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11 Suppose that the average annual estimated population standara cost of automobile insurance is 8850 with an deviation of S245.a. What is the probability that auto...

Question

11 Suppose that the average annual estimated population standara cost of automobile insurance is 8850 with an deviation of S245.a. What is the probability that automobiles a simple random sample of size 150 insured will have a sample mean insurance cost greater than $9502Answer:What is the probability that a simple random sample of size 50 insured automobiles will have a sample mean insurance cost within $25 of the population mean?Answer:12. The owner of a local restaurant wants to estimate the

11 Suppose that the average annual estimated population standara cost of automobile insurance is 8850 with an deviation of S245. a. What is the probability that automobiles a simple random sample of size 150 insured will have a sample mean insurance cost greater than $9502 Answer: What is the probability that a simple random sample of size 50 insured automobiles will have a sample mean insurance cost within $25 of the population mean? Answer: 12. The owner of a local restaurant wants to estimate the average amount spent per customer for dinner at the establishment Data was collected related to 81 customers_ a. If it is assumed that the population standard deviation for the amount spent is $5, what is the margin of error in estimating at the 90% confidence level the average amount spent? Answer: b. If the sample mean amount spent is $24.83,what is the 90% confidence interval for estimating the average amount spent per customer for everyone who eats dinner at the establishment? Answer: . .



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?

All right. So this question gives us some information about spending on fast food and party wants us to compete the margin of error for a 95% confidence interval. So we know that margin of error equals since we only have sample standard deviation, we have to use a tea distribution. So it's t star times our sample standard deviation divided by the square root of our sample size. So we need to find our T star. So to do this, we can draw normal curve, and we know that 95% of the area it must be between negative t star and T star, so we can compete the area in each tale, which is 0.0 to 5. So to find our t star value, we just do inverse t of our area 0.975 andr degrees of freedom, which is just the sample size of minus one, which this turns out to be 1.990 So now we have everything we need to computer margin of error. We have our tea star value times air sample standard deviation divided by the square root of our sample size. And this gives us a margin of error of 1 22 point 369 Now it wants us to compete the confidence interval. So our interval is our estimate, plus or minus our error. And we know both of these. So our mean was given his 18 73 closer minus our standard margin of error from the last question. And this gives us a range of 17. 50 0.63 as are low and 1995 point 37 is our high. So we're 95% confident that the population mean lie somewhere in this interval. No, for part C, it wants us to estimate the total expenditure from this sample. So for our estimate, we estimate that a single person spends on average X bar equals 18 73 and that's in dollars. And we have 80 people in the sample, so we have $1873 per person times 80 people and that gives us a total expenditure of 1 49 a 40 for this whole group because we use our point estimate and multiply that by the number of people to get an expected total. Then Part D says that this is skewed, right? So do we expect the median to be greater or less than the mean So help? Let's draw a right skewed distribution. So this is right skewed. So for a right skewed distribution, our median would be here. But our mean would be over here because even though crest of this hill up here is the median of the data, we have these data points in the tail over here that are artificially dragging up the value of mu, so we'd expect the median to be less than the mean.

Number 16, which is a each note. I think that new one is wonder than or equal to YouTube. Each one is that new one is bigger than you two. So the critical value stated in table four corresponding to probability off open 99. So that is equal to 2.33 So the distance statistic in that if they quinto x one more, one is six to bar minus me one minus for you too over square root off signal only squared often anyone waas signal two squared over. And to which approximately equal toe 0.21. So in the very on the this is it within the rejection reason which is the old values greater than 2.33 so as to point someone is smaller than 2.33 So we fail to reject the null hypothesis. So there is no sufficient evidence to support that claim.

