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*Chapter 10, Section 10.3, Problem 031 An insurance company wants to know if the average speed at which men drive cars is higher than that of women drivers_ The com...

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*Chapter 10, Section 10.3, Problem 031 An insurance company wants to know if the average speed at which men drive cars is higher than that of women drivers_ The company took random sampl of 26 cars driven by men on highway and found the mean speed to be 71 miles per hour with standard deviation of 1.95 miles per hour; Another sample of 16 cars driven by women on the same highway gave mean speed of 68 miles per hour with standard deviation of 2.35 miles per hour Assume that the speeds at which al

*Chapter 10, Section 10.3, Problem 031 An insurance company wants to know if the average speed at which men drive cars is higher than that of women drivers_ The company took random sampl of 26 cars driven by men on highway and found the mean speed to be 71 miles per hour with standard deviation of 1.95 miles per hour; Another sample of 16 cars driven by women on the same highway gave mean speed of 68 miles per hour with standard deviation of 2.35 miles per hour Assume that the speeds at which all men and all women drive cars on this highway are both normally distributed with unequal population standard deviations Let /1 be the mean speed of cars driven by all men on highway, and //z be the mean speed of cars driven by all women on the same highway: Construct 95% confidence interva) for the difference between the mean speeds of cars driven by all men and all women on this highway: Round your answers to two decimab places_ miles per hour Test at 29 significance level whether the mean speed of cars driven by all men drivers on this highway is higher than that of cars driven by all women drivers_ We can conclude that the mean speed of cars driven by all men drivers on this highway is that of cars driven by all women drivers_ Suppose that the sample standard deviations were 1.55 and 3.25 miles per hour; respectively. Redo parts and Discuss any changes in the results_ Round your answers to two decimal places_ The new confidence interval miles per hour We can conclude that the mean speed of cars driven by all men drivers on this highway is that of cars driven by all women drivers Because the standard error and critical values are both hypothesis test the confidence interval The conclusion for the



Answers

What’s wrong? A driving school wants to find out which of its two instructors is more effective at preparing students to pass the state’s driver’s license exam. An incoming class of 100 students is randomly assigned to two groups, each of size 50. One group is taught by Instructor A; the other is taught by Instructor B. At the end of the course, 30 of Instructor A’s students and 22 of Instructor B’s students pass the state exam. Do these results give convincing evidence that Instructor A is more effective?
Min Jae carried out the significance test shown below to answer this question. Unfortunately, he made some mistakes along the way. Identify as many mistakes as you can, and tell how to correct each one.
State: I want to perform a test of
$$H_{0} : p_{1}-p_{2}=0$$
$$H_{a} : p_{1}-p_{2}>0$$
where $p_{1}=$ the proportion of Instructor A's students that passed the state exam and $p_{2}=$ the proportion of Instructor B's students that passed the state exam. Since no significance level was stated, I'll use $\sigma=0.05$
Plan: If conditions are met, I'll do a two-sample $z$ test for comparing two proportions.
$\bullet$ Random The data came from two random samples of 50 students.
$\bullet$ Normal The counts of successes and failures in the two groups - $30,20,22$ , and $28-$ are all at least $10 .$
$\bullet$ Independent There are at least 1000 students who take this driving school's class.
Do: From the data, $\hat{p}_{1}=\frac{20}{50}=0.40$ and
$\hat{p}_{2}=\frac{30}{50}=0.60 .$ So the pooled proportion of successes is
$$\hat{p}_{C}=\frac{22+30}{50+50}=0.52$$
$\bullet$ Test statistic
$$z=\frac{(0.40-0.60)-0}{\sqrt{\frac{0.52(0.48)}{100}+\frac{0.52(0.48)}{100}}}=-2.83$$
Conclude: The P-value, $0.9977,$ is greater than $\alpha=$ $0.05,$ so we fail to reject the null hypothesis. There is not convincing evidence that Instructor A's pass rate is higher than Instructor B's.

