5

Suppose a study is conducted to estimate the difference betweenmiddle-income shoppers and low-income shoppers in terms of theaverage amount saved on grocery bills p...

Question

Suppose a study is conducted to estimate the difference betweenmiddle-income shoppers and low-income shoppers in terms of theaverage amount saved on grocery bills per week by using coupons.Random samples of 60 middle-income shoppers and 80 low-incomeshoppers are taken, and their purchases are monitored for one week.The sample results indicated that the mean amount saved permiddle-income shopper was $5.84 with a sample standard deviation of$1.41. The sample mean saved using coupons for low-income

Suppose a study is conducted to estimate the difference between middle-income shoppers and low-income shoppers in terms of the average amount saved on grocery bills per week by using coupons. Random samples of 60 middle-income shoppers and 80 low-income shoppers are taken, and their purchases are monitored for one week. The sample results indicated that the mean amount saved per middle-income shopper was $5.84 with a sample standard deviation of $1.41. The sample mean saved using coupons for low-income shoppers was $2.67 with a sample standard deviation of $0.54. Use this data to construct a 98% confidence interval to estimate the difference between the mean amount saved with coupons by middle-income shoppers and low-income shoppers



Answers

Construct the confidence interval estimate of the mean. Listed below are measured amounts of caffeine (mg per 12 oz of drink) obtained in one can from each of 20 brands (7UP, A\&W Root Beer, Cherry Coke, $\dots,$ TaB). Use a confidence level of $99 \% .$ Does the confidence interval give us good information about the population of all cans of the same 20 brands that are consumed? Does the sample appear to be from a normally distributed population? If not, how are the results affected? $$\begin{array}{rrrrrrrrrrrrrrrrrrr} 0 & 0 & 34 & 34 & 34 & 45 & 41 & 51 & 55 & 36 & 47 & 41 & 0 & 0 & 53 & 54 & 38 & 0 & 41 & 47 \end{array}$$

Brought him 17 number off sample is equal to 14. Sample standard deviation in 3.9. The confidence clever is a 0.299 and the alpha is one minus C over to which is over four or five. So the critical value from the table off the high school distribution extra square off one liners, all for over two is 3.565 on Chi square, off over two is equal to 29.8 819 The boundaries off the confidence. Confidence in the interval for the standard deviation is, uh, end minus one over chi square off Alpha over two times s, which is 2.5751 and the other boundary in minus one off Chi Square for minus all over two times asked equal to 7.4474 eso The boundaries for the variance is the square off this value, which is 2.5751 square 7.4474 squared, which is 6.631 and 55.46 14

80 two different mind. The critical value for table six degrees of freedom is on minus one, which is 11 minus one, so it's equal to 10. So the chi square off one minus alfa over to is able to 4.865 and the chi square of all for over two is equal to 15.987 So the boundaries for the standard deviation is the square root of n minus one over X squared off Alfa over to times of thunder deviation, which is equal to 86.2 or seven, and the square root of n minus one over chi square off one minus off over two times. Standard deviation is equal to 156.273 eso the boundaries for the variance, which is the square off the number off the standard deviation. So the variance is, um, the square value off these values so is equal to seven for one 3.647 and 244 to 1.251

