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Question5: (20 points)(A) A major store is interested in estimating the mean amount its credit card customers spent O their first visit to the chain'$ new stor...

Question

Question5: (20 points)(A) A major store is interested in estimating the mean amount its credit card customers spent O their first visit to the chain'$ new store. Seventeen credit card accounts (n=17) were randomly sampled and analyzed with the following results: and S (Standard deviation) = 22. a) Construct a 95% confidence interval for the mean amount its credit card customers spent on their first visit to the chain'$ new storeb) Interpret the results (the interval) you got.

Question5: (20 points) (A) A major store is interested in estimating the mean amount its credit card customers spent O their first visit to the chain'$ new store. Seventeen credit card accounts (n=17) were randomly sampled and analyzed with the following results: and S (Standard deviation) = 22. a) Construct a 95% confidence interval for the mean amount its credit card customers spent on their first visit to the chain'$ new store b) Interpret the results (the interval) you got.



Answers

Constructing Confidence Intervals Use the data set to (a) find the sample mean, (b) find the sample standard deviation, and (c) construct a $99 \%$ confidence interval for the population mean. Assume the population is normally distributed. If convenient, use technology. GPA The grade point averages (GPA) of 15 randomly selected college students $$\begin{array}{llllllll} 2.3 & 3.3 & 2.6 & 1.8 & 0.2 & 3.1 & 4.0 & 0.7 \\ 2.3 & 2.0 & 3.1 & 3.4 & 1.3 & 2.6 & 2.6 & \end{array}$$

And this problem we're given three different scenarios and we want to know when can we use this E interval, a procedure to find a confidence in the world for these um different different situations. So the first one for a It says that our sample size is 10. Mhm. So on this one we're not going to use we're not going to use the z integral. Right? Because the sample size is less than 30 and that's our standard right there. So if our sample size is Um equal to or greater than 30 then we can use the c. interval. If it's not greater than 30 then we're not going to use this interval. So let's look at our other choices to be has a sample size of 72 sorry, 75. So on this one, yes we could use the the interval because the sample size Is greater than 30. And then for our last one answer choice C has your choice but section see our sample sides is 20. So in this case we're not going to use the sea interval mm because the sample size is less than 30

Just Siles are the points that mark off the lowest 10% and highest 10% of a density curve. So this problem is asking us to find the death styles of the standard normal distribution. So first thing we wanna do is we want to look at the standard normal distribution. The standard normal distribution is are bell shaped curve, and our units are the Z scores. So with the standard normal curve, your average is zero, and your standard deviation is one. So all of these numbers on this axis are your Z scores. And what we're trying to find is the bottom 10%. So where would the bottom 10% B So there's 10% in this tale and then where would the top 10% bay? So that means there's 10% in this tale. And by doing that and keeping in mind that the full standard normal curve has an area of one, that means there's 80% in here. So we're trying to find this cut off that separates the bottom 10% and this cut off, which separates the top 10%. So to find the bottom cut off, we're going to use your inverse norm feature on your calculator, and in doing so, you need to talk about the area in the left tail, the average and the standard deviation. So to find the bottom or the left death style, we're going to use inverse norm. The area in the left tail is 0.10 The average of the standard normal curve is zero, and the standard deviation is one. So I'm going to bring in the graphing calculator and we're going to hit second. There's and we're going to do number three inverse norm 0.10 comma zero comma one. And we find that one of the death styles is approximately negative. 1.28 Now, to find the other deaths style, I'm going to change colors for you. So to find this death style right here, we're going to do Z equals in verse, norm. And this time going to the left would be 80% plus 10%. So we would have to use 0.90 for the going into the left tail comma. Zero comma one. So again, I'm going to bring in my graphing calculator and we're going to do second. There's inverse norm 0.90 comma, zero comma one and we find that the other deaths style is positive. 1.28 So for part a, the death styles of the standard normal curve are Z equals plus and minus 1.28 part to be part B is asking us what the death styles are for a distribution about the heights of young women. So again we have our curve, and this time our horizontal axis will be the heights measured in inches. And we know that the mean height is 64.5 and we know that the standard deviation of the heights of women would be 2.5 inches. And we want to find the death styles of that distribution. Well, keep in mind that the Death Siles are still in the same place at Z scores of negative 1.28 and positive 1.28 which will separate the 10% the lowest 10% and the highest 10%. So we have all the information we need to go ahead and find the corresponding raw pieces of data, or are ex values associated with each of those Z values So we have a formula for a Z score that reads Z equals X minus mu, divided by sigma. So for this ex score, we're going to say negative 1.28 equals our unknown X value, minus the average 64.5 over the standard deviation of 2.5. And if I find my cross products meaning I'm going to cross multiply, I will get X minus 64 5 is equal to 2.5 multiplied by negative 1.28 And then if I add 64.5 to both sides, I will get X equals 64.5 Plus what I get when I multiply 2.5 times negative 1.28 So this lower death style is going to be 61.3 inches. To find the upper death style. I'm going to use the same formula, and I'm going tohave Z equals. And in this case, my Z is positive. 1.28 So I'll have one positive 1.28 equals X minus 64.5 over the standard deviation of 2.5. Again, I'll do my cross products so I'll have X minus 64.5 equals 2.5 multiplied by 1.28 And if I add 64.5 to both sides, I will have 64.5 plus 2.5 times 1.28 yielding the second death style to be 67.7 inches. So for Part B, the death Siles of the distribution with a mean equal to 64.5 and a standard deviation of 2.5 will be 61.3 inches and 67.7 inches.

