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1.Aowner of a steak restaurant is considering opening new branch in Region X and consulted you if she should open Or not: It is known that this kind of luxury resta...

Question

1.Aowner of a steak restaurant is considering opening new branch in Region X and consulted you if she should open Or not: It is known that this kind of luxury restaurant will be successful if the average monthly household income of the region is above 20,000 TL. random sample . of 100 houscholds was oblained and their mean monthly income was [9.000 TL Assune thc avcrage houschold income is nonnally distributed with the population standard deviation of 2000 IL(a) Do you advise the restaurant owne

1.Aowner of a steak restaurant is considering opening new branch in Region X and consulted you if she should open Or not: It is known that this kind of luxury restaurant will be successful if the average monthly household income of the region is above 20,000 TL. random sample . of 100 houscholds was oblained and their mean monthly income was [9.000 TL Assune thc avcrage houschold income is nonnally distributed with the population standard deviation of 2000 IL (a) Do you advise the restaurant owner to open this new restaurant? Derive acceptance interval to make your decision (Open O not to open). Use the most commonly used level of a . (6) By using the collected data. You nOw calculated the sample standard deviation of household income as 2.500 TL Calculate 959e confidence interval for populatlon staadard deviatiom and discuss the result by comparing it with the empirical value of population variance. (c) If only [0% of such kind of restaurant is successful in general out of 150 restaurants in Ankara. what is the probability that the sample proportion of the restaurant to be successful is greater than 209 0?



Answers

Opening a restaurant You are thinking about opening a restaurant and are searching for a good location. From research you have done, you know that the mean income of those living near the restaurant must be over $85,000 to support the type of upscale restaurant you wish to open. You decide to take a simple random sample of 50 people living near one potential location. Based on the mean income of this sample, you will decide whether to open a restaurant there.
(a) State appropriate null and alternative hypotheses. Be sure to define your parameter.
(b) Describe a Type I and a Type II error, and explain the consequences of each.
(c) If you had to choose one of the “standard” significance levels for your significance test, would you choose A
0.01, 0.05, or 0.10? Justify your choice.