Following a solution number 46. And we're gonna look at how outliers affect certain datasets whenever sample sizes are different. So we're giving a simple data set where the population mean is 50 and the standard deviation for the population is 10. and we're asked to find a 95% confidence interval for that Data set. So I'm using technology here just to save some time. So I'm using a T- 84, but you can use whatever technology or the formulas if you want. Um and if you go to stat and then I'm gonna go ahead and say edit now. L one, I'm gonna use both columns here, L one and L two but L one is the first data set and you see actually is the the exact same. So these 12 data values are exactly the same as L two until you get to a certain point. But this is the small data set that we originally start with. So L one Only has 12 data values. So there they are, went ahead and punched him in and um we're gonna go to stat tests and it's the seventh option here, the Z interval since we know the population standard deviation and keep it ever done on the data. Uh Since we have an actual data set, Sigma's 10 to list in this case is going to be L one. That will change that later whenever we get to the second part of this question and then the sea level frequencies one always and then the sea levels 10.9 fox that asked for a 95% confidence interval. And whenever we calculate now keep this X. Bar in mind 50.25 I'll leave it up here. But Um here's our interval here, 44.592 and 55908. Let's go and write that down. That's really just the first part, is a simple confidence interval. We don't even need to interpret it because they don't think there's a word problem associated with this, But it's 44 592 On the lower bound, and then the upper bound is 55.908. Okay, so that's our Confidence interval. Now, it says, Okay, let's pretend like the 41 and that data set is changed to a 14, and we need to confirm that it is in fact an outlier. Okay, well let's see what they're talking about. So stat edit And it's saying that this 41 here is changed to a 14, so let's go and change that and let's see what happens. I'm gonna go ahead and calculate the one bar stats here. Now there are a couple of ways that you can determine if something is an outlier. You can see if it's two standard deviations above or below the mean. So if you were to do that, you remember the standard deviation was 10 now it gives you something little different here, but we're assuming that it's 10. Um So two standard deviations would be 22 times 10 is 20, so 48 minus 20 is 28 14 is significantly less than that. So anything above or below? Two standard deviations from the mean, but there's a better way, it's the inter quartile range method, So what we do is we take the inter quartile range which is Q three minus Q one, we multiply it by 1.5, and that gives us a number to work with, So let's go and do that first. Um Yeah, so the inter quartile range, remember as Q three minus Q ones, that's 53 minus 435 in this case which is 9.5. And then we're gonna take 1.5 times 9.5. And that gives us 14.25. So this is a little bit more robust than the two standard deviation deals because we don't know if this is normal actually, so so this is good, so 14.25 And then what we do is since it's on the lower bounds of 14 is smaller, we're gonna take Q. One which is 43.5, we're gonna subtract that 14.25 And that gives us 29.25. And since 14 is less than 29-5 than it is in fact an outlier. So it's an outlier. In both cases it's definitely two standard deviations below the mean and it's 1.5 times the intercourse, our range below the first quarter. So in both cases it's an outlier. So that's the confirmation. Now, we're going to find the new Um 95% confidence interval. Okay, so let's go back to stat tests and it's that seventh option? The z interval. Okay, so everything again is exactly the same. But remember that 41 is now 14, so it should change it a little bit and it does. So 42.342 and 53.658. Let's go and write that down and then let's compare. So 42 .342 All the way up to 53 .658. And it asks what you know, how do these compare? Well, if you look at the first confidence interval to the second confidence interval, everything's been shifted down a couple units. So that confidence interval gets pulled just like the mean would be it gets pulled towards that outlier. So the confidence level or the confidence interval ci with the outlier is shifted down or pulled down towards the outlier? Okay. It's also, you know, a little bit bigger um because that variation