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Follow-up to report on gas consumption consumer group conducted study of SUV owners to estimate the mean mileage for their vehicles A simple random sample of 86 SUV...

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Follow-up to report on gas consumption consumer group conducted study of SUV owners to estimate the mean mileage for their vehicles A simple random sample of 86 SUV owners was lected and the owners were asked to report their highway mileage. The results that were summarized from the sample data were x=18. mpg and $ = 5.2 mpg: Based on these sample data compute and interpret 9996 confidence interva estimate for the mean highway mileage for SUVsThe 99% confidence interval is mpg mpg (Round t0 one

follow-up to report on gas consumption consumer group conducted study of SUV owners to estimate the mean mileage for their vehicles A simple random sample of 86 SUV owners was lected and the owners were asked to report their highway mileage. The results that were summarized from the sample data were x=18. mpg and $ = 5.2 mpg: Based on these sample data compute and interpret 9996 confidence interva estimate for the mean highway mileage for SUVs The 99% confidence interval is mpg mpg (Round t0 one decimal place as needed. Use ascending order)



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The fuel economy information on a new SUV's window sticker indicates that its new owner can expect $16 \mathrm{mpg}$ (miles per gallon) in city driving and $20 \mathrm{mpg}$ for highway driving and $18 \mathrm{mpg}$ overall. Accurate gasoline records for one such vehicle were kept, and a random sample of mileage per tank of gasoline was collected: $$\begin{array}{llllllllll}\hline 17.6 & 17.7 & 18.1 & 22.0 & 17.0 & 19.4 & 18.9 & 17.4 & 21.0 & 19.2 \\18.3 & 19.1 & 20.7 & 16.7 & 19.4 & 18.2 & 18.4 & 17.1 & 17.4 & 15.8 \\17.9 & 18.0 & 16.3 & 17.5 & 17.3 & 20.4 & 19.1 & 21.0 & 18.1 & 19.0 \\19.6 & 18.9 & 16.8 & 18.2 & 17.6 & 19.1 & 18.0 & 16.8 & 20.9 & 17.9 \\17.7 & 20.3 & 18.6 & 19.0 & 16.5 & 19.4 & 18.6 & 18.6 & 17.3 & 18.7 \\\hline\end{array}$$ a. Determine whether an assumption of normality is reasonable. Explain. b. Construct a $95 \%$ confidence interval for the estimate of the mean mileage per gallon. c. What does the confidence interval suggest about SUV fuel economy expectations as expressed on the window sticker?

Roll. Nine deals with fuel efficiency of cars and trucks and the question Give us the sample size of 40 a sample mean of 28.1 mpg a population standard deviation of 4.7 MPG. And it's asking us to estimate the confidence interval of the true mean mileage with a 95% confidence level, we would need the following to construct our confidence interval. Next, we need to determine the Z value that's associated with a 95% degree of confidence. And that's 1.96 plugging in our values we would end up with. Now we need to simplify 20.1 plus and minus 1.5. Now we can construct our confidence interval and that would be 28.1 minus 1.5. True mean it's between 28.1, close 1.5. Therefore, my true mean mileage is between 26.6 months per gallon and 29.6 since my sample size had only one decimal point. My interval should also have just one decimal point. Look

