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Esimate Differerce 2 Means (2) O1b Two surveys were done regarding credil caid debl Survey #1: Five years ag0 the average credil card debt was 58318 ae-Survey #2: T...

Question

Esimate Differerce 2 Means (2) O1b Two surveys were done regarding credil caid debl Survey #1: Five years ag0 the average credil card debt was 58318 ae-Survey #2: The average credit card debt fOr recent year was 59205. Assume sample sizes Ol 35 were used and the standard deviations of both samples were 51928Construct the 95% Confidence Intetval estimale for the dillerence Ihe amount credit card debt based on the IwO surveys above.Slect to one TRUE statement below:Answer179002 $-16.3081790 $W1 42

Esimate Differerce 2 Means (2) O1b Two surveys were done regarding credil caid debl Survey #1: Five years ag0 the average credil card debt was 58318 ae-Survey #2: The average credit card debt fOr recent year was 59205. Assume sample sizes Ol 35 were used and the standard deviations of both samples were 51928 Construct the 95% Confidence Intetval estimale for the dillerence Ihe amount credit card debt based on the IwO surveys above. Slect to one TRUE statement below: Answer 1790 02 $-16.308 1790 $W1 42 < 16 308 -1790 16 308 -1790 $p1 W2 $-16,308



Answers

Number of Credit Cards Among individuals who have credit cards in $2010,$ the mean number of cards is 3.5 according to the Federal Reserve Bank of Boston. Treat the individuals who have credit cards in the Sullivan Statistics Survey as a simple random sample of credit card holders. The $n=160$ individuals have a mean of 3.3 cards and a standard deviation of 2.3 cards. Do the results of the survey imply that the mean number of cards per individual is less than $3.5 ?$ Use the $\alpha=0.05$ level of significance.

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.

The following is a solution to number 22. And this study's credit card debt for college students and a school administrator asked a certain amount of college students if they have $2000 or more credit card debt. And we want to know the minimum sample size necessary in order to be 2.5% points to have a margin of error of about 2.5% points. So what we do is we use this formula. I wrote this formula here and this is in your book. But you can also derive it from the margin of error formula it's in equals P. Hat times one minus P hat. So the sample proportion times one minus the sample proportion times the critical values East are divided by the margin of air quantity squared. And all this information. We're actually most of the information we're given in the problem. So that margin of error remember was 2.5% point. So we're gonna put 0.0.25 there. And then as for the confidence they give us the 94% confidence level, and that's actually not a typical confidence. And we'll see. You may not have that Z score memorised. And if you don't that's fine, you can use a chart where you can use a calculator. I prefer to use a calculator. I think it's a little easier. So if I go to second distribution, this is on a T. I 84 but you can use any sort of calculator you want or excel or any sort of software You so wish and you're gonna go to inverse norm and we're gonna change that 2.94, the mean is zero. The standard deviation is one always whenever we're using inverse norm and it's the center. So anytime you're working with a confidence interval, that tail or just look concerned with the Middle 94%. Now, if you have an older version of the calculator, you would actually have to put in 397 Because that makes up the 6% different. And if you have an option of changing the tail, you know, you need to go left or right. But if you have a very old T. I. T. For it just defaults to a left tail test. But hopefully you have a new one because it makes it a lot easier. And whenever we paste that thing, we get these numbers here, so negative and positive and it's the same number about 1.881 we're just gonna use the absolute value of that. So 1.881 is RZ star 1.881. Now we square the result anyway, so it wouldn't matter if you use positive or negative, but typically speaking you're just gonna use the positive And were given in the problem that 34% of college Students have credit card debt of $2,000 or more, which means that 66% do not. So 1 -14 gives you that 66%. So now we're just gonna plug that in the formula. So it's P. Hat which is .34 Times 1 -4. Hat which is 66 Times The Z. Star which is 1.881 divided by the margin of error of .25. And then we square that. And that should give you um A number C It's a pretty big number actually 1270.3. But for in for sample size we can't have you know, .3 of a person. So you're always going to round up. Always always always round up to the next hole number. Don't round off so early on in our math career, we ah Are we learn? You know, Okay, we got around that down to 1270 because .3 isn't above .5 and statistics that doesn't work that way. You always go um with the next hole number. Whenever you're looking at sample size, it's the more conservative estimate. So 1271 would be the right answer there. Now whenever you don't have any prior estimates um it's the same formula except instead of P hat and one minus P hat being what what we had up here, you're gonna change those 2.5 and 0.5. And the reason why you do that uh basically it's going to give you the biggest number possible. So 0.5 times 0.5 is bigger than any scenario that you can have. And this gives you, you know, the most conservative estimate the maximum size of end that is necessary in order to get that 2.5% point difference. So we do that, we plug it in and you should get 14 15 .3 and once again we're gonna round that up, not off, but up to 1416. So sample size necessary is 1,416 people. So another way to double check that we, you know, punch this in the calculator. Right? The in whenever you don't have prior estimates should always be larger than the end. If you do have prior estimates and that's a typical thing with statistics, the less information, you know, the more conservative or the safer you gotta be uh when selecting your sample size.