is gonna be a little bit better, but the main thing is that shifted down because that mean is not robust, it's it's pulled towards that outlier and the mean remembers the point estimate. Okay, so that's part c. So now we go back and we're actually going to change change that back, so stat Edit and then let's go and change this back to 41. Okay, so the next part Were given another data set. Now let's go ahead and write this down. So the X bar the mean for the smaller data set, where it was just 12 data values was 50.25. That's what it asks for here. So 50.25 and then the large data set. That's where I put it in L. two now. Already populated in there. I already punched it in. But here are it's you know significantly bigger. I don't remember how many Data values. 30 or so. Yeah. So 36 data values. All right. So that's the difference there. So let's just make sure that it's the same mean. So let's change this to L. two. Okay? So 50.25. So the sample mean is exactly the same. So 50 point 25 So that's all we do there. So the sample means the same and you may think that the confidence interval needs to be the same now but we're gonna see here in a second it's not so we're gonna make a 95% confidence interval for the larger data set. So if we go back to stat Tests and again it's that 7th option. The z interval. Okay, this is all good. Except this time. The list is L two. Okay. And then we calculate and it gives us this 46.983 and 53.517 46,983, I'm sorry. Oh yeah, that's right. 46.983 2 53.517. Okay, So what does that mean? Well, it's if you look at this confidence interval of 46 to 53 compared to that first confidence interval, 40 40 55 it is a narrower confidence interval now, it's the same confidence level, we're still 95% confident. But what changed? Well in went from 12 to 36. So what does that mean about the width of the confidence interval within? Well, as in increases the width of the confidence interval decreases because that margin of error is going to decrease because you're dividing by a bigger number. So as in increases the margin of error decreases, which means the width of the confidence interval will also decrease, and that's shown there. So the X bar did not change, right? So nothing changed their with those point estimates, the only thing that changed there was the margin of error. Okay, so this last few parts, we basically just do what we just did. So, um we're just working with the second data set and we're gonna change that 41 to 14 for the second dataset. Okay, and again, we're going to confirm that it is in fact an outlier. So let's go back to couch one of our stats. Okay, so 49.5 as the means, so if you think of two standard deviations, that would be 29.5 um and 14 is less than 29.5. Or you can do the inter corte range, that's a little more robust. So I'm going to the inter corte range, that's Q three minus Q one, which is 13. Okay, so we're confirming that it's an outlier city. Inter quartile range is 50 6 -43, which is 13 And I take 1.5 times 13 to give me 19.5 and then I take Q one, which is 43 minus that, 19 5, That gives me 23.5. And since that 14 is less than 23, it is in fact an outlier. So, again, it's an outlier. In both cases it's two standard deviations Below the mean, which in this case is fine because the sample size is big enough, but the intercourse, our range is usually the more preferred method. Now we're gonna make a new 95% confidence interval. Okay, so we go back to stat tests and it's that seventh option. Okay, so all this is good. So remember I had that 41 change to 14 calculate I get 46.233-52.767. So let's write that down and then we can compare real quick. So 46 233 All the way up to 52 767. So that's the new 95% confidence. And so how did that change this here? Well, yeah, 46.9 and it got pulled down to 46.2 and this one is 53.5 and that got pulled down to 52.7. So it did get pulled down a little bit, but not nearly as much as whenever the sample size was 12. Remember this one it got pulled down, you know, a full 2 2.5 units, whereas over here it just got pulled down. You know, I don't know, just a little over a half a unit. And the reason for that, the reason why it didn't get pulled down as much is because that sample sizes larger, the larger the sample size, the less influential those outliers can be. So what happened here? The outlier pulled the confidence interval down a bit but not nearly mm as much scroll down a bit as when And equals 12. Okay, so the larger the sample size, the less influential those outliers will be.

