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CNNBC recently reported that the mean annual cost of auto insurance is 1026 dollars. Assume the standard deviation is 242 dollars. You take a simple random sample o...

Question

CNNBC recently reported that the mean annual cost of auto insurance is 1026 dollars. Assume the standard deviation is 242 dollars. You take a simple random sample of 89 auto insurance policies_Find the probability that a single randomly selected value is at least 975 dollars. P(X 975)Find the probability that a sample of size n 89 is randomly selected with a mean that is at least 975 dollars_ P(M 975)Enter your answers as numbers accurate to 4 decimal places_

CNNBC recently reported that the mean annual cost of auto insurance is 1026 dollars. Assume the standard deviation is 242 dollars. You take a simple random sample of 89 auto insurance policies_ Find the probability that a single randomly selected value is at least 975 dollars. P(X 975) Find the probability that a sample of size n 89 is randomly selected with a mean that is at least 975 dollars_ P(M 975) Enter your answers as numbers accurate to 4 decimal places_



Answers

The mean annual cost of automobile insurance is $\$ 939(C N B C,$ February $23,2006)$ Assume that the standard deviation is $\sigma=\$ 245 .$
a. What is the probability that a simple random sample of automobile insurance policies will have a sample mean within $\$ 25$ of the population mean for each of the following sample sizes: $30,50,100,$ and 400$?$
b. What is the advantage of a larger sample size when attempting to estimate the population mean?

The following is a solution to number 15, and this looks at auto insurance rates And we're comparing them 2002. So in 2002, the mean, uh Premium for Auto Insurance Rates with $774. and we're comparing it a few years later in 2005 with This data. So we sampled 35 individuals and the sample mean was 735, which is lower It is definitely lower, but we don't know if it's low enough and it's got a sample standard deviation of 48.31 and we're testing at the 1% or the point a one alpha level to see if there's been a change In auto insurance rates from 2000 to 2005. So first off, let's say what kind of test we're doing. We are doing a T. Test. And the reason why we were doing a T. Test is we're testing a population mean. So it's either going to be the Z or the T. Test and it's a T. Test because we don't know the population standard deviation. So if we were given sigma, the population standard deviation, um then we could use the Z. Test but were only given the sample standard deviation. So we have to use the T. Test. So there's a five step process to these hypothesis tests. The first part is To state our two hypotheses. So the alternative or started the null hypothesis is that the mean is 774 back from 2002. And the alternative is that it's different. So we don't know if it's less than or greater than. We're just wondering if it's different few years later. Okay, the next two steps, we're gonna use a calculator. Now you can use formulas and charts and what not, but if you have technology, I really suggest you use it. It can cut down big time. This way, you don't have to use formulas. Um So I'm gonna use the T. I. 84 because it works out pretty nicely. And if you go to stat and then air over two tests, we're gonna do this t test right here. Some option to summary steps should be highlighted. Community is 774. X bar is 735 That's the sample mean the sample standard deviation, that's that. Sfx is 48 31 And the sample size is 35. And then that alternative hypothesis is not equal to. So then whenever we calculate, that gives us everything we need, so negative 4.776 And then the p value is 3.347 times 10 to the negative fifth. So I always have students that say, okay, the p values 3.3, which is bigger than alpha. No the P value is never ever ever bigger than one. P value stands for probability. It's always gonna be between zero and one. So if you see this three point whatever, just know. It's written in scientific notation is a very very small number. This is pretty much 3.3 times 10 to the negative fifth is basically zero. So know that it's a very small a number. So we're gonna go and write those down. So the T value remember was -4776. And then the p value of 33 times 10 to the negative fifth. So that people is very important because what we do Is we explicitly compare it to the alpha and like I said this P value it's almost zero, which is less than .01. So whenever the P values less than alpha than we reject. H. Not. Okay, so we're rejecting the null hypothesis that the mean is $774. So the final statement or the final step is to conclude. So we're going to say that there is convincing evidence to suggest that the mean expenditure for auto insurance is different from The 2002 amount, So it is different than that $774. So that is a. T. Test five step process using the P value method to test if auto insurance has changed or not.

