interval and that's kind of what what party is talking about. So in playing our words will save the width of the interval decreased when the confidence level decreased, which makes sense. The less confident you are, the narrower the interval can be. Yeah.

So we're looking to see if there's a difference between the grocery expenditures on a credit card versus the dining out. And so we have a sample of 42 people, and we have that. The main difference that they got and subtracting grocery caught expenses minus dining expenses that that mean of those 42 people came out to be 100 and 50 with a standard deviation of $1123 and we want to right some hypotheses. We'll assume that that means difference is equal to zero, and alternately, it's not equal to zero. You can hear the ring Doorbell just went off, and so again, picture wise we're assuming the mean is zero. We're getting 850 so we're getting the grocery expenses higher than the dining expenses, and we want to find this area plus this area. If that difference had been in the reverse direction to find our P value and so were you. There are two ways we could set this up this sample size. We could say that this is large enough to assume approximate normal distribution, but I'm going to do it with a test statistic of a T value with 41 degrees of freedom. Again, you could do it with the Z value as well. Do it depend on what your textbook says and we're going to take that 8 50 minus the mean We're assuming the standard deviation over the square root of and and when we get that test statistic, we end up getting 4.905 very large. And so if I find what the likelihood is of a T value with 41 degrees of freedom being greater than or equal to 4.905 And I used my software, my, um, normal, my T c d a button and I got a P value for this that was extremely small. It was 1.5 times 10 to the negative fifth power, and that is definitely smaller than any significance level you're going to use. Let's say we were going to use, like, a kind of a nit picky significance level of 1% so we would have, um, strong evidence to reject now strong evidence to reject yeah, the novel and conclude there is a difference. Now where do we think that difference is it appears as though the groceries have a higher mean right and I would have expected that are kind of hoped. That and our point estimate for the difference. Our point estimate point estimate for the difference is 850. And if we want to calculate a 95% confidence interval again, that's going to depend on whether you're going to assume a T distribution or a Z distribution. But we're going to take 850. And let's say I show you both for a T and for a Z we would take plus or minus. And then we look up a T star value for 41 degrees of freedom and that comes out to be, I believe, this value. And then we would take our population while our sample standard deviation divided by the square root of and and we would get an interval that goes from about $500.5 up to $1200 now. On the other hand, if we use a 95% confidence interval and we assume a normal population and think that sample size is large enough to allow us to do that, then we would have the 8 50 plus or minus, and we would use 1.96 times the this. I prefer this T value. It's going to be more accurate in this setting. Uh, and when you do that, you end up getting 510.37 to 1189. Yeah, and calculator just turned off 89.6. And again, this is the one I would use because we do have the ability to use the t value because it was a sample standard deviation, and so it's a better, better statistical calculation.

All right, This question gives us some information about Continental Airlines working hours. And first part wants us to find the margin of error for a 95% confidence interval. So remember that error equals closer minus a specialty star value. We're dealing with t star because we don't have access to the population. Standard deviation, times, the standard error. So the only thing we don't know right now is T star. So to find that we can either use the table or drawing a normal curve for a 95% interval, we know that the middle of the curve needs to contain 0.95 area. So that means there's 0.0 to 5 in each tale. So from there, we know that our T star is inverse T as we want an area to the left of 0.9 75 And how many degrees of freedom we have is just one minus the sample size or one less than the sample science. Sorry, which in this case, Artie star value is one point 984 All right, so now we can plug in to our formula, so we know that error, ankles plus or minus Artie Star value that we found earlier 1.984 times our sample standard deviation over the square root of our sample science which is in this case, one point 68 64 in either direction. So now it wants us to generate the interval. So Barbie wants the 95% interval. And this is pretty easy to do now because we know that our interval is just X bar plus or minus our margin of error, which we already calculated. So that's 49 closer minus 1.6864 Which gives us an intern colonel from 40 7.3136 two, 50.68 64. So that's our 95% confidence interval for the population means, and now it compares it to a different airline. It says that mule equals 36 for United, and if you look at that mean 36 as well, outside are 95% confidence interval for the mean. So we can say with 95% confidence that are continental mean is bigger. So the wording is we're 95% confidence that the mean, let's specify here that the population mean for Continental is between. And then this is our interval from part B. I'm just not gonna rewrite it because there's a lot of decimals and that will just because a bunch of unnecessary time, because we have the information here and then since 36 is well outside this interval, yeah, is reason to believe the populations have different means. This explains high labor costs. All right, So just to summarize question, asked if if this confidence interval provided any insight into why Continental has the highest labor costs and since our interval says that the mean for the population is 95% confident that it's between this range, which is about 47 to 51. And United has a population mean of 36 which is not in our confidence interval, we can say with 95% confidence that the two populations don't have the same mean so they work more hours at Continental, which explains the highly

What's up, stat? Cats. My name is Aaron, and in this video we're gonna be performing an F test using the traditional method of hypothesis testing. So what we want to know is does the variants in daily museum attendance differ between the winter and summer months? So we want to know if they differ. And we are told that the sample size for each is 30 and that the standard deviation of the winter months is 52 standard deviation of the summer months is 65. We are also told that are awful. Level is 0.5 So stop A is stating our hypotheses and identifying our claim. So are no hypotheses is always gonna be that the variances air equal And our alternative hypothesis is that the variances are not equal. This is our claim because we want to know if the daily attendance varies between the summer winter months and because our variances air not equal for alternative hypothesis. This tells us that we're doing a to tell test. So some of our awful level being 0.5 when we look at our H table to find our critical value, we're actually gonna use 0.25 age table to get our critical value. Step B is getting are critical value. And because both of our sample sizes are the same, we already know that the degrees of freedom for each of our numerator and denominator are gonna be the sample size at minus one. So they're both gonna be 29. So let's go ahead and look at the 0.25 h table. So 29 for our denominator and 29 for a numerator. And as you can see, we don't have 29. We have 30 or 24 we can't go up. So we have to use 24. So are critical. Value is 2.15. Step C is finding our test statistic. And to do that we have to find out which variants is greater, so we can put that in the numerator. So let's go to this Excel spreadsheet where I copied over the standard deviations that we were given. So to find the variants, all we do is we square the standard deviation so we can see that the variants for the summer month is greater than the variance of the winter months. So we're gonna dio summer variants divided by winter variance. And that gives us 1.56 for our f test statistic. Step D is making our decision. So our f calculated, which was our F test statistic, was 1.56 and our F critical value waas 2.15. And what we want to ask is is our calculated is are calculated greater than our ask critical. And if it is, then we can reject are no help, no hypothesis. So is 1.56 greater than 2.15? No, it's not. So we fail to reject are no hypothesis. So Step E is summarizing our results And because we failed to reject Arnold hypothesis, we can't support our claim, which was our alternative hypothesis. So there is not enough evidence two suggests that the variance and daily you see, um, attendance is different between summer and winter. Alright, guys, in this video we performed an F test using the traditional method of hypothesis testing we failed to reject are null hypothesis, which means we couldn't support our claim, which was our alternative hypothesis. All right, I hope you guys learned a lot and I'll see you next time


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