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Last ycar; local television station determined that 65% of the people who watch news at H:OOpm watch its station. The station management believes that the current a...

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Last ycar; local television station determined that 65% of the people who watch news at H:OOpm watch its station. The station management believes that the current audience share may have changed In an attempt t0 determine whether the audience share had in fact changed the station questioned random sample of 180 local viewers and found that 103 watched its news show.Are the necessary conditions met t0 conduct hypothesis test for p the true proportion? Show all appropriate workState the appropriat

Last ycar; local television station determined that 65% of the people who watch news at H:OOpm watch its station. The station management believes that the current audience share may have changed In an attempt t0 determine whether the audience share had in fact changed the station questioned random sample of 180 local viewers and found that 103 watched its news show. Are the necessary conditions met t0 conduct hypothesis test for p the true proportion? Show all appropriate work State the appropriate null and alternative hypotheses_ Conduct test of the hypotheses identified above: Calculate the test statistic Calculate the corresp _ onding P-value IL Make statistical decision using level of significance of 0.05. Justify Your decision: IV. Interpret your decision in the context of the problem



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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. Use the P-value method unless your instructor specifies otherwise. Use the normal distribution as an approximation to the binomial distribution, as described in Part 1 of this section. Trials in an experiment with a polygraph yield 98 results that include 24 cases of wrong results and 74 cases of correct results (based on data from experiments conducted by researchers Charles R. Honts of Boise State University and Gordon H. Barland of the Department of Defense Polygraph Institute). Use a 0.05 significance level to test the claim that such polygraph results are correct less than $80 \%$ of the time. Based on the results, should polygraph test results be prohibited as evidence in trials?

In this problem, we're going to be testing the claim that men and women have equal success in challenging calls from made by the referees in the U. S. Open and the year 2006 10 x. So we have two samples. One sample is made up off 2441 men, and the second sample is made up of 1273 women. So the proportion off calls that were overturned for men is 1000 27. Oh, well, 2441 and the proportion for women's called the schools that Robert turned was 509. What? 1273. And to to test the claim that men and women have equal success in challenging calls we're going to uh huh used to approaches the hypothesis test on the confidence interval medal. So when they have prosthesis test, we're going to use the critical the Significance Level album off 0.5 And in this case, the null hypothesis is p one equals Sorry. P one is equal to P two. On the alternative hypothesis, his P one is not equal two p two Therefore, this is going to be a two tails test and the critical value for that is plus or minus 1.96 And therefore we now need to work out the values, the critical value theory, test statistic, that value by substituting the values. Okay, so when we substitute and values correctly, the value off zed clear value of that he is 1.23 three, so we can compare the critical value on the completed value said, drawing the critical regions. Here we have the critical region and when we look it test statistic, we notice that it is not within the critical region and for that reason we fail to reject. I'm not hypothesis. When we fail to reject the non hypotheses, it means that there is enough evidence there's sufficient evidence to support the clean that the men and women have equal success in challenging the colts. So, in other words, the proportions don't seem to have any significant difference between them Now. When we compare when you test using the confidence interval method, then we need to substitute the values in the formula for getting e correctly on. When we do that, we get E 0.0 333 and for the confidence interval limits, the lower limits is negative. 00 123 yeah, and upper limit 0.5 43 And this confidence is have one includes zero, which means that there does not appear to be a significant difference between the proportion off men. Men's caused men's calls that get overturned, the proportion off women's cause that get overturned. So this is in agreement with hypothesis test that shows that there is not enough There's enough stuff. They sufficient evidence to support the claim that the men and women have equal success in challenging counts. So, patsy, the question requires us to to tell whether it appears that men on women have equal success in challenging the Colts. Based on the results, we see that both men, and even appear to have the same proportions off course that are off a ton

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

Okay, So the given problem explains about touch therapy, and from the study is observed that among 280 trials, the touch therapists were correct. 123 times. Therefore, n is equal to 280 on the X is equal to 123. Okay, so here the researchers use a fair coin to test the sample. So the probability of getting heads on tails scene on bullets your 0.5. So the solitude guests now the next step be the best point estimate of the population proportion is the sample proportion be hot, which is equal to X over end, which is their frequent 2 to 80. Okay, saying 1 23/2 80 which is equal, Does you have put 439 now in part C of this problem is to construct the 99% conference level for the given data. Right, Andi, The 99% confidence interval for the population proportion P is given by portion P is equal to plus or minus e. So first it's compute the value of the margin of error e using the 99% conference level so the Formula 50 use is e CFO to times p cure. You are. So now we know that c ALF over to is zero. Quit. 23 Um, 2.583 maybe. And we have r p portion pee here. You know, que there's one man is p. So it's therefore equal to one minus 0.439 which is your point 561 So now, playing all these values back into this equation, we get e is equal to 0.76 Okay? And now, subsisting the values we can find the confidence interval P. Okay, He bless her, minus e which is equal to, as your point is, your 439 plus or minus Europe 0.76 over. So then we will get the range of 0.363 Did your 0.515

No. Okay, so we know that the number of patients given sustained care is 198. Among that, 82 0.8%. They're no longer smoking after one month. So 198 times 82.8 percent. Well, give us 163.94 Okay, so we can round that 264. Yeah. Andi making, then Right down. Arnel Hypothesis, which is given that P is 1 80 80% since the test claims that 80% of patients stopped smoking after giving state care and our culture bosses will therefore be P is not equal to your right June at a given level of significance. Which is your prince? Your one. Okay. Me? Yeah. So we will get a Z statistic value off 0.99 on the critical value for the normal area. Table movies equals plus or minus +257 eight A p value. 20 above output is just your 200.32 Sure. Yes. On. Since this is greater than our little of significance, we can include that there's not sufficient evidence to support the rejection of the claim that 80% off patients stop smoking when sustained care. What about


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