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B) A manufacturer of bus engine claims that the life of the bus engine is normally distributed with standard deviation equal to 2.1 years: A random sample ofl12 of ...

Question

B) A manufacturer of bus engine claims that the life of the bus engine is normally distributed with standard deviation equal to 2.1 years: A random sample ofl12 of these engine has standard deviation of1.8 years Test the hypothesis that the standard deviation of the life of the bus engine is smaller than 21 years at 5% significance level (5 marks)[15 MARKS]

b) A manufacturer of bus engine claims that the life of the bus engine is normally distributed with standard deviation equal to 2.1 years: A random sample ofl12 of these engine has standard deviation of1.8 years Test the hypothesis that the standard deviation of the life of the bus engine is smaller than 21 years at 5% significance level (5 marks) [15 MARKS]



Answers

A car manufacturer claims that the miles per gallon for a certain model has a mean equal to 40.5 miles with a standard deviation equal to 3.5 miles. Use the following data, obtained from a random sample of 15 such cars, to test the hypothesis that the standard deviation differs from $3.5 .$ Use $\alpha=0.05 .$ Assume normality. $$\begin{array}{llllllll} 37.0 & 38.0 & 42.5 & 45.0 & 34.0 & 32.0 & 36.0 & 35.5 \\38.0 & 42.5 & 40.0 & 42.5 & 35.0 & 30.0 & 37.5 & \end{array}$$ a. Solve using the $p$ -value approach. b. Solve using the classical approach.

You know that someone takes the bus five days per week to their job. We're told that the waiting times until they can board the bus or a random sample from a uniformed distribution on the interval from 0 to 10 minutes in part a were asked to find the probability density function and then the expected value of the largest of the five waiting times. So to find the probability density function first we have that because X, which is the waiting time, is uniformly distributed on the interval from 0 to 10 minutes, we have ffx is equal to 1/10 minus zero, which is 10 so 1/10 and therefore cumulative distribution function for X capital F of X is X over 10 begin to roll this and therefore we have that g five of y. This is the largest of the five waiting times by a formula from this section is equal to five times why over 10 so ever. Why, to the five minus one is fourth power times lower case F of y, which is simply 1/10 which in turn is equal to five times wider, the fourth over 10 to the fifth. Yeah, and this is only for y between zero and 10. This is the probability density function now to find the expected value. The expected value of y five largest of the times is the integral over the possible values of why so from 0 to 10 of why times the probability density function of why? Which is five times wire the fourth over 10 to the fifth de y And evaluating this integral, we get 50 over six, which is approximately 8.33 and the unit is in minutes. Expected waiting time. Largest waiting time is 8.33 minutes. Next in part B were asked to find the expected value of the difference between the largest and smallest times. Well, again from part A. We have that the expected value of why one could be found by finding G one of why, and this could be found using a formula from this section. And so, if you are to carry out this computation, we find that the expected value of Y one is 10 6 or approximately one point 66 minutes, and therefore the expected value of range y five minus y one this is a linear combinations. This is the same as the difference of expected values. Affected value of y five. I asked the expected value of Y one and using the previous exercise we have 56 minus 16th. So we get 46 which is approximately 6.67 And this is also in minutes. This is the expected value of the range in Part C were asked to find the expected value of the sample Median waiting time. Well, because we have five samples, the median waiting time, the five samples is going to be. Why three. And we have again from a formula in this section that the pdf of white three is G three of y, which is given by five factorial over two factorial times. One factorial tends to factorial times capital f of why the CDF. This is going to be squared times a pdf f of why to the first power times one minus the CDF of y. You mean second power which putting in calculations from part A. This gives us 30 times why squared times 10 minus y squared all over 10 to the fifth. And this is also for why lying between zero and 10 And so it follows that using integration or by symmetry, we find it. The expected value of Y three is going to be five course. You could also do this by direct integration in the unit. Here, again is minutes. Finally in part D were asked to find the standard deviation off the largest time. We're looking for the standard deviation of Y five. Well, first, it's find the second moment of Y five. This is the integral from 0 to 10 of why squared times the pdf of y five, which, when there was five times why did the force over 10 to the fifth de y and evaluating this integral we get 500 over seven, therefore by the shortcut formula for variance the variance of y five. Well, this is going to be the second moment of y five minus the first moment of y five squared. And so we have. This is 500 over seven minus. And from the first part, this is 50/6 squared and calculating the obtained 125 over 63 which is approximately 1.984 from which we obtain, but the standard deviation of why five is the square root of this and is equal to approximately 1.409 minutes.

