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Question 22 (1 point) State your conclusion to the hypothesis test A national organization has been working with Utilities throughout the nation t0 find sites for l...

Question

Question 22 (1 point) State your conclusion to the hypothesis test A national organization has been working with Utilities throughout the nation t0 find sites for large wind machines that generate electricity. Wind speeds must average more than 22 miles per hour (mph) for site to be acceptable. Recently; the organization conducted wind speed tests at_ particular sile. Based on sample of n=33 wind speed recordings (taken at random intervals), the wind speed at the site averaged 22.8 mph_ with sta

Question 22 (1 point) State your conclusion to the hypothesis test A national organization has been working with Utilities throughout the nation t0 find sites for large wind machines that generate electricity. Wind speeds must average more than 22 miles per hour (mph) for site to be acceptable. Recently; the organization conducted wind speed tests at_ particular sile. Based on sample of n=33 wind speed recordings (taken at random intervals), the wind speed at the site averaged 22.8 mph_ with standard deviation of mph: To determine whether the site meets the organization s requirements, perform the following hypothesis test at the 1% significance leve and state your conclusion Ho: p = 22 HA: p > 22 At a = Ihere sufficient evidence exceeds 22 mph_ conclude tho true mean wind speed at Ihe site Ata = .01, there is nqt sufficient evidence lo conclude the truc mcan wind specd at the site exceeds 22 mph We are 9g% confident Ihat the site meets the organizalion's requirements We are 9g% confident that the site does not meet the organization'$ requirements_



Answers

Test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim.
In a study of treatments for very painful "cluster" headaches, 150 patients were treated with oxygen and 148 other patients were given a placebo consisting of ordinary air. Among the 150 patients in the oxygen treatment group, 116 were free from headaches 15 minutes after treatment. Among the 148 patients given the placebo, 29 were free from headaches 15 minutes after treatment (based on data from "High-Flow Oxygen for Treatment of Cluster Headache," by Cohen, Burns, and Goads by, Journal of the American Medical Association, Vol. $302,$ No. 22 ). We want to use a 0.01 significance level to test the claim that the oxygen treatment is effective. a. Test the claim using a hypothesis test. b. Test the claim by constructing an appropriate confidence interval. c. Based on the results, is the oxygen treatment effective?

Problem. 17. No hypothesis is that new one is bigger than or equal from YouTube, and they returned to five processes. That me one is more than that you. So the critical values are the negative before corresponding to probability on opening off. So now that is equal to negative 1.645 So there is actual reason they contain all values smaller than being at the 1.645 in order to return minds tested statistic, which is X one bar minus X to bar minus me one minus Muto over square rolled off. Single one is squared over in one from Sigma squared over, into which approximately equal to negative 1.94 So and in the 1.9 for a smoother than 0.645 this means that 1.645 this means that way rejected on on high processes. So there is sufficient evidence of support. Okay,