We want to start this problem by taking the Sullivan Statistics Survey # one. About how many credit cards do you currently have and put it into a frequency distribution. So our data includes how many credit cards. And as you scan that list you're going to find that there were people that said they had one credit card or two 34 567 89 10 and 20. Now the frequency for each of those there were 23 people that said one credit card. There were 44 that said to 43 said three 18 said they had four 13, said they had five, seven people said they had 64 for seven. Three said they had 82 said they had nine two said they had 10 And one said that they had 20 credit cards. If I would add up the frequency column I'm going to discover that there were 160 pieces of data in this data set. So Part A. Is asking you to determine the mean number of credit cards. So in order to calculate the mean of our X. Variable, we are going to use the formula some of X. F. Divided by N. And N. Is the same thing as saying the sum of F. And we just discovered that to be 160. So as I look at the top part of this formula, it's telling me that I need to create an additional column in my chart and I'm going to call that column X times F. So when I multiply one times 23 I get 23 2 times 44 I'm gonna get 88. three times the 43 will get you 129, four times 18 gets you 72. And I'm going to keep that pattern going throughout the entire column And five times 13 gets me 65. So now when I add up that column, the sum of my x f, I end up with 329, So my average will be 329, divided by 160, And that's going to give you three 30 6 to 5. And just for the ease of working with the number, we're going to say 3.3. So that was part a of this problem. Now, part via this problem is asking you to determine the standard deviation based on this raw data, so therefore we're going to have to create additional columns. And when it comes time to determine standard deviation, your formula that you would use would be that the standard deviation or sigma X is equal to the square root of the sum of x minus mu squared times f all over. And so that means we need to add a couple rose to our chart. So we're going to need x minus mu column. We're gonna need an x minus mu squared column. And then you're gonna need an x minus U. Squared times F. Column. And I'm going to let the graphing calculator help me get the values in those columns. I'm going to bring in my graphing calculator and I've already preloaded the data. So I'm going to go sit on top of list three And I'm going to tell it to take every number in list one And subtract the mean from it. So we'll subtract the 3.3. And in doing so you're going to get these values, you're gonna get negative 2.3 negative 1.3 negative 0.3 Positive seven and then 1 7-7, 3.747 5767 and 167. Now, if we were to average those numbers up, we would end up not getting us some valuable information. So what we do now is we square all these numbers and I'm going to Move over in my calculator. I'm going to sit on top of L four and I'm going to tell it to take everything in list three and square it. And you're going to see the following numbers. So we have 5.29 1.69 point yeah. 0949 2.89 7.29 13 69 22.09 3249 44.89. And then finally 278 89. And now we have to take every number in that column and multiply it by its corresponding frequency. So I'm going to again sit up on top, tell it to take every number that we just had enlist for and multiply by the frequencies that we had recorded in list too. And you're going to get the numbers 1216, 7 74 36. And these are rounded values 387 8.8, 2 37 57 51, 3 54-76 66-7 64, 9, 8 89 78 and 200 78.89. And now we have to add up that list in order to utilize the formula. So The fastest way to add up that list, which is list five, is to quit out of the table. And I'm gonna use 2nd stat. Move over into the math operations and ask the calculator to sum up list five. So the numerator is 852. We've already decided that N, Which was found in the previous formula Was 160. So we're going to take the square root of 852 divided by 160. And you are going to get a standard deviation of approximately 2.3, 1 part C is asking you to determine a probability distribution for the random variable X. So I'm going to move my calculator out of the way and we're going to create now a new chart, a different chart and this chart is going to have the same X. Column. So I'm going to move down and we're going to start the columns again and again. We had X values of 123456789 10 and 20. And then this time it's a probability distribution. So that means the next column is going to be the probability of selecting someone and finding out they have one credit card or two credit cards. So I want you to remember what probability is probability of an event occurring is always favorable outcomes. Mm hmm. Divided by possible outcomes. So if we go up to the top There happened to be 23 favorable outcomes for one credit card and the possible could have been 160. So We're going to say 23 divided by 160. And then for two, The favorable outcomes is 44. And the possible again was 160. So we'll get 44 out of 160 and we're going to keep that going for each of the X values. So three would have been 43 Out of 1 60 four, was 18 out of 160 five was 13 out of 160 six has a probability of seven over 1 60 seven has a probability of four over 160 eight had a probability of three over 160 nine has a probability of two out of 160 10 had a probability of two out of 160 and 20 had a probability of one out of 160. Now, for the sake of the next part, which is going to be drawing a probability hissed a gram. I'm going to convert each of those to a decimal approximate value. So the approximate value for 23 out of 160 is 1 4.14 .28. And again these are approximate values. Then we have 27 0.11 0.8 0.4 This is about .03, This is about .02 and these are close to 0.1 So the next thing is to create a hissed a gram. So let's create our hissed a gram. So to create a hissed a gram, we are going to have our vertical axis and our horizontal axis. And we are going to have to figure out the best way to number our vertical axis. So, since the highest probability is approximately 28%, we're going to have to get as high as 28. So what if we did 246, 8, 10, 12, 14, 16, 18, 20 22 24, 26 28 and 0.30. So, the first tower has to go as tall as 0.14. So, our first tower, it's going to be okay that here Our next tower has to get as tall as 28. The next tower has to go as tall as 27. The next towers 11. Mhm. The next hour, That's to get to about .08. The next tower has to get to about .04. Then we've got .03 then we've got point 02 