So the question here is about data analyses and more specifically, we're gonna look at the seas and more specifically, we're gonna look at some statistical studies here. So we're looking at statistics, coal Um studies in this particular regard. So I'm here were given the dataset essentially representing the length of life in years of um 30 similar uh fuel pumps here. So um for part it wants us to conduct a stem and leaf plot just based on our values here for the life of the fuel pumps here. So I'm gonna give you a sample of what it should look like and then I'll let you do the rest of this. So this is going to be called the stem and that's going to be called belief. So the stem here is essentially just going to generally be the first point before the decimal place. So for example, in our dataset, we're going to have something like um in our one step we're gonna have zero and this can be three. Uhh the next one can be five. And before that that can be, for example, To hear. So what this represents is going to be 0.2 is 0.3. here And then we can go on with the ones. For example, we can have 1.2, so we can have two here, 1.3, we can have three here, We're going to have 1.0 ahead. So five again, this is gonna be one point oh, 1.2 and so on. And so you can just do the same thing with this data set and you can pick up something similar to that for me here. It asked us for a relative frequency distribution. So for example, if you split it between 0 to 11.1 to two and so on, essentially what you want to plot is how many times has occurred. So for example, from 0 to 1, let's say it occurs maybe like six times between 1.1 to 2, it can occur like maybe let's say five times. So you want to reflect, you want this thing to reflect the true data values here and lastly for part C here, it wants us to calculate the sample mean range and standard deviation, so the mean is essentially going to be the average and um when we put into our calculated this is going to be essentially 2.92 Um the range is essentially how what what's the difference between the biggest smallest value here and in this particular case it's going to be 6.3 and last week for the standard deviation, the standard deviation is calculated using a formula here, and in this case it's going to be 2.29. So these are the answers to the question.