All right. So this is the Washington we have. We have the diamond minutes for 30 visits. We have been given a table on there saying that timing off 30 visits has been recorded. Now, what we want is to provide a point estimate for the population. Mean drive through times. Now, what is the point? Estimate? Now, this is our part here. Okay? What is the point? Estimate? Point estimate is going to be our expert expert is nothing but the sample. Mean sample Mean is going to be our point estimate. All right. How do we calculate expert eggs? Bar is going to be nothing but submission off eggs divided by N And is what in this case and is 30 right? Submission of X means what? 0.9 plus 1.0 plus what is the next 1 to 1.2. And so on all the 30 numbers divided by 30 this is going to give me my point estimate, which happens to be expert now. Luckily, what we do in our real life examples whenever we apply statistics, we never do these things by hand. We are always going to use some kind of a statistical tool or software. And that is what I've used aware you can see that I've simply input. I just simply put in the values in this history Graham maker. And it has given me all of these values over here on the left. So what does it mean? The mean, you see, turns out to be 3.8. So this thing actually turns out to be 3.8. This is my expire, and this is going to be my point estimate. So this is my answer to the first question. Okay, moving on to part B. What is part we have to say now, at 95% confidence, what is the margin of error now? What exactly is the formula for margin of error margin off at a What is the formula for this? In this case, we see that sigma is not known, right? Sigma is unknown. Sigma is unknown, right? So the formula that we use over here is t Alfa by two into s by route and this is going to be my margin off error. P is what the critical value off the piece distinct. What is my Alfa by two I have 95% rate. My confidence interval is 95%. Which means my Alfa is 0.5 or my Alfa by Do will become 0.25 Okay, So what is going to be mighty Alfa by two? I'm going to use a critical value. Calculate over here. So what is going to be my alphabet To 0.25 So I input 0.5 over here. What is my degrees of freedom? Degrees off? Freedom is always end minus one. What is my end? In this case, there are 30 observations. Right? So my degrees off freedom is going to be 29 and my critical value is 2.4 Faith. So this value is 2.0 for faith. Okay, What is in in is 30. Great. Now what is s s is the sample standard deviation. This is the sample standard deviation. How do I calculate this? My sample standard Division s is given by route over X minus mute square upon N minus one. But there is a submission here. This is the submission, son. Okay, this is the sample standard deviation. So if I just give you an example, What this is going to look like is, uh, just a moment. Have you made a mistake? See, the point is that we're always going to solve this using software. So that's why even I don't have a lot of practice doing this. I am a professional data scientist, but still, So Okay, let us just look it up. Sample standard deviation formula. Okay, so this is a formula is absolutely correct. So this is submission, and then it is brewed over the entire thing. Thank God. So, for example, let us take the first one. This is 0.90 point nine minus what was our new muse? What? Or let us write this as X bar. In this case, this is X minus X bar. What is X bar? Where is expert? Yes, 3.8, minus 3.8 whole square. Okay. Plus, the next number is 1.0. So this is going to be one minus 3.8 whole squared. Plus, now, this will go on all the way after the last number. That is nine minus 3.8 whole script upon and is a sample size 30 30 minus one is 29. So this is 29 this is a root. All right, So what is this going to be? I can see that in this case, this has been calculated for us again. What is the standard deviation? 2.257 to 5. This is two point 257 to fight 257 to 5. This is my standard deviation sample. Standard deviation. Okay, now, what is the formula for margin of error? We have all the things that we need. So this is going to be two point zero for five multiplied by s, which is 2.257 to 5 by route N. Okay, bye. Route off 30. So let us use a calculator to find this. Do 0.45 multiplied by two points. 25 75. Divided by root off 30. So this is zero point 840.84 2776 Let me just write this as three 0.843 This is going to be my margin of error. Let me start. This is eat. Okay, This is my margin off. error. And I think this was a second question. Okay, The third question says we have to find a 95% confidence in double estimate, 95% confidence interval estimate at 95% level. We already have the margin of error. So what is going to be my confidence interval? My confidence interval is expert plus minus my margin off error. Okay. What is my ex bar? That is the sample mean, this is 3.8. So this is 3.8 plus minus. What was either? I got 0.8430 point 843 Okay, so if I use a calculator for this, 3.8 plus 0.843 is 4.643 on 3.8, minus 0.843 which is 2.9572 point 95 seven. All right, so this is my confidence interval at 95% confidence level. Okay, this is my confidence interval. And this is the answer to my third part. The fourth one is discussed. This communists. Okay, if I go to this history Graham, I can say that my data that I have with me is a little bit skewed, right, But okay, The sample size is very small. It is only 30. I am sure that if we take more and more samples like, let's say if the sample size were to be 500 or 1000, this would be close to normal. Right now I can see that it is cute to the right. Okay, things is skewed to the right. All right. And by looking at this, what I can say is these guys are actually doing a pretty good job. If I just look at this history. Graham, I can say that you no longer wait. Times are very rare. People are the people that really have lower rate things. So, yeah, they're doing a very good job. If I look at this hissed a gram, this is a little bit skewed to the right. Which says that. OK, these guys are efficient. You just Okay, So this is how we go about doing this question. And this is our answer

We have data sets A and B on the right. We want to utilize these data sets to test the claim that there is no difference in the underlying distributions for A and B it out the equals 0.1 significance. This question is testing an understanding of non parametric tests, particularly how to take perform the ranks on tests we proceed through steps A through G to solve. So first we see the outside of hypotheses. These are alpha equals 0.1 hypotheses. H. And distributions are saying hk distribution are different and be we can be the test at the sampling distribution which is normal and check the requirements which are met. So their ranks are for A and B. As follows That we have and one equals 11 and two equals 12. You are sigma are 1 30 to 16.2 from this, we have our equals and summer range from a 1 73 so Z equals ar minus. Moreover signal articles 2.53 thus received complete the P value. As for normal distribution two PZ greater than 0.114 Thus, we conclude, indeed, that we fail to reject nation on since P is greater than alpha, which means we lack evidence for AJ.