So we start from the givens and is equal to 20. See, which is the confidence Interval is equal to 95%. Starting by step one. We are interested in estimating the population mean, um, you at Ah, the 95% confidence level where the mu is the main angle off the deformity in the population off all such peaches. Okay, step two. We are planning on the calculation on one simple tea interval for population means, um you know, so we need to satisfy the three conditions for the random variable. The first condition is the random condition. Yeah, this condition is satisfied because the patient's originated from a random sample. The second condition, which is 10% condition. And this condition is also satisfied because the sample off 37 patients are lead, then 10% off the entire population off all patients with a deformity on the big two. The third condition, which is normal and large sample. This also condition is satisfied because the normal probability bloods, uh, does not show strong could feature, and thus it is safe to assume that the simple originated from normal distribution. It's so since all the three conditions are satisfied, we will turn to step three. So the mean is the sum off all values divided by the number of values. So we will. Some old values, which is 15.8, lost 13.6 plus 10.6 plus 19.1. Yes, until we get to the end 20.9 over the number off values, which is 20. So the final answer will be 18.48 The variance is the some off the square deviation from the me, divided by in minus one saw the variance as or the standard deviation is the square root off the variance which is equal toe each value minus the mean square. So 15.8 minus 18.48 squared plus 13.6 minus 18.48 It's quit and and so on. Okay, this is over toe end minus one, which is 20 minus one. So the standard deviation is equal to three point 1158 Tow This remind the T value by Lincoln in the rules, starting with the degrees of freedom. So the degree of freedom is ableto end minus one, which is ableto 20 minus one. She's 19 and in the column. Controllable Confidence interval. 95% in the table. So T Alfa over two is equal to two point 093 So the marginal error, which is T Alfa over two times s over square root Off N, which is 2.93 times 3.1158 over square root off 20. She's 1.4 58 82 So the boundaries for off the confidence interval they become, it's bar, which is immune minus the error, which is 17.218 and X Bar plus p, which is equal toe 19.93 882 So still four, we conclude that 90% confidence we are 90% confident that the true mean fuel efficient for as the vehicle is between 17.218 and 19.93 eight 82 Thank you

In this question were provided with the regression analysis for car wait and fuel efficiency For part A were asked if there is strong evidence of an association between the weight of the car and its gas mileage, and we're Esther right and appropriate hypothesis. So the hypothesis, no hypothesis, would be that the slope of the regression line is zero, which is saying that there is no association between car weight and fuel efficiency. And so the alternative hypothesis is that the slope is not equal to zero, which is to say that there is an association. And so if we want to look for evidence of the association between car weight and fuel efficiency, the regression analysis is predicting this slope for the regression line. And the P value is less than or equal to 0.1 which is a very small P value. So we would conclude that there is strong evidence to suggest an association between the weight of the car and it's gas mileage and for B were asked if the conditions for inference are satisfied. So there's some conditions that we need to check. One is the linearity assumption the independence assumption. So for linearity. If you just look at the scatter plot that's provided in the question, the one where they have the line drawn through it. Ah, it looks like pretty straight data Like it. It makes a pretty nice straight line. So we could. That speaks to the linearity assumption in terms of independence. Ah, the only thing we have to look at is the scatter plot of the residuals, and we're looking for any clumping or patterns that might exist in this. We want to see kind of randomness. It looks fairly random, so that's pretty good for equal variance. Assumption. We again look at the residuals plot, and we want to see that the deviations above and below the red line do not change as you move from left to right or from right to left. And it's pretty good. It does slightly fan out or thickened as you move towards the right, like at about 30 mpg. There's a bit more variance than there is at, say, 15 or 20 but it's not too bad. It's it's almost. It's almost constant, you could say. So that's OK. And for the near normality assumption. If you look at the history graham of the residuals. It's a pretty nice bell shaped plot. It's got a nice single mode in the middle, and it's fairly symmetrical. So you could say that all of these conditions appear to be met? Not perfectly, but it's it's pretty good. And finally for Part C, whereas too, test the hypothesis and state the conclusion. So again, we're looking at these values here. So we have a negative slope and we have a T ratio of 12.2 and negative 12.2 and a P value of less than or equal to 0.1 So based on that, we would reject the no hypothesis and stated that there is evidence there is sufficient evidence to suggest that there is an association between car wait and and fuel economy or MPG.

Hi This question. We need to find the prominence interval for the many and percent so offers. Here it was a point or two and here's the formula. So you plus minus C over to assemble standard deviation. Devine was squirrel defend. I used the school color on. I found that this fellow, uh, 35,534 to this one and what I really want to show is. And when you calculate this one trying to calculate that the marginal error sometimes hope me on developed chick that's matching the the confidence of terrible that's you have.


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