To start this problem off, we've got to talk about what the empirical rule is. The empirical rule is a rule that governs bell shaped curves or bell shaped distributions, and as long as your data is classified as being bell shaped, the center would be where your average would be found. It would also be where your median and your mode is found. And the empirical rules states that if you were to look at your data and look within one standard deviation above and one standard deviation below your average, you will find 68 of your data. Then if you were to go two standard deviations out in each direction, it would account for 95 of your data. And if you were to go three standard deviations out in each direction. Yeah, It would account for 99.7 of the data. So therefore we can break our bell curve up into various sections. So since on the left side and the right side are symmetric, We can take the 68 and we can break it up into 34 to the left of the mean And 34 to the right of the mean. And if the four sections 12 three Four have to total up to 95 and we've already accounted for 68%. That means there's 27 left to be split between these two sections, which would then make this section .135 And this section .1 35 And if from here To hear has to give us 99.7 and we've already accounted for 95, that means there's 4.7 left over to be split evenly between the other two little sections. So if I take 4.7 and split it up, I'm going to get 2.35. So we could say we've got .0235 in this section And .0235 in this section. So we're going to use that data to answer part B in this problem. The other thing that is mentioned in this problem is asking you to calculate a Z score. So is the score is a way of standardizing every bell shaped set of data so that we can make comparisons. So what happens is the average Equates to a zero Z score. So the average would have a Z score of zero, and if you're one standard deviation to the right, it would be a Z score of one. If you're two standard deviations to the right, we would have a Z score of two. If you're three standard deviations to the right, it will be a Z score of three. If you're one standard deviation to the left, it would be a Z score of negative one, two. Standard deviations to the left of the mean would be a Z score of negative two and three. Standard deviations to the left of the mean would be a Z score of negative three. And we have a formula that can help you transition any data set or data point into its corresponding Z score, and you would do that by saying whatever the raw score is, subtract the mean and divided by the standard deviation. So now that we have a little bit of the um theory, let's answer our question. So, in this particular problem, we are talking about a certain brand of automobile tires and you are told that they're mean life span of these tires would be 35,000 miles. The standard deviation of that particular brand of tire is 2250 miles. And it tells us that the lifespan of tires have this bell shaped distribution. So we're going to draw a bell shaped curve for our data And we are going to place our mean in the center. So we're going to put 35,000 miles in the center and we are going to select three different, Um, randomly selected tires and we find that their lifespans are a 34,000. So 34,000 would be like right in here, We're going to select a tire that had 37,000, well 37,000 is going to be above the 35,000. And we're going to pick one with 30,000. So 30,000 is gonna be back here and we want to calculate the Z scores for each of these lifespans. So to find the Z score for the 37,000, We'll do 37,000 minus the average, which was 35,000 Divided by the standard deviation, which was 2250. And when you do that we get a Z score of approximately 0.89 Now we're going to do the Z score for the 34,000. So we're going to do z equals 34,000 -35000, Divided by 2250. And we will get a Z score of approximately -44. And then we're going to do the Z score for 30,000. So we'll do z equals 30,000 minus 35,000 over the standard deviation. And we're going to get a Z score of approximately -2.22 and we want to determine which of these is unusual. So any time you have a Z score greater than two Or less than -2, it's classified as unusual. If you have a Z score that's greater than three or less than negative three, it's very unusual. So therefore We had a Z score that was less than -2. So this life span of the tire is unusual. So that was part A. Now in part B. We want to do something very similar and we want to use the empirical rule to find the percentile. So again, for part B, I'm going to draw a bell shaped curve. We're going to place that average in the center and we want to Start with the 30,500 miles. So 30,500 miles would be back here. So we're going to calculate its Z score. So we do 30,000 500 -35000. And we're going to divide it by 2250. And in doing so we are going to get a Z score of, let me bring in our calculator. So we get 30,500 -35000, Divided by 2250. And we're getting a Z score of -2. So when you look up at our curve with all the bell values filled in, we could see that From the center up would be 50 of our curve. And then with the Z score of -1, we have a 34 in here and we have a .135 in there. So our goal is to figure out the percentile and the percentile, anytime you're dealing with a percentile, you want to know what percent did worse. So in essence we're trying to find this area in here. So what we'll do is we're going to think about the fact that entire curve is 100% or if we add up the areas it would be one. So I'm gonna start with the entire curve, I'm going to take away the 50%, that is in the Right half of the curve. I'm going to take away the 34%, that's between a Z score of zero and a negative one and I'm going to take away the .135, which is the area between the Z score of -1 and -2. And I'm going to be left with a .025 as the area down here in the left part of the graph. So therefore a tire that lasts 30,500 miles is in the 2.5%ile Or is at the 25%ile. Let's look at the next one. The next one we're gonna draw are bell shaped curve. Again, we're going to place our average in the center And this time we want to look at a tire that lasted 37,250 miles. So we're going to go to 37 250 mi. So again, we want to find its disease score. So we'll do 37,250 -35000 mi, divided by the standard deviation. So I'll bring in my calculator. So we're going to do 37-50 -35000. And we're going to divide that by 2250 and we're getting a Z score of one. So we want to find the percentile. So again, I want you to think about the fact that in the lower half of the bell Would be .50. That's between a Z score of zero and all the way into the left half. And if this is a Z score of one, that means there's 34 in there as well. So when we're talking percentile we're talking about the percent less. So we're talking about the entire curve to the left of that boundary line. So we could say a tire that last 3007 or sorry, 37,250 miles is at the 84th percentile Because when I add up the .34 and the .50, I'm getting a .84. And then the final part of part B is to talk about a tire that lasts exactly 35 1000 miles. So if I think of the bell shaped curve, Here's 35,000 miles. So I want to know what percent is less than that. Well, that would be half of the curve. So I could then say, a tire that lasts 35,000 miles is at the 50th percentile. So just to recap, the empirical rule governs bell shaped curves and these would be the percentages between various data sets. R Z scores would be a way of standardizing various data data sets that have different means and different standard deviations and we can convert all data into corresponding Z scores by using that formula of X minus mu over sigma. It's unusual when you have a Z score less than negative, two or greater than positive too. And it's very unusual when you have Z scores greater than three or less than negative three.