Okay for this problem we're looking at, um, some weight values, overweight men. So we're looking at a random sample of 60 and we're going to find confidence intervals. So it's, you know, it's really no over here, so we know that we have, ah, sample of 60 came. And we know that the number of men so from our sample are X bar next cream you, the sample mean equals two £30. So these men's on average were £30 based on the 60 men, and the standard deviation of the sample equals to £4.2. And what we want to know is we want to know what the point estimate ISS A. What's the point Estimate? Well, the point estimate means the population are best prediction for the population is the same thing as the sample mean, so we know our best estimate is £30. Based on the data we have, that's pretty part B asks us to find you 95% confidence interval of this. So maybe a conference in every will see, I what? We're gonna use our technology for this. We're gonna use the calculator we see up over here So when you stop and tests and we're going to get into a paged work here going to get into a ah zi interval so and we have not dated. But we have statistics. Silveria, look at our statistics here, so use what we kind of written down already. So we have been told already that the, uh, 4.2 the standard deviation that was given to us. So plus or minus 4.2 of the typical deviation from the mean and the mean value off what we were told was that £30 overweight. But these men were, and this is based on a sample size of 60 and the confidence interval of the confidence level will question be asked us to find wants us to find the 95% confidence level. So by doing this, the car this is going to tell us how wide oven interval we need to be to be confident that we're going to capture the Truman mean weight of these men. So the confidence interval for this means that the weights are between 28.9 for 95% confident that the true mean is between 28.9 and £31.1 to be 95% certain. Guess there's 95% confidence interval and then Part C asked. Justifiable. What if we want to be 99% sure 99% confident that were captured, the true meaning? Well, a good thing to know is that when you up your certainty, it's going to make it a little whiter in terms of, uh, the spread of what our prediction is. So we're gonna do the same thing. Test it is the interval and our data's deliver. Next, we just put it in 74.2 30. But we use me to change this to 8.99 when the 99% certainty certain along for 1% error in our prediction and so we can see that the for the 99% confidence interval the range of values. The true prediction is between 28.6 and £31.4 and party is the best question. It says which one is whiter? I've kind of mentioned it before, but we'll answer it out. So, uh, the 99% confidence terrible is wider because we want to have a more certainty. So look at the question here was that you have always larger and why, Uh, that was 9% confidence. Intervals is whiter because we want more certain TV. Want more certainty about possible values. So our spread of our prediction is whiter way confidence that holds for means of two different confidence levels.

Mhm. So we have three questions on this one, A. B. And C. Would we be using a One of the Z. Formulas? Yes. For the first one it's moderate size and it has no outliers. 20 is of moderate size, it has no out leaders. Outliers and it's normally distributed. So yes we would use that and let her be again it's moderate size but it must be normally distributed basically. It can be awful little but um the fact that it's extremely skewed means absolutely not and the letter C. Is a large sample. So that's a yes because it's generally used for anything. Um And there's no outliers on this, so we have nothing to worry about.


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