Now, what do we have in this question? They're saying let X be a random variable that represents that every daily temperatures in the month of July in a small town in Colorado. Okay, now the ex distribution has a mean off 75 F. Okay, So the mean, the mean is 75 F. Okay. And the standard deviation sigma is approximately 8 F. Okay, Now, a study was conducted over a span of 20 years. That is 620 July days, and we are given a table off those increase from that study. Okay, So what are the columns that we have? Let's just look at the columns this we're also going to use in order to find the chi square statistic. Okay, so here it has, given that new minus three sigma is less than equal to X, which is less than noon minus two Sigma mil minus two sigma. Okay, similarly, here. I think that was me. Minus two Sigma minus sigma. Then there is mu. There is mu plus sigma, and there must be mu plus two sigma. Okay, these are Sigma's, this Sigma Sigma Sigma. Okay. Less than equal to less than equal to less than equal to less than equal to less than equal. Okay, then these are the low elements. And now we will write the upper limits. This is going to be Sigma. This is going to be immune. This is going to be mu plus sigma. This is going to be mu plus two Sigma, and this is going to be mu plus three sigma. Okay, so what is happening over here is we are having arrange the frequency. The frequency count for the number of days when the temperature waas between I mean minus three Sigma and me minus two Sigma. Right? When that when we can say that. Okay, if I see this normal distribution. Okay, let me just draw this normal distribution here. This is the mean. Okay, this is one standard deviation away. This is two standard deviations away. And this if I extend the graph, this is going to be three standard deviations of it. Right? So all of these categories are basically the frequencies off. How many values lie? Let's say between the first categories. View minus three. Sigmund U minus two Sigma so mu minus three. Sigma is 123 That is this point to me. Minus two signals this point. What is the frequency that lies between this? The number of values that live in this region, then in a moral values that, like between U minus two sigma and you mind a sigma, right? That is this region. So in this way, this is the study off all the different regions. All right, so this is the first column. The second column is actually the values. What a Sigma Sigma is eight. Right? So what is happening over here is now. They have just given us the values. When excess between 51 2. 59 between 1559. Then when excess between 59 to 67. Then when excess between 67 to 75. Then when excess between 75 to 83. Then when access between 83 to 91 Then when excess between 91 to 99. Okay, so these are the temperatures, right? What does the rains when X minus three sigma We do. We get 51 degrees, and when we do X minus two sigma, we get nine degrees. So what was the frequency or what was the number of days where the temperature waas between these two values? Right. So this is just a study off the number of values that are between two and three. Standard deviations away to the left, then one and two standard deviations away to the left and so on. Okay, now we are. I think we are also given the expected percent from normal have. Okay, now, what is happening is they're expecting this to be perfectly normal, right? This distribution to be perfectly normal. So this is the third column. Okay, So what should be the values? If it isn't be perfectly normal. It should be 2.35%. 2.35% for the first category. For the second one, it should be 13.5%. Okay, then it should be 34%. Then again, 34%. Then it should be 13.5% again. And then it should be 2.35%. Because the normal distribution is symmetry. We can see that the values here are symmetric, right? 2.35 13.5 34. Then in the decreasing order, 34 13.5 and 2.35 Okay, now this is expected. Okay, but what are the observed values? What are the observed observed values now? The observed value that there were 16 days when the temperature was between two and three. Standard deviations away to the left. Then for the second category, it was 78 days. Then it was 212 days. Then it was 221 days, then was 81 days. Then it was 12 days. Okay, now this addition is given to us a 6. 20 between that. This is our sample size in this. Is our sample size in Okay? No. What is going to be the expected value? In order to find the Chi Square statistic, we also need to find the expected values expected values E. Now, what is the formula for? This was the part A. We understood what the top of the first three columns are saying. Now, how do we find the that is the expected values this is given by the sample size. The sample size, which happens to be in in our case, which is 6 20 multiplied by the probability off each category. The probability or the proportion, right? The probability off each category. Okay, so let's say if this world this distribution work, uh, to follow a normal distribution, then this would be are expected probabilities in column three. Okay. Okay. So let's find the expected values for the first category. There will be now, use my calculator. Okay, So the expected value for the first categories. 2.35% of 6. 20. So point zero 235 the two into 6. 20. This is 14.7 41 57 14.57 Then it is starting 0.5% of 6. 20 0.135 in 26 20. There is 83.7, 83.7. Then we have 34% of 6. 26. 20 in two 0.34 This is 210.8 210.8. Then this is again 31. 34% of this shoot. Again. We 210.8. That this is 13.5% which is 83.7 again. 83.7. And then this is 2.35% which is 14.5 14.57 Okay, these are expected values. Now, in order to find the chi square statistic I need to find for all the cells, I need to find the value off. Oh, minus C. That is the observed value minus the expected value. Whole square upon the expected value. And then I need to sum them all up so that I get the guys were statistic for my entire problem. Okay, so this will be the column for individual for individual chi square values. Okay. All right. So let us just do what we saw in the formula. The difference between observed and expected. So this is 16 minus 14.57 We square this square 1.43 and divide this by 14.57 So this is 0.14 zero point went four, then the difference between 83.7 and 78. So this is 5.7 square and divided by 83.7. The 6.3 18 So I can write. This is 180.39 Then the difference between 212 and 210.8, we square this 1.2 square and divide this by 210.8. So this is 0.68 But I can write this at zero point 007 Okay, then the difference between 21 210.8. We square this and divide this by 210 point eight. This is 0.493 organizes 0.5 0.5 0.5. Now the difference between 83.7 minus 81. We square this and this is divided by 83.7. This is 0.87 0.87 And then there is a difference between 14.57 minus 12. We square this 2.57 and divide by 14.57 So this is 0.4533 0.4533 Now I have the individual Christ question districts. Now all I have to do is add them all up. So this is 0.14 plus 0.39 plus 0.7 plus 0.5 plus 0.87 plus 0.4533 So this is 1.5603 This is 1.5603 All right, so my high square value over here is 1.560 three. All right? Now, in order to find the P value, what I need to do is find the degrees of freedom. Degrees of freedom. DF is given by the formula. Number of categories, number off categories minus one. Okay, how many categories do we have? You see here the first If you look over here, we have 1234566 categories, right? So what is going to be our answer for degrees of freedom? It is going to be six minus one or I write this as Fife. Now, you can either use a chi square table to find the value, or you can use a chi square calculator or any statistical software, so I'm using it online. To hear 1.56 is my price square statistic, which is one point 5603 1.56 zero. Create on The abuse of freedom is five now, what is my Alfa when Alfa in this case is 0.1 Right. My Alfa is 0.1 as the developed significant is 1%. So this is 0.1 And when I calculate this, I find that my P value is 0.90 My p value is 0.90 Now I can see that my P value is greater than Alfa. Hence I say that I will fail to reject mine. L hypothesis. I fail to reject my null hypothesis. H not Okay now what was the hypothesis? What was my hypothesis here? Mine l hypothesis Waas. Okay, the I think we forgot to write the null hypothesis. My little hypothesis would be that the distributions are the same, right that the normal distribution O r Let's just say the average daily july temperature follows a normal distribution panel Hypothesis would be that the daily July temperature temperature follows a normal distribution. Okay, this waas my null hypothesis. What would be my alternative hypothesis that the normal distribution doesn't fate The daily July temperatures. All right, So what exactly I am I going to say I say that I failed to reject my null hypothesis, meaning that I I do not have enough statistical evidence. Enough statistical evidence to suggest that the daily July, the daily July temperatures and the normal distribution Okay, uh, do not have enough sufficient enough statistical evidence to suggest that the normal distribution that the normal distribution Let me just write it like this. Uh, just a moment. Okay. That the normal distribution that the normal distribution doesn't fit doesn't faked. Yes, the daily July temperature distribution. Okay, distribution. This line is going to be a answer. So I will say that I do not have enough statistical evidence to suggest that the normal distribution does not fit or doesn't fit the daily July temperature distribution. And this is how we go about doing this question.