This question were asked to identify the population and sample based on different scenarios and tell whether or not they can be used to create a confidence interval. So let's first start off by just defining the variables. So we have P, which is the population proportion and P hot, which is the symptom, the sample proportion. So in taste A our population will be all the cars, whereas the sample size is going to be the cars stopped at the certain checkpoints and like we said, p as the population proportion. So in this case, is all cars with safety problems, and P hot is a sample proportion. So these air the cars that are actually seen with safety problems. So we can further calculate P hat by using the numbers given. And we know that there are 14 of 134 cars stopped have at least one safety problem, so that number ends up being 0.1045 Weaken further transfer that into percentage form and we get that it's 10 points 45% as RP hot volume and were also asked whether these methods can be used to create the confidence interval. So when a sample of data is representative, then it can be used to create a confidence interval and in this case it is because it's sampling all cars. For case be, we are going to find the population and sample once again for the population. We have the general public and for the sample, it's people that are logged into the website. We can further define them as P being the favor. The people in favor of prayer in school where us The sample proportion are the people that voted in this poll who favor prayer in school, we can calculate p hot with the given values were at 488 over 602. We get 0.81 and making that into a percentage value, we get 81% toe. Decide whether or not the sample can be used. We can Onley consider people logging into the website for this case. So in a way it's a bit biased and non random. So you're unable to apply the methods to create the confidence interval in Casey. The population is the parents at school and the sample is the parents expressing opinions through the question here. So the population proportion are all parents who favor the uniforms, whereas the sample proportion P hat are the respondents favoring uniforms. We can calculate the P hot value based on the numbers given 228 over 380 and that gives us a value of 0.6 that can be converted into a percentage of 60%. And since there were 1245 surveys sent home but only 380 returned, there is a complication of non response bias. And so you would use these methods with caution if creating a confidence interval. And the last part D were given a population of students at college, and the sample size is the 16 31,632 College admits the population proportion are all the students who will graduate on time and P hot the sample proportion as the students graduating on time that year. So based on the given values, we have 1388 over 632. Actually, this number is supposed to be 16 32. Sorry, and so based on that we get a value of 0.85 and that could be converted to 85% based on this value, and the sample data for this case was pretty representative. So since it is representative, you can apply these methods to create a confidence interval.

So we're looking at people wearing seatbelts and not wearing seatbelts Children and finding out how many days they stay in the I. C. U. And so we're going to assume that those students with a seat belt have equal stay in the hospital as those who do not wear a seatbelt. And alternately that the seat belted Children have a mean stay less than those who were not seat belted. And so we're assuming that the difference between these two seat belt minus not seatbelt is zero and we're actually getting something that's negative and this will be our p. Value. So let's look at the data and let's get our test statistic and we have sample sizes to sample sizes. One sample size was 1 23 and the other was to 90. So that was the for the seat belt and this is for the non seatbelt. And so we're going to use degrees of freedom of 122 to be conservative. And they're relatively large sample sizes anyway. And so let's find our test statistic and we have our 0.83 minus are 1.39 And then we're going to divide that by the standard deviation of the seat belted group squared, divided by the sample size, and then the standard deviation of the non seat belted Children divided by the sample size of it. And when we do that calculation, we get the test statistic comes out to be negative 2.330 And so now we want to find, and that's so that's what this value is. We want to find the likelihood, if the difference is actually zero or the means are equal, how likely is this number or more extreme to come up? And so I'm going to use my uh my T c D E f to find this mighty CDF. And I'm going to use negative uh minority has negative one times 10 to the 99th hour in it. And then our upper value is going to be that negative 2.330 My degrees of freedom is again the 1 22. And I will paste and let that do the calculation. And I have a value of a p value of about 1%. And that is less than my 5% significance level. So I definitely have sufficient evidence to reject the null meaning that we believe that the Children that had seat belts on have a smaller uh stay in the uh in ICU than uh those that are non seatbelt. And again you write my sons for that. So now we want to find the confidence interval and the appropriate confidence intervals. Since this was a one tailed test and we had 5% in this tale, the confidence interval has 5% down here, 5% up there. We would need to look at a 90% confidence interval. And so my table does not have a T. V. Star value for 122 degrees of freedom. So I'm going to use my inverse T. Value my inverse teeth. And on the inverse T. I want the area below it to the point oh five. So I'm going to put in the 0.5. It will be finding basically this low limit which will be symmetrical. What the upper one and then our degrees of freedom are that 1 22? And we get that value to be Yeah. Uh And this was 1 22 comes out to be negative 1.657 And so now let's get that confidence center, let's look at the difference first. So we need this difference the 0.83 minus 1.39 And that difference comes out to be negative 0.56 plus or minus. And we have that T star value 1.657 because we have one down here and went up here and then we have this same uh same standard deviation we had here. So let's see if I get 1.77 squared and that is a 3.6 squared mm And our sample sizes are 1 23 and 2 90. Okay, And let's get that margin of air, Yeah. Mhm And 1.657 times square root of 1.77 squared divided by 1 23 plus 3.6 square divided by 2 90. And that margin of air comes out to be 0.398 and two. And it keeps going on but I'm just going to store that is X. And so let's get those two values. We have the negative 20.56 minus the X. Value. And that comes out to be negative 0.958 And then we can change that into an addition sign. And we find out that that is still negative and notice that that does not include zero. So we would again from this interval see that the difference does not appear to be zero as he appears to be negative. So it does appear as though um that the seat belts make a difference seatbelts beautiful. They're good that they appear to have a lesser chance of a lesser stay in the hospital in the ICU for Children.

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.


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