A classic story involves four carpooling students who missed a test and gave an excuse that they had a flat tire on the makeup test. The instructor asked students to identify which tire went flat. If they really didn't have a flat tire, would they be able to identify the same tire? So the author decided he was going to run a an experiment, and the author asked 41 other students toe identify the tire that they would select, and some of them said left front. Others said right front. Some said Left rear and others said Right rear. And the data that was collected was that 11 people said Left front 15 said Right Front eight said left rear, and six said Right rear. Now that totals up to Onley 40. Even though he asked 41 students, one of the students said That spare tire, So we're gonna leave that out of our data. So this author is making a claim and he believes that the results fit a uniformed distribution. So in order for us to test this claim, we're going to have to construct are null hypothesis and our alternative hypothesis, and you're no hypothesis when you're trying to determine whether observed data fits something that is expected, the statement of fit becomes your null hypothesis. So therefore, our claim is going to be our null hypothesis. So our alternative hypothesis is going to be that the results do not fit a uniformed distribution. Mhm. And the hypothesis test that we're going to run is going to be a chi square goodness of fit test, which requires us to generate a chi square test statistic for our data. And we will have to apply the formula some of observed, minus expected quantity squared, divided by expected. So let's go back up to our data, and the information that we've collected would be classified, as are observed data. Yeah, so now we need to calculate are expected values. And if there were 40 people involved and we would expect a uniformed distribution, that means we would expect each response to get 10 people so we would expect tend to say left front, tend to say right front, tend to say left rear and tend to say right rear. So we're now ready to calculate our chi square test statistic, so we're going to add on to our chart An additional column and we're going to call that column oh minus e quantity squared, divided by e. So we'll take the observed value minus the expected value. We'll get a difference of one. We're going to square that So it's still 1/10, which is that expected value. So we'd get 0.1 and then we'll do 15 minus 10. We get a result of five when we square it. It's 25 divided by the expected value of 10. Yields a value of 2.5. Do the same thing for left rear eight. Minus 10 is negative two. But when we square it, it's positive. Four, divided by the expected value of 10, resulting in 0.4 and then six minus 10 would be negative. Four. When we square that we get positive 16/10 or 1.6 now to calculate the Chi Square test statistic, we will have to add up these values. And when we add up those values, our Chi Square test statistic is 4.6 now. Another component of the hypothesis test is to calculate our P value, and R P value is referring to the probability that Chi Square is greater than the test statistic we just found. Now, to get a better sense of what that's about, we're going to draw a chi square Distribution and chi square distributions are skewed to the right, and their shape is dependent on the degrees of freedom, and our degrees of freedom can be found by doing K minus one. And K represents the number of categories which you've separated your data into. If we go back to our chart, you can see we have separated our data into four different categories, corresponding with the four different tires on the vehicle. So if K is four, then our degrees of freedom will be three. Not only does the degrees of freedom indicate the shape of the graph, but the degrees of freedom also is equivalent to the mean of that chi square distribution. So on our picture we could place the mean, which will be found slightly to the right of the peak on the Chi Square axis. Now, for us to find our P value, we're trying to figure out what's the probability that Chi square is greater than 4.6. So we're trying to determine this shaded region, and in order to do so, the most effective way is to utilize your chi squared cumulative density function in your graphing calculator, and any time you use that function, you have to provide the lower boundary of the shaded area, the upper boundary of the shaded area and the degrees of freedom. So for our data, the lower boundary is the test statistic, the upper boundary. If you imagine that curve continuing infinitely to the right, you're going to get to some extremely high values of Chi Square. So we're going to use 10 to the 99th Power to represent our upper limit, and our degrees of freedom was three. So let me show you where you can find that chi squared cumulative density function. So I'm bringing in my calculator and I'm going to hit the second button and the variables button and select number eight. We're going to put the low boundary of the shaded area, the upper boundary of the shaded area, followed by my degrees of freedom, and I end up with a P value off approximately point 2035 Now there's one more component of a hypothesis test that we could find and that is called your chi square critical value. And to determine that chi square critical value, we're going to look in the back of your textbook. You will find a chi square distribution table and down the left side of the table, you'll find degrees of freedom and across the top of the table, you will find levels of significance, and levels of significance are denoted by the Greek letter Alfa. And we want to run this hypothesis test at a level of significance of 0.5 So we will locate 0.5 across the top of the chart and our degrees of freedom down the side of the chart and where the to correspond or meet up is your chi square critical value, and they meet up at 7.815 So let's recap the three components that we have found so far we have found the Chi Square test statistic to be 4.6. We have found the P value to be point 2035 and we have found our chi square critical value to be 7.815 So what do we do with these values to make a decision about that claim. So when it comes time to make your decision, you can either utilize the P value or you can use the chi square critical value. You do not need to do both. I'm going to show you both and then you can make a determination of which method you prefer. In order to use the P value. You're going to compare your level of significance to your P value. And if your level of significance is greater than your P value than your decision is to reject the null hypothesis. So let's run our test. So our level of significance was 05 and we found RPI value to be 0.2035 So we could say that Alfa is not greater than the P value. So therefore our decision will be fail to reject the no hypothesis. Let me show you how you can use the critical value to arrive at that same decision. To use the critical value, I recommend drawing out another chi square distribution and placing your critical value on that curve. And by placing that on the curve, you have broken your graph into two parts. You have the tail which we're going to define as the reject, the null hypothesis region. And then the other part of the graph is going to be determined or called our fail to reject the null hypothesis region and you're then going to look at your calculated Chi Square test statistic and we found our Chi Square test statistic to be a 4.6. So if 7.8 is right here, then 4.6 would fall back here, which is in the fail to reject region. So again, we end up with the decision that we will fail to reject the null hypothesis. So if we go back to our statements, we are not able to reject this statement. So that's saying it could be true. It might be true. It might not be true, but our evidence don't support throwing it away. So therefore, our conclusion there is insufficient evidence to reject the claim that the results fit a uniforms distribution again. We don't have enough to throw it away, but we don't We're not saying we support it, either. It's just the data is inconclusive and that concludes your hypothesis test