In question five. We have information about a certain type of test that young people can take in their scores. Scores range from 0 to 500 and we have a simple random sample of 840 people from a large population. We have a mean score of 280 and a standard deviation of 60 in part A. We were asked to describe the shape, center and spread of the sampling distribution in this case, Well, because we have a large population that indicates that our shape is approximately normal. Best central limit The room that comes into play here, our center. Because we have a sampling distribution, the center of our sampling distribution will ultimately match the population. Mean so in this case, our center of our sampling distribution. Thanks, Bar 280. And then the spread in this case s are standard deviation is 60. But in order to calculate our spread of a sampling distribution of explore, we have to take our population, senator vision and divide by route in, which is gonna be 60 about about fruit 8 40 That gives us a spread of 2.702 in Viet SS to sketch the shape of the sampling distribution with 12 and three standard deviations on either side of them. We're gonna do this. So we have a normal curve we have are mean, which is 280 and we're gonna put 12 and three standard deviations. And he said so one to the right would be approximately 2 82.1 to keep going to 84 0.2 one more to 86.3 and to the left of the mean be approximately 2 77.9 to 75.8 and finally to 73.7 in si says, According to the 68 95 99 7 rule, about 95% of all values of X bar lie within a distance. Film of the mean of the simply distribution wants to know what is him and then shade the region on the axis of the sketch that is within him of the mean. So in this case, um, we're dealing with the 95% of value, so the 95% value is gonna represent the two standard deviations. So to center. Deviations outside the mean on either side will put us at this to 75.8. Put us at to 84.2. This is the 95% region. According to the empirical rule of the 68 95 99.7, the ultimate loses Looking for him. Um um, which distance between the mean? So this essentially two standard deviations. So to center deviations in this case on either side is approximately 4.2. So we're going to say that I am is approximately 4.2, and in the region of shaded the boat and then Part D says, whenever explore falls in the region, you shaded the population mean utilizing a confidence interval X bar plus from honest for what percent of all possible values does the interval capture mu. So this is the idea of a confidence interval. So we're doing with this 95%. Um, we know that about 95% of the time we will capture the true I mean, if we're creating these complex intervals. So the whole idea is if we're trying to create multiple intervals using different samples 95% of the time, we will capture that true value within our confidence interval


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