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Most air travellers now use e-tickets. Electronic ticketingallows passengers to not worry about a paper ticket, and it coststhe airline companies less to handle tha...

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Most air travellers now use e-tickets. Electronic ticketingallows passengers to not worry about a paper ticket, and it coststhe airline companies less to handle than paper ticketing. However,in recent times, the airlines have received complaints frompassengers regarding their e-tickets, particularly when connectingflights and a change of airlines were involved. To investigate theproblem, an independent agency contacted a random sample of 20airports and collected information on the number of comp

Most air travellers now use e-tickets. Electronic ticketing allows passengers to not worry about a paper ticket, and it costs the airline companies less to handle than paper ticketing. However, in recent times, the airlines have received complaints from passengers regarding their e-tickets, particularly when connecting flights and a change of airlines were involved. To investigate the problem, an independent agency contacted a random sample of 20 airports and collected information on the number of complaints the airport had with e-tickets for the month of March. The information is reported below: 13 11 10 14 12 18 16 16 10 11 16 17 10 17 18 17 12 11 13 18 At the 0.01 significance level, can the agency conclude that the mean number of complaints per airport is less than 16 per month? a. What assumption is necessary before conducting a test of hypothesis? (Click to select) The population of complaints follows a normal probability distribution. The population of complaints not follows a normal probability distribution. The population of complaints follows a uniform probability distribution. b. Not available in Connect. c. Conduct a test of hypothesis and interpret the results. H0 : μ ≥ 16; H1 : μ < 16; (Click to select) Accept Reject H0 if t < . (Round the final answer to 3 decimal places.) The value of the test statistic is . (Negative answer should be indicated by a minus sign. Round the final answer to 2 decimal places.) (Click to select) Reject Do not reject H0. There is (Click to select) enough not enough evidence to conclude that the mean number of complaints is less than 16.



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Airline Flights: No-Shows Based on long experience, an airline has found that about $6 \%$ of the people making reservations on a flight from Miami to Denver do not show up for the flight. Suppose the airline overbooks this flight by selling 267 ticket reservations for an airplane with only 255 seats. (a) What is the probability that a person holding a reservation will show up for the flight? (b) Let $n=267$ represent the number of ticket reservations. Let $r$ represent the number of people with reservations who show up for the flight. Which expression represents the probability that a seat will be available for everyone who shows up holding a reservation? $$P(255 \leq r) ; P(r \leq 255) ; P(r \leq 267) ; P(r=255)$$ (c) Use the normal approximation to the binomial distribution and part (b) to answer the following question: What is the probability that a seat will be available for every person who shows up holding a reservation?