In this question. We have information about the average number of miles driven by a licensed driver. So this the average that were given is 14,000 90 miles, driven by the average driver with a standard deviation of 3500. So part of the question is asking us if somebody drives 16,000 miles, what is their Z score? So let's review the formula for finding a Z score. Z is equal to X minus X bar over s where X is the individual data value. X bar is the mean and s is the standard deviation. So we can simply substitute each of these pieces of information into the formula and calculators. He's so Z will equal 16,000 minus 14,090 divided by 3500. And when we compute that subtraction then divided by 3500 we find that the Z score is 0.55 In other words, a person who drives on average 16,000 miles is 0.55 standard deviations above average for the amount of drivers, uh, the amount of miles driven by licensed drivers. Part V says, What if the driver only drives 10,000 miles. So what is the Z score for somebody who only drives 10,000 miles? So somebody who's driving less than average. So we would expect when we put this into our formula on we do our calculations, we would expect this answer to come out to be negative because the amount of miles driven is below average. And in fact, when we do that when we do our calculations, it turns out that this is a Z score that is negative 1.17 So this person who drives on average 10,000 miles drives 1.17 standard deviations less than the average driver imports see, and has us sort of switching gears, working in the other direction. So it wants to know. Then what would the Z scores be? Not Does he scores? What would the data values be for people who have driven Brownie, whatever amount of miles where we end up with either 1.6 0.5 or zero as R Z score. So we just have to work backwards. So let me bring that formula back down. We write it over here, so we have it so X minus X bar over us. So we're going to substitute the pieces of information that we know into this formula and saw for what we don't know. So these are Z scores, so we'll start with the 1st 11.6 is equal to X. That's the piece. I don't know. That's what I'm looking for, minus the mean divided by the standard deviation. And then we need to use algebra to work backwards to solve this equation. So we will multiply both sides of this equation by 3500 and then when we're done with out, we're gonna add so 3500 times 1.6 is 5600. And that leaves us with our data value, minus the mean And then to finish solving this equation, we're gonna add the mean to both sides, and we find out that the data value that we are looking for is 1 19,000 690. Okay, so for 0.5, we're gonna follow exactly the same procedure except 0.5 will be substituted in for see, we have the same mean and the same standard deviation. So the right side of the formula doesn't change it all. And so when we follow that procedure, same thing multiply both sides by 3500 than at 14,090. We find that this X this data value is equal to 15,000 840. And that makes sense because we have a number that is above the mean but only a little bit of off the meat on Li like half a standard deviation on Li like 1717 50 above the meat. Okay, I'm and finally our last one. What if Z is equal to zero and we don't even need to do any calculation here of C equal to zero means what? That tells us that we're looking for the mean what data value is zero standard deviations above or below the mean And that is, in fact the mean. So that would mean that the data value would have to be 14,000. So somebody who drives 14,090 miles on average drives the mean amount of miles. In other words, there's the score would be syrup and no Cappy, no calculations are involved. Just a little bit of reasoning


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