Alright, this problem is about 84. Used, uh, cars. It's a sample so are in value is going to be 84 and they gave us some more data on the sample. They told us the sample mean was $6425 and they told us the sample standard deviation was $3156. This problem does come in four different parts. And when we get to part, see, there's parts within parts C. So let's tackle part. A party is asking us which distribution should you use for this problem and you should be using the student's T distribution. And the reason we are using the student's T distribution is because we do not know the population standard deviation. We do have a standard deviation that was provided us, but it is referring back to the sample of 84 used cars. So because we do not know the population standard deviation, we will use the student's T distribution for Part B. It's asking us to define the variable X bar inwards, so X bar is going to represent the mean cost of the 84 used cars. Now for part C. There are three parts to it, and I'm going to tackle them a little out of order. So in part C, it's asking us to come up with the confidence interval to sketch the graph and to calculate the error bond are bound in order to come up with the confidence interval. We do have to find the error bound first, so let's tackle that. So we're trying to come up with a confidence interval at the 95% level. And in order to do that, we're using the formula that says Arrow error bound of the mean equals T sub alfa over to multiplied by the standard deviation of the sample divided by the square root event or the square root of the sample size. So because we're trying to find the 95% confidence interval, we're going to put 0.95 in the center of that bell shaped curve. The Alfa, in this case is three remaining part of the bell, and if 95% is accounted for, that means there's 950.5 unaccounted for, so we need to find the T value standard T value associated with Alfa over to so Alfa being 0.25 Alfa over two is going to be a 20.25 And that's referring to how much is in this left tail and the area that's in the right tail. And in order to calculate this t value, we're going to have to use our graphing calculator and we're going to access the inverse T function. And when you use that function, it does ask you for the area of the curve that's in the left tail. So it's gonna be 0.25 And it also asks you for the degrees of freedom, and our degrees of freedom is always going to be n minus one. In this case, we were using 84 um, in our sample. So therefore are degrees of freedom is 83. So I'm gonna bring in the graphing calculator and show you where in verse t is. So you're going to hit the second button, the distribution or variables button and number four. And it's going to ask you what is the probability or the area in the left tail, which is 0 to 5, and it asks you for your degrees of freedom and your degrees of freedom in this case was 83. So therefore our T value is negative. One point, uh 989 So what is that? Referring thio That negative 1.989 is referring to the lower boundary of this confidence interval. And because the T distribution is symmetric, then this upper bound is going to be positive. 1.989 So, in our formula, the error bound of a mean we're going to use the 1.989 for the tea value. Our s value in this problem was 3156 and our sample size was 84. So therefore our error bound is 684.91. So when I said we were going to do the problems out of order, what we have just found is the answer to part three of letters C. So now let's go back and tackle part one off the part C and that's asking us to actually come up with the confidence interval. And it's I always like to dry picture to kind of reference this confidence interval. So what we will do is we're gonna put the average in the center, and we're going to add the error bound for the mean and subtract the error, bound for the mean in order to get the low boundary and the upper bound the boundary of the confidence interval. It's giving us that wiggle room so our confidence interval at the 95% level is going to be X bar, minus the error bound and then x bar, plus the error bound. So in terms of the numbers of this problem, it's gonna be 6000 425 minus 684.91 and 6425 plus 684.91. So that is going to result in a confidence interval of $5740.9 and an upper boundary of $7109.91. So what we have just found is the confidence interval, the 95% confidence interval for this data and it is part one or I of part C eso There's one more part in part C. It asks you to sketch a graph. So we're going to go back to our bell shaped curve and we're going to put the low bound on the left side of the confidence interval. We're going to put the upper boundary on the right side and right in the center, we're going to include our average, which is 6400 25. So this refers to the 95% confidence interval, and that picture is going to be part two of part C in this problem. All right, There's one last part to this problem Part D. And in part D. You are asked to explain what a 95% confidence interval means for this study. And it would mean that if we were to sample n equals 84 many times. So we're gonna grab a sample of 84 cars, then we're going to grab another sample of 84 cars. Then we're gonna grab another sample of 84 cars. We're gonna do that many times than we would expect to see 95% of the confidence intervals that we have generated. They're going to capture the true mean. So again, I like to draw an image to go along with this meaning as well. So the picture that I would draw would look like this. So we've got our true mean and in this case, we really don't know what it iss. All right, we have a sample mean but we don't know the true mean. So our true mean is right here, and it's saying that 95% of the different confidence intervals that we would create would capture it. So maybe we'd have one confidence interval here. And we have another confidence interval here and another confidence interval that spans this with and there's another confidence interval. But spans this myth this with 95% of the confidence intervals going to capture that true mean. But there is going to be 5% of the time that we don't capture it. So, for example, we might end up with the lower boundary being higher than the true mean. So notice this last one this interval here did not capture that true me

Mm hmm. For the solution, Step one. Find the 98% confidence interval now next step from the given information. The sample size is 50. The value of Sigma is dollar 5.92 and the sample mean is dollar 18.21 The formula for confidence interval is as follows that X bar miners in brackets that a by two multiplied by sigma by underwrote and is less than mu is less than X. Ba plus in brackets had a bye to multiply by Sigma, divided by underwritten their expertise is denoted as simple means next step. Therefore, this confidence level can be calculated as follows that by substituting the values we get 18.21 minus that 0.2 by two multiplied by nine point 5.9 to develop 100 15 less than muse. Less than 18.21 plus 0.2 by two multiplied by 5.92 divided by 100 50. By solving this, we get 18.21 minus in bracket 2.33 multiplied by 0.837 It's less than amused. Less than 18.21 plus and back at 2.33 multiplied by zero point 837 So from there you exist between 16.26 and 20.16 Thus the 98% confidence interval for the mean, the for all cars is dollars 16.26 earned 20.16 So this is the solution. Step by step in detail with definition. Please go through this. Thank you. Right. No.

Question 83 now a three. We don't know anything about the distribution, and N is not greater than 30. So this wish gives a little pause. Except for that, they're saying the normal price of gas is 4 59 and we want to find the probability that the gas price for 16 as an average of X greater than for 69. So it's a fairly remarkable question in that our answer is gonna be extreme. So when I do my normal CDO, I'm looking at a lower band of for or You nine, the positive infinity nine or 25 9 as my mean and white one divided by the square root of 16. And when I do that, my answer gives me 3.17 fines, 10 to the negative five. So I would have moved the decimal five places to the left. Why three war or I would write it as played 00 three you person. So it's a fairly small percent, almost zero, which is a So the only reason we didn't say that is because it is so extreme


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