Question. We're using an airline example and we're having that the probability of a no show is 16%. And we're having what happens on the next 42 reservations. And again, assuming Lee that these reservations are each independent begin. And we know that that's not really a justifiable situation because you have people that go together in pairs or families that travel together or business people that are traveling together. And if we find N times P, we find that end times B. Which would be the mean of our distribution is equal to 6.72 And I happen to store that in my calculator as alpha A. That would just be helpful for me later on. And the standard deviation of that distribution is a square root of N times p times one minus P. And I did that calculation and I got 2.376 And I happened to store that in my calculator as X. Just so I don't have to type all that. And again, so when I write them down, I'm going to refer to him with those letters. Now we know that N. Times P and end times one minus P. And the one minus P would be that 10.8 for their both greater than a wrinkle to five. So this distribution is approximately normal, centered at 6.72 With this standard deviation, I could draw the picture of the distribution, but we have quite a few things we want to find And we want to know what's the probability if 42 reservations are made, what's the probability that there are exactly five, no shows. And so to do that normal uh that continuity correction and find this with a binomial excusing with the normal probability, we need to go half a unit below four point 5.5 a unit higher 5.5 and find that area between those two numbers. To use the normal approximation. Now we need to convert these two Z values and I'm going to write this one down and then I won't won't write down all the rest. We have 4.5 minus are mean provided by our standard deviation and then that will be a Z value and then we'll take that 5.5 minus are mean and then divided by our standard deviation and let me quick do that on my calculator here. So I have that 4.5 minus four mean, which I stored as alpha A divided by the standard deviation. And that Z value rounds to a Z value of negative 40.93 Mhm. And then the other z value which again I can just arrow back up and change that 4.5-5.5 I get negative .51. Now, I'm going to end up using my normal CDF with these two values, you can also use your table very easily and look up But negative .93 would be my low number and negative .51 would be my upper number and leave my mean and standard deviation 01 respectively. And when I do that I get .1288. You can look these values up in your table as well, defend that area between next we want to find from uh nine people out of the 42 all the way up to 12 people and it says inclusive. So we want to include those values now For using the continuity correction. This needs to bump down to 8.5 and this one needs to bump up to 12.5 and then I'm going to do that same calculation right here, just substituting 8.5 and 12.5 in of those values. So let's see what we get for the Z values. So let me just arrow back up And we will change that value to an 8.5 And that z value becomes .75 .75. Is that CLUCR 3/4 of a standard deviation higher than the mean. And this 12.5. Let me do a little insert there to get that value in 12.5 And that's the value becomes 2.43. Now, once again, you can look those values up in your table. I'm going to use my normal CBF to find those values and I'm going to use this .75 as the lower and there's 2.43 as the upper And then find that area between. And that comes out to be .2191. And likewise you can look those values up in your table. Part C. Now we want to find the probability that at least one no show. Well in order to calculate that as a Z. value again we are going to use the continuity correction and that means I have to go down .5. so we're going to use 0.5 and use that value in place of our 4.5. So I can go up in do that little change and we want zero wait five. That's the value. I'll leave it in blue. Nice to see a color different black all the time Is negative 2.62. And then you can look that up in the table. But remember you are finding a value that here is negative 2.62 and you're finding this. So if you're gonna look it up in the table, it's actually easier to look up positive 2.62 and find the area below And positive 2.6. To looking it up on the table, I get .99 56 And now we're almost done. We have one more to do. Let's go green. And we want to find the probability that at most two now for our continuity correction, we need to bump this up. So this is what we'll use in place of uh this value right here and we'll convert that to a Z. Value. So let's find out what that Z value is. So I just have to arrow up, Change that to a 2.5. And we get that Z value Is negative 1.7 and that would round to eight. So I'm just going to click look that up in my help and my table and not using my software. And when I do .8 that comes out to be .0375. Mhm. And I think we're all done. Yeah.

So we're told that tickets for a particular flight are $215, each. The plane seats 120 passengers, but the airline will overbook will purposely sell more than 120 tickets because not every paid passenger will end up showing up. So we're going to have D. Denotes the number of tickets the Arab line is going to sell and will assume that the number of passengers that actually show up for the flight X, follows a distribution. Um So ultimately we want to write the program to simulate the scenario um and we know that they wish to determine the optimal value. So running our program, we record the average profit from each run and it appears that the target value T. Is going to equal 142 tickets. So that would be our final result. Once we develop that program.

