Cradei6) A vehicle owner interested in her vehicle's fuel economy calculated the gas mileage in miles per gallon (mpg) each Commnunication time she filled her...

Mileage tests are conducted for a particular model of automobile. If a 98$\%$ confidence
interval with a margin of error of 1 mile per gallon is desired, how many automobiles
should be used in the test? Assume that preliminary mileage tests indicate the standard
deviation is 2.6 miles per gallon.

Yes, all right. First problem. We're back to our situation with trucks and cars talking about highway and city fuel efficiency. So in the previous problems about this, there were 632 total vehicles that were tested for both highway and city fuel efficiency. And then the differences were found. And the main difference between our of all of those all 632 right, because there were 316 cars 316 trucks. The mean difference between highway and city Fuel efficiency is 7.37 mpg, with a standard deviation of 2.52 MPG. And what we want is a 95% confidence interval and an interpretation of that. So let's start by finding the interval and then we'll interpret it if you click stat and then go over to tests. We don't want to test. Now we want an interval tests. We're gonna go down to tea interval and we're going to the stats because we have those. Unless you wanna enter 632 Ah, fuel efficiencies, fuel efficiency differences which I don't want to dio I've detected in. But we have our our X bar, or MU, is 7.37 That's our average difference. Standard deviation 2.52 and value 6 32 We want 0.5. That's a 95% confidence interval. And let's calculate. All right, there's our interval. So what we then can conclude from this is you could say, Well, say we are 95% confident rat, uh, mean difference and fuel efficiency. A little bit of highway minus said you just so we know which order it's and several different change things is between of important one so 32 mpg and 7.5668 rounds per gallon. So that is our confidence interval and our interpretation, So we're good to go on this problem.

All right. So For this question, we are given a number of car ratings and we are MPG MPG for 14 different sized cars and we want to find a range in the sample standard deviation for this dataset and all I've actually done here is I've sorted the data from smallest to largest. Um, this does two things. Firstly, it allows us to group similar terms, and secondly for finding the range, it becomes very straightforward because the range is simply the biggest to the smallest. So we can very quickly and easily see that the biggest, 17, the smallest is 11, which gives us a range of six history. All right. Um, now, when we're trying to find the mean, while the mean relies on the some divided by the total number of terms, so add them all up and divide by the total, Which is 14. Why do we need to know the mean? The right. The problem is asking for the standard deviation? Well, the standard deviation relies on the squared differences from the mean. So if we're going to find the standard deviation, we need to first find the means. So let's add it all up, divide by 14. That's going to give us a means of 14.1 miles. Which makes sense. Right? It's right here in the middle, very close to fortune. All right. So when we are finding the sample standard deviation two Square root expression where the numerator is the sum of the squared differences from the mean? So what does that mean? I mean, uh, it means that for each of our terms to start with 11, work our way up to 18. We are going to find these square difference from the mean. So some trafficking first At the next one. Right track. I mean where it I actually noticed um for several of our terms here, so 13 is copied twice. I think that's your two times. 13 minus the mean. Weird Because we're going to have two copies of those and actually 14 is listed five times. So we can do five times 14 -14.1 very 15 is listed three times. Right? So three times 15 -14 squared And then we just have 16 -14 1.1 sq and 17 17 min 14.1 sq. Um the reason why I used abbreviations is just why I use the multiplication. Just to show you that you can really group right? Group all those 14th together. It just makes your calculation easier. If you wanted you could have listed all of those out and you would have had 14 terms and that's fine. But if we can if we can shorten that better. All right. And that's our numerator in our denominator, it's going to be simply end minus one. Well, Ennis 14 so and minus one is 13. All right. Let's take out our calculators. Let's get to work. Um This ends up being in our numerator, It ends up being 30.93. So the numerator is 30.93 approx. And we are going to divide by 13. Right? The Denominator 13 You work all that out. You will get 1.5 approximately and were told to round To one decimal place. That's what weird. Miles. Uh huh. Alan.

