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Q1)Chi-Square Text for Goodness of Fit (11 points)Two employees at an ice cream storewere discussing the popularity of the three basic ice creamflavors: Vanilla, Ch...

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Q1)Chi-Square Text for Goodness of Fit (11 points)Two employees at an ice cream storewere discussing the popularity of the three basic ice creamflavors: Vanilla, Chocolate, and Strawberry. Employee A thoughtthat the three flavors are equally popular but employee B didn’tthink so. To settle their debate, they decided to tally the numberof customers asking for those flavors for a whole day. There were45 customers asking for one of those flavors that day: 12 asked forvanilla, 23 asked for chocola

Q1) Chi-Square Text for Goodness of Fit (11 points) Two employees at an ice cream store were discussing the popularity of the three basic ice cream flavors: Vanilla, Chocolate, and Strawberry. Employee A thought that the three flavors are equally popular but employee B didn’t think so. To settle their debate, they decided to tally the number of customers asking for those flavors for a whole day. There were 45 customers asking for one of those flavors that day: 12 asked for vanilla, 23 asked for chocolate, and 10 asked for strawberry. A Chi-Square Test for Goodness of Fit using a = .05 as the significance level. What is the variable in this test? What type of variable is it (nominal, ordinal, or continuous)? (2 points total, 1 for each answer) State the null and alternative hypotheses in words (2 points total: 1 for each hypothesis) Calculate X2 statistic (3 points total: 1 for final answer, 1 for setting up the correct OFs and EFs, 1 for correct calculation process) Calculate the degree of freedom and then identify the critical value (2 points total, 1 for df, 1 for critical value) Compare the X2 statistic with the critical value, then report the hypothesis test result, using “reject” or “fail to reject” the null hypothesis in the answer (1 point total, .5 for each answer) Explain the conclusion in a sentence or two (1 point)



Answers

Please provide the following information. (a) What is the level of significance? State the null and alternate hypotheses. (b) Find the value of the chi-square statistic for the sample. Are all the expected frequencies greater than $5 ?$ What sampling distribution will you use? What are the degrees of freedom? (c) Find or estimate the $P$ -value of the sample test statistic. (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis that the population fits the specified distribution of categories? (e) Interpret your conclusion in the context of the application. Meteorology: Normal Distribution The following problem is based on information from the National Oceanic and Atmospheric Administration (NOAA) Environmental Data Service. Let $x$ be a random variable that represents the average daily temperature (in degrees Fahrenheit) in July in the town of Kit Carson, Colorado. The $x