The following is a solution video to number 25 and this looks at a very small data set. In fact, there are only four data values of the amount that it costs for repairs for chevy cavaliers. And the first part of this asked for a point estimate for the mean and the point estimate for me, and it's just gonna be the sample means. So X. Bar. And we're also assuming that the sigma, the population standard deviation we know it is $220. So um the way you find X bar is just, you know, finding the average the arithmetic mean. So you add them together and divided by four Or you can use technology and I'm gonna use technology. So if you go to staten edit their your data values. So one of them was $225, 462 7 29 and 7 53. And then if you go to stat and then couch and it's one of our stats and we're gonna make that L. One And we calculate that. And we get this this x bar here. 542.25. Okay. So the point estimate is 542 points $25. Okay. And then in part B were given a box plot and a normal probability plot. And we're basically just checking to make sure that the distribution of the population is approximately normal. And looking at those two, it looks it appears to be approximately normal. So the short answer yes, we can use the Z interval here, even if the sample size is so small, because um the box plot appears there are no outliers and the normal probability plot looks like it's about linear. So, so in short, we're just going to say yes, it's approximately normal. The box plot says that there are no outliers. And since the normal probability plot is approximately linear, so the normal probability plot is within range will say, Which means that none of those data values are kind of outside those lines. And then the next couple of parts, we're gonna go ahead and use our inference procedure and we're gonna find the 95% confidence interval. And I'm gonna go back to technology and do that. Now. You can use the Formula film, but technology is much faster, so I'm going to do that. So if you get a stat test and we're gonna use the Z interval since we know what the population standard deviation is, So it's the seventh option. And usually we leave it on stats because we're giving summary stats, but this time we're actually giving a data set, so I want to leave it on data. Now. The sigma Is was given to you in the problem is 220 for your list. Go and change that to L. one unless you put it in different column and then put it as L. Two or L. Three or whatever. But I put in L. One and then the frequency is one and then the sea level, the confidence level is 95% of 950.95 Then we can calculate. And then this front first line here that gives us the 95% confidence interval. So 326.65 and 757.85. It's between those two. So it's gonna write that down before I forget it. So 326 0.65 2757 0.85 And then we also need to interpret that's the way we interpret that. We'll say we can be 95% confident that the mean repair cost mm for all chevy cavaliers is between mhm $326.65 cents and $757.85. So between those those two, Okay then the third part or four fires should say the fourth part is a 90% confidence interval. We're actually going to compare these two. Now I'm not gonna write everything down. Um Now we're gonna do the same thing. So we go to stat and then test and we're still using that Z. Interval. So go to the seventh option there. Z interval not the T. And everything stays the same except the sea level. We're gonna change 2.9. And we calculate that we get these numbers. Let's go and write these numbers down and we'll just compare it real quick. So 90% confidence interval between 361 0.32 Let's draw a little line here so we see it. Okay? And then 7 23 point 18 and then whenever we interpret this, I'm not gonna write everything down because it's basically the same thing. So we can be 90% confident everything else stays the same and then, you know, confident that the mean repair costs for all chevy cavaliers is between between $361.32 And $723.18. Okay, So everything else, you know, basically stays the same. So let's look at these two confidence intervals, the 92 the 95. So how do they compare? Well, the 90% confidence interval is actually a little bit narrower, so the lower bound is higher than the 95% and the upper bound is lower than the 95%. So it's a smaller confidence interval, so the width decreased as the confidence level decreased And the next part of that is is this reasonable? Yes, it is. If you just look at those critical values, so if you look at the margin of air, the critical value for the 95% confidence interval Is 1.96, whereas for the 90% confidence interval that Z star is 1.645. So we're multiplying by a smaller number whenever it's 90% confident. But also, you know, if you just think about logically the more confident you you have to be, the wider you need to make that interval. Otherwise we would just want to make it, you know, 99.9% confident every time. And so that that's the trade off. If you want to be less confident, well then you can narrow down that confidence interval and you can give you better information. But if you want to be more confident, then you do need to widen that confidence interval up just a little bit. So yes, it is reasonable. Um, as the confidence level decreases, you do, you will have a narrower confidence interval, which is a good thing.

21 eight The confidence interval export postive or negative all over two times. Sigma always square Roto eso as explore is 30. Was the war negative that off over to 1.96 times 12 over square, not 49. So the interval is 26.64 to 33.36 eso The marginal off error is it's the difference between, um, the limits over two, which is 3.36 So for question being, um, using using, um and is equal 200 instead of 49. So the interval is 27.648 to 32.352 with marginal of error is 2.352 For questions, see, using n is equal toe 2 to 5. So the interval is 28 0.432 31.568 through its modern off is 1.568 which is the differences between the two values over to For question D. We know that a Zen increases the marginal off decreases and for questioning as error increases. As you see here, the interval limb decrease the source


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