In this problem, we're going to be running a complete hypothesis tests from beginning to end um and what we are going to do first of all is we have to kind of see what the um what the outcomes are. So what we're mainly looking at is first of all are null hypothesis is kind of what we would expect to happen. And based off of that first sentence, we would expect the average to be 19.8 MPG. However, we have gotten a sample that is higher than that and we want to know is it significantly higher? So our alternate hypothesis is that the true average is greater than 19.8. Now, some things we have to take into account here, we need to make sure it's a random sample, which it does tell us. We have a random sample of um 50 passenger vehicles. It doesn't actually say random that we're going to assume that it is also, we have to make sure our sample, Our sample size is large enough and since 50 is greater than or equal to 30, the central limit theorem tells us that our sampling distribution is normal, therefore we can use the Z scores to calculate things. So we need to calculate our Z score Are observed, sample main is 21, The expected sample mean is 19.8 and our standard deviation of our sampling distribution Is going to be 2.45 divided, bar The Square Root of 50, which will give us our standard deviation of our sampling distribution. I am doing All the parts at one time when I get a Z score of 2 0.86, which is not very high. Okay Um so then what we have to do is we have to compare it to the Z score at the 20% significance level um which can be found from the table in the back When the tail probability is in this case 80, Okay. The tail probability is 80 Because it's a 20% significance test and we're looking for that value. So if we look until we find a .8 on the table, it looks like it looks like it appears at about a Z score of .085. So that is the Z score with that left tail probability. So it's a 4.085. Just point 85. And I just found what I did was I found .8 on the chart, the area to the left was .8. And then I matched the Z score up with that. So anything greater than .85 would be rejection region in terms of um this particular ah in the terms of this particular problem. So here we go. Since our Z score we found was greater than our critical value for the rejection regions ours is gonna fall right here. Okay, we would reject the null, basically falls right off that, right at that cut off which is um no, you know, a significant level of Point of 20% of Is not that common. Um it's very high, so it doesn't tell you a whole lot. So we would reject the notion that it actually is higher than that. I don't personally like using a 20% significance level, but that is what we have there. All right, so then part B. We are going to calculate the p value And to do that, we're gonna do a normal CDF of our z score of .86 All the way to infinity with zero and 1 as our And we get .194, which again is right there at that significance level. Um but it is Less than .2 so we still reject. So again, if the Z score is greater than the Z score that matches the rejection region, you reject the null. If the p value is less than the significance level, it's almost like we're working backwards there. One of them, we know the area and we're finding the Z score, the other one. We are we know the Z score and we're finding the area and comparing the areas versus comparing the um values on the Z score curve. Um I personally like part B and C a little bit better, but a does kind of build up that conceptual understanding

Following is a solution to number 18. This looks at the 2005 Honda accord average gas mileage and or miles miles per gallon, and uh we sampled 42,005 Honda accords and we found that the average or the sample mean for gas mileage or MPG is 23 MPG, with a sample standard deviation of 1.5. And were asked to construct and interpret a 95% confidence interval based on that sample. So we need to use a T. Interval because we don't know what the population standard deviation is, We only know what the sample standard deviation is, so we're given s but we don't know what sigma is. So since we only know s we're gonna use the tea interval And I'm gonna use the IT4 for this, because it's it's pretty nice to use technology whenever available. But you can use any sort of technology want or you can certainly use the formula, but you should get the same answer. If you go to stat on T. I 84 you air over two tests and it's this eighth option down here, the T interval and you have a choice between data. If you're given a long data set, you would use that. But we're not we're giving summary stats. So keep summary stats highlighted or press enter over it. And then X bar was 23 S was 1.5 in was 40 and the sea level is 400.95 So then we can calculate And this top band here that gives us the confidence interval. So 22.52 and 23.48. All right, so let's write that down to 22 0.52 And 23.48. Okay, So that's our our confidence interval and then we need to interpret it. Now these interpretations all kind of follow the same pattern, the same structure, which is like so, so we're going to say we are, or we can be 95% confident that the mean gas mileage or MPG For 2005 Honda accords Is between 22.52 and 23.48 MPG.