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.


Similar Solved Questions

5 answers
PointsSESSCALCET2 6.6.013.Determine whether the integral is convergent or divergent:17xeconvergentdivergentIf it Is convergent; evaluate It. (If the quantity diverges enter DIVERGES.)Need Help?RaadItMatch ItIalk to Jutor
points SESSCALCET2 6.6.013. Determine whether the integral is convergent or divergent: 17xe convergent divergent If it Is convergent; evaluate It. (If the quantity diverges enter DIVERGES.) Need Help? RaadIt Match It Ialk to Jutor...
5 answers
17 . Which of the following rows identifies the structure that Is the site of spemm production and the gland that produces an alkaline secretion that neutralizes the acidity of the vagina?Site of Sperm ProductionGland that Produces an Alkaline SecretionRowseminiforous tubulesprostate glandseminal vesiclestestesseminiferous tubulestestes prostate glandseminal vesicles
17 . Which of the following rows identifies the structure that Is the site of spemm production and the gland that produces an alkaline secretion that neutralizes the acidity of the vagina? Site of Sperm Production Gland that Produces an Alkaline Secretion Row seminiforous tubules prostate gland semi...
1 answers
Samples of six different brands of diet or imitation margarine were analyzed to determine the level of physiologically active polyunsaturated fatty acids (PAPUFA, in percent), resulting in the data shown in the accompanying table. (The data are fictitious, but the sample means agree with data reported in Consumer Reports.) $$egin{array}{llllll} ext { Imperial } & 14.1 & 13.6 & 14.4 & 14.3 & \ ext { Parkay } & 12.8 & 12.5 & 13.4 & 13.0 & 12.3 \ ext { Bl
Samples of six different brands of diet or imitation margarine were analyzed to determine the level of physiologically active polyunsaturated fatty acids (PAPUFA, in percent), resulting in the data shown in the accompanying table. (The data are fictitious, but the sample means agree with data report...
5 answers
The University of Central Arkansas Wants 40 investigate alcohol use among the students. UCA surveys students about their alcohol use The results are illustrated in Table 3 Is alcohol use at UCA unequally distribuled?Table % UCA s Surver on Collepe DrinkingUsed Didn TotalState the null and research hypothesis:Conduct a chi-square test at the & 01 level:
The University of Central Arkansas Wants 40 investigate alcohol use among the students. UCA surveys students about their alcohol use The results are illustrated in Table 3 Is alcohol use at UCA unequally distribuled? Table % UCA s Surver on Collepe Drinking Used Didn Total State the null and researc...
5 answers
Use the fact Ihat 0.01 integers and b # 0.and 0,001 999convert each of Ihe following to the form where = and are0.46 6560.46 . (Simplily Your, answer; Type an intogar Or simiplilied fraciion )656Simplify your answer: Type an Integer or simplilied (raction )
Use the fact Ihat 0.01 integers and b # 0. and 0,001 999 convert each of Ihe following to the form where = and are 0.46 656 0.46 . (Simplily Your, answer; Type an intogar Or simiplilied fraciion ) 656 Simplify your answer: Type an Integer or simplilied (raction )...
5 answers
2002 2007 L when dd It occut? 11 abou wnal%as ndicular? 1 1 uE4 V 1 1 1hotesi
2002 2007 L when dd It occut? 1 1 abou wnal%as ndicular? 1 1 uE4 V 1 1 1 hotesi...
5 answers
2= 3 +3 4 nzi h?
2= 3 +3 4 nzi h?...
5 answers
Identify the hypothesis and conclusion of each statement.If $x-3=7,$ then $x=10$
Identify the hypothesis and conclusion of each statement. If $x-3=7,$ then $x=10$...
5 answers
Replace the Cartesian equation with equivalent olar equation (x+3)2 + (y _ 5)2 = 0.Graph the polar curve_T =sin(0)_Find the slope of the curve r = cos(20) at 0 = 2
Replace the Cartesian equation with equivalent olar equation (x+3)2 + (y _ 5)2 = 0. Graph the polar curve_ T = sin(0)_ Find the slope of the curve r = cos(20) at 0 = 2...
5 answers
30 points) Prepare the force based equation system for the spring system consisting of equal masses and 5 springs as in lecture 15)- Find the eigenvectors of this system and visualize all vibrational modes.This question is to be written by hand on paper. Your solution must be in question2 pdf format_ Yon can use Microsoft Lens O similar programs to generate YOUr question2 pdf: Wrong file formats get Opts. Disoriented pages in the pdf get -lOpts each.
30 points) Prepare the force based equation system for the spring system consisting of equal masses and 5 springs as in lecture 15)- Find the eigenvectors of this system and visualize all vibrational modes. This question is to be written by hand on paper. Your solution must be in question2 pdf forma...
5 answers
Question90 adults with gum disease were asked the number of times per week they used to floss before their diagnoses_ The (incomplete results are shown below: #of times floss per week Frequency Relative Frequency Cumulative Frequency0.08891444 1667 1333 14445o0889Complete the table (Use decimal places when applicable)What the cumulative relative frequency for flossing times per week?Submit Question
Question 90 adults with gum disease were asked the number of times per week they used to floss before their diagnoses_ The (incomplete results are shown below: #of times floss per week Frequency Relative Frequency Cumulative Frequency 0.0889 1444 1667 1333 1444 5o 0889 Complete the table (Use decim...
5 answers
8. Given% -1+4J ud9 - 1+7A Nrte Ech compkex nuabar in polr fon (2 points)B. Find 2122 (leave answer in polar for) (5 points)Find(lcave answer polar fon) (5$ points)Find (=1) Ucutl nnsw crPolar fon) points)
8. Given% -1+4J ud9 - 1+7 A Nrte Ech compkex nuabar in polr fon (2 points) B. Find 2122 (leave answer in polar for) (5 points) Find (lcave answer polar fon) (5$ points) Find (=1) Ucutl nnsw cr Polar fon) points)...
5 answers
O/e 2 85 Wh 8 1 V 8 3 7 1 } L 1 1 11 { iL ; 1 8 D 3 2 } } 8 # 3 &f;n"{a3 7 !"1'2' '"
o/e 2 85 Wh 8 1 V 8 3 7 1 } L 1 1 11 { iL ; 1 8 D 3 2 } } 8 # 3 & f;n"{a 3 7 !"1' 2' '"...

-- 0.024503--