In this exercise, we're going to be testing the hypotheses. The passenger car owners were late License plate lose at a higher rate than owners of commercial trucks. So we have data concerning two samples. One sample is, uh, off the passenger cars on another sample is for commercial tracks. So we're told that out of the 2049 passenger cars, 239 cars were elated the license plate laws, while out of 334 commercial trucks 45 violated the license plate laws. So we're going to test this hypothesis at the 0.5 significance level. Andi, uh, this being a one tailed test, the critical value is going to be 1.1 point 65 one 0.645 And our another hypothesis is going to be P one equals p two. Which is to say that the proportions are the same. Vous is the alternative report. This is P one is greater than p two, which means that there is a higher proportion for the passenger cars violating the license plate lose. So for us to get the test statistic, we need to substitute the values into the formula to get set. And in our case, P one heart is obtained from uh X one, which is 239 who have violated the laws divided by 2049. And for P two hot. We're going to halve the proportion 45 those who violated divided by the total sample size, which is 334. And when we substitute these values into the that statistic, this is what we have. So for P one hut, we have zero 0.117 minus p too hot, which is 0.135 And so that Z supposed to be minus zero. Then you divide that by the square root off PBA. When you walk out the value P bite 0.119 multiplied by Cuba, which is zero 0.88 one. And we need to divide that by, uh, n one, which is 2049. And after that we added to 0.11 90 speedball time 0.8 81 Cuba divide by end two, which is 300 that you for, And when you simplify, you get the value off that the completed values that is negative 0.942 So the critical value here is going to be the negative critical value. And in this case, when you compare there using this craft, you'll see that we come. The critic The rejection that's the critical region is at negative 1.645 When we shared this region, we'll be able to see the rejection region. So when we place the calculated the test statistic in this carve, it will be before the critical value that is negative 0.942 And for that we make the conclusion to feel to reject denial hypothesis because the test statistic is not within the critical region. Yeah, so this claim is so there is not sufficient evidence to support the claim that the car owners were lit license plate lose at higher it than owners off the commercial trucks. Now we need to create a 95 a confidence interval for this, And to do that, we need to get imagine off era E by substituting the values into the formula here and e his zero point 03 29 and when we substitute the values will get that they call confidence Interval will be. Uh huh. Negative 0.509 It's less than P one, minus P. Two. She's less than zero 0.1 49 and you can notice that zero is within the confidence interval. In other words, zero they enter. The confidence interval contains zero. And because the confidence interval contain zero, there is not significant is not there is a significant difference between there is not a significant difference between the two proportions. In other words, there is not enough evidence to support the claim that the car owners of they're left license plate lose at a hair. Written the owners off commercial tracks. So both the hypothesis test on the confidence interval. I agree that there is not sufficient evidence to support the claim that car owners valued lesson tracks. Leave clues, uh, at a higher rate

In this problem we're going to be looking at to smoking cessation programs. One of them is the sustained cab, and the other one is the standard camp in the sustained care. We had 198 smokers going through the program on out of those 198 smokers, 51 no longer smoke after six months and for this standard care program among 189 smokers, fatty one no longer smoking after six months. So we're going to use the 0.1 significance level to test the claim that the rate of success for smoking cessation is greater with the sustained care program are compared to them standard care program. So the first part of the question we're going to test the claim using a hypothesis test. After that, we're going to test it using the confidence interval. And then we're going to see whether there's a difference between the two programs, uh, in terms off the proportions. Okay, so let's begin and state the hypothesis on. In this case, the non hypothesis is P one equals P two, which implies that the two proportions have are equal and the alternative hypotheses is P one is greater than P two. So this means that according to the hypothesis that we have a higher proportion off people who no longer smokey the sustained care program compared to the standard care program. So this being, uh, one tales test, the critical value will be 2.33 So we need to substitute the values into the test statistic, and we have the following proportion. For those in the sustained care, we have 51 longest, smoking out off 188. And for the standard care, we have 30 the longest, walking out off the 199. And to get the calculated value of that, we need to substitute these values into the formula. And here p one hot in decimal form is 0.258 and we need to subtract p too hot, which is 0.151 Okay, then the difference. We subtract zero from the difference, which is assumed to be, uh, zero, according to the Nile hypothesis. So then, from there we get the square it off. P one p bar Cuba, developed by N one plus p. Baquba Weber. And to So PBA is given by the sum of all the, uh all the X X one and next to divided by anyone and into So what we need to have here is the sum of 51 and that he divided by the summer of 188 and 199. And when you walk that out together, P bob is 0.204 divided by N one, which is 188 on, we have to multiply the PBA, Cuba and Cuba is one minute 0.204 which is going to be 0.796 So the new Morita is repeated in the next fraction. So it's 0.204 time 0.796 divided by N to an end to is 199. And when we simplify the value off the test statistic that he's 2.6 five now, we can compare the calculated value of that and the critical value off that and you can see. But the critical value is 233 so we can share the right side and the calculated value of that is within the critical region, which is 2.65 And since that is the case, we have to make the conclusion to reject the null hypothesis. No, this means that there is sufficient evidence to support the clean, that the rate of success is for the smoking cessation is greater with the sustained care program compared to them the standard care program. Next, we're going to test the same claim by constructing, uh, confidence interval. And in this case, we're going to construct 98% confidence interval. And for that we need to get the margin of error. E is in the formula given, and when you walk that out you will get e equals 0.9 0.9 35 And when you substitute into this formula for P one heart minus B p one hut minus be too hot minus e, you will obtain the following confidence interval. So to be 0.1 35 it's less than P one main US P two, which is less than zero point 2005 now. In this case, we noticed that the confidence interval limits do not contain zero. And that means that there is significant, a significant difference between the two proportions, as we have had really seen in the test off hypothesis, meaning we could reject the null hypothesis off having the proportions being the same. So because the interval consists off positive numbers only, it appears that the success rate for the sustained care program is greater than the success rate for the standard care program. Lastly, in Passy, we're going to see whether the difference between the two programs have practical significance. So if you check the the percentages, for example, for P one it is 25.8%. And for P two hot, this is 15 0.1%. And based on this sample, the success rates off the programs our 5th 25 points, 8% and 15.1%. And that difference does appear to be substantial. There's a there's a difference. So the difference between the programs does appear to have practical significance, because it's about a difference off 10% between p of P one heart and P to heart


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