And second here let me just make a .03 a little bit higher. Then we had .02. Then we had .01 and .01. And before we do that last one let's number our Access going across the bottom. So this was tower one credit card to 34 567 eight 9 10. And then we should hold a space for 11:12 13 14 15 and so forth. So 20 is going to be what all the way out here? Okay so it's separated from the rest. So now if I had to describe the shape of this I would describe it as being skewed to the right or you can define it as being positively skewed. So that was part D. Let's go on to Part E. Part is asking you to determine the mean and the standard deviation of the credit cards. But we want to use this probability distribution. So in order to do that for part E, we need to apply the formula that says that the mean is equal to the sum of X times P of X. So that means we've got to create an X times P. Of X. Column. So I'm going to take the X. Value times the P. Of X. Value. And I'm going to get 23 over 160 and then I'm going to take the X. Value times the P. Of X. Value And I'll get 88 over 160. And again I'm going to keep that hole pattern going. So this is gonna be 129 over 160, Then I'll have 72, over 160 65 Over 160 42 over 160 28, over 160 24, over 160 18, over 160, 20, over 160 and 20 over 160. Now the formula says to add up that column, so if I sum up that column, I'm going to get 529 Over 160. So therefore my average using the probability distribution is 529 Over 160 which is approximately equal to 3.3. So let's take a moment. So we found the average or mean To be 33 when we use the probability distribution and a little bit earlier we found the average To be 3.3. Using the frequency distribution. Now let's go back and let's solve part. Um f actually, let's finish party, determine the standard deviation from the probability distribution. So that means we are finding sigma X. And from a probability distribution we are going to total up what you get when you do X minus U squared times p of x. So I'm going to let the calculator do some work for us. So I'm going to bring in our graphing calculator again and I'm going to go to my tables And I'm going to clear out list five. I'm going to clear out list for and I'm going to clear out list three And for this three I'm going to tell it what this formula says to do. So I need to create a P of X column In my calculator. So therefore in list three, I'm going to type in all my p. of X values. So I've got 33 divided by 160 and I'm going to leave them in as many decimals as possible. Try again. That should have been 23 Divided by 160. There we go. And you can see now the connection that it was close to 14. Then I'm going to do 44 divided by 160 And 43 divided by 1 60. So these are all my p. of x values. Okay, so now they are done. So now what I need to do is I'm gonna sit on top of L4 and I'm going to tell it to take the quantity of all the exes, Enlist one, subtract the average of 33 square that And then multiply it by the p of X. is that we just put in list three. In doing so you're going to get these approximate values so create a new column. And again, this is going to be called X minus view squared times P of X. So I'm getting approximately .76, about 46 Then I'm getting about .02 0.6 0.23 0.32 0.34 0.41 About 4 1 again 0.56 And then a 1.74. So I've got to total up that column. And when I total up that column, I will be finding my standard deviation after I take a square root. So we'll have to take the square root of that. So I'm gonna say let's add up list four. So I'm gonna say second, quit second stat. Move over to the math menu some up list four. And in doing so, I'm getting five and 13 40th. So I've got to do the square root of five and 13 40th. So so I'm going to then say square root of that value. And I'm getting my standard deviation based on the probability distribution to be approximately 2.31. And I just found standard deviation to be 231 And if we go back to earlier in the problem, we had standard deviation Being 2.31. So what this is showing is that we could either take data and find mean and standard deviation using a frequency distribution or we can find the mean and the standard deviation using the probability distribution. Now we're ready to move on to part F and in part F, you are asked to determine the probability of randomly selecting an individual whose number of credit cards is more than two. Standard deviations from the mean. Okay, so more than two standard deviations from the mean? So keep in mind what the mean is. So the mean in this case Was 3.3. So we want to talk about being more than two standard deviations away from that, so I'm going to either Add two standard deviations to it, so I would be adding two of the standard deviations, which would end up giving me A about a seven let's do it this way, let's say 3.3 plus 4.62, which would be a 7.92. So I'm looking for the probability that x Is greater than a 792. So as I go back and I look at my distribution here, I want to be Greater than having 792 credit cards. So that means there were three people, There were two people, There were two people and there was one person. So 567. There were eight total people out of the 160. So then I could say my favourable is eight Out of 160 which is 0.5 The next question is is this result unusual? And you could say yes this is unusual because we say that any probability that is Less than or equal 2.05 or five is unusual. And we got right on the money .05. The final question says to determine the probability Of randomly selecting two individuals who are issued exactly two credit cards. Mhm. So we want the probability that we pick someone And they say they have only two credit cards and then we pick someone That also says they only have two credit cards. So we'll start with the probability of two credit cards. So we're gonna come back here favourable is 44 Over possible, which is 160. So we'll say 44 Over 160. And then we have to multiply By the probability of getting another person. Well if you're picking from the same batch of people, well now there's only 43 people left that have only two credit cards. And there's a total of 159 left in the pool of people to select from. And if I were to multiply those out, I'm going to get an approximate probability of .074 37 And then it says to interpret the results. So our interpretation of the results will be if we surveyed two individuals 100 times, we would expect seven of the survey. Yeah. Two result in two people. And why did we say seven and 100 times? Because this here means seven or seven out of every 100 times. And that concludes this problem.