So in this particular problem were asked to do several things. So first of all, we need to identify Alpha, the significance level is 1% 0.01. And we know that the null hypothesis is that P is 0.301. And then the alternate hypothesis would be that P is less than now to figure out what kind of distribution it is. It's going to be standard, normal. And a quick check Is N. Times P. Greater than five. And indeed it is. So to 15 time zero point 301 is indeed greater than five because it equals 64.7 about And then his end times Q Greater than five. So if P is 301, Then 1 -301 will give us Q. So Q will be 0.699. So when I multiply those together To 15 times 0.699, I get approximately 150. So yes, that is indeed greater than five. Now I need to find the test statistic which in this case will be P hat and they're asking us to find the Z value. So I'm going to do this all at once with my calculator. Probably got it worked out but let me walk through it with you. So stat tests. This is a one proportion Z test. So number five. Now the probability of success is .301. We're told that our is 46. So in the calculator that's the x the total is population is 215. Were testing if it's actually a less than cursor down to calculate and the information they need. It's all right there. So I have this on the other screen. So P hat Is .21 depending on rounding here 214 and Z is -2.78. And in this particular situation I'm also given the p. value Which is this one. I know you got all these peas, you got the little P. For probability and you've got row and you've got your P. Value 27 So that's part C. Were asked for to find the P value. So if we're if we're actually shading this then we make our normal curve And here's negative 2.78. And I'm shading to the left. So that's if we had to shade it. Now I need to check this. So is this P value less than greater than or equal to my significance level? And I can see that my P value is less than or equal to. So that means I need to reject the no and then how do I write that out? How do I explain that? I would say something like At the 1% level of significance. The sample data indicate that the population proportion in the revenue is less Than 0.301. So this next part is asking for your opinion. And if you're doing a multiple choice question, the answer might be a little bit different than what I worded here. But This indicates the fact that P is in fact if he is in fact less than 0.31, Then that indicates that there are not enough numbers that start with one. So, yes. So what does that mean as a stockholder? Well, as a stockholder, that could mean that the value of your stock is inflated for the FBI. That might be a red flag to investigate. Because, according to Bedford's law, there should be a certain amount of Values that start with one. So for a stockholder it could mean that your stock is not worth as much as you think it is. And then for FBI this could be an indication that there's fraud. Now. Finally, just because we reject the null hypothesis doesn't mean that we have proved anything. So we did not prove H. Of zero, which was the fact that the probability should be this. Mhm. All we did was take some sample data and because the sample data let us to reject the null, then there could be too few numbers with leading digits of one. So you need to investigate more. So it's not an indication that this is actually false, but it's an indication that more investigation needs to be done. Maybe another sample or maybe a larger sample.

So this problem has several parts to it. First thing I need to do is I need to identify the significance level and from the story problem at 0.01. My null hypothesis is that P is indeed 0.301. And then my alternate is that P is actually greater than We're told. We have a sample size of 228. R equals 92. and again the probability is 0.3 01. So that means Q is 0.699. This will be a standard normal distribution. Just checking his NP greater than five. And yes, it is because 228 times 0.301 is indeed greater than five because it is Approximately 68.6. And then when I test end times Q 2 28 times 0.699. that gives me an approximate value of 1594. And that is indeed greater than five. Now I need to find P hat and because we're doing standards normal, I'm looking for the Z value. I'm not going to do this by hand. I'm going to use technology so I've already got it done here. But let me work through it with you stat Test. This is a one proportion z test. So number five The probability piece of 0.301. My ex this goes with the r value that was 92. And my sample sizes 2 28. And we want to test greater than so I'm going to click on calculate and there's the information I need so I got this on the other screen so my P hat is approximately point 40 and then rounding let's say we'll go three places and then my Z value is 3.37 So if I'm drawing this on the curve Here's 3.37 and I'm shading to the right and then here's my P. Value. So for part C. I need to check is my P. Value less than greater than or equal to α. So this either the -4 means it's a very small number. That's a form of scientific notation. So zero. Then the six will make the three round up. So .0004. And is that greater than less than or equal to? It's less than. So that means we will reject the null hypothesis. And then what does that mean in this in context of the problem? So at the 1% significance The sample data indicates that the proportion of numbers in the revenue file with a leading digit of one exceeds zero 301 So at the 1% significance level the sample. So we reject them all. Okay so let's interpret that If p. is in fact larger than 0.301. What does that tell you? Does it seem that there are too many numbers in the file with leading ones? And the answer would be yes. I guess I could have take this one out too. Huh? It indicates There are too many numbers leading with one. So what does that mean? As far as I could this indicate that the books have been cooked, chances are you're not writing numbers that are too big. So it could be that there was an error somewhere. Um The IRS of course should investigate more because it is very unusual. Could be some kind of as it says in the book. Could be profit skimming. So they take The extra and then right in the books one. But the bottom line is the FBI should investigate. So it could be a mistake. Or it could be that somebody is actually writing lower numbers in than what are actually true. And then finally, what does it mean to reject the null in this situation? So it's really important to know that by rejecting the nol, we have haven't actually approved then all to be false. The data did lead us. So that indicates that too many numbers start with one. So more investigation is needed.


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