In this problem solving party first. So as we all know that the sample size formula is given by an is equal to p multiplication one minus p. Multiplication 04 by two by E square. So just putting all the value here in this expression so I can write 0.34 multiplication one minus 10.34 multiplication 1.88 by 0.25 square and on solving it, I get the final answer Ridge 1269 now going to solve part B. So for part B, I can write the formula of sample size yet and is equal to P multiplication one minus p multiplication Zadar for wide to buy e square. So simplifying it further I can ride, evaluate 0.25 multiplication 1.8 it by 0.25 Holy square And on simplifying it I get the answer at 1414 So this is the answer for part B. I hope you understand it


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Questiou#2:Show that the iterative procedure for evaluating N by using secant method can be written 711 "- x P-1 ~-)+N (x -Xn- x} -x3-] n+l Then use it to find the third approximation x, of the square root of 9 using the initial approximations 10 =2 41=25,...
5 answers
Jg JI .8 (abii 3) If & = 2 - Vr-y+3 'y(0) = -1, theny(2)36
Jg JI .8 (abii 3) If & = 2 - Vr-y+3 'y(0) = -1, theny(2) 36...
5 answers
Question 8 (3 points) A 3.5 kg block is at rest on a 27 degree inclined plane: Determine the normal force (in N) acting on the block by the plane_Use 9.8 m/s? for g. Enter your answer accurate to one decimal place:Your Answer:Answer
Question 8 (3 points) A 3.5 kg block is at rest on a 27 degree inclined plane: Determine the normal force (in N) acting on the block by the plane_ Use 9.8 m/s? for g. Enter your answer accurate to one decimal place: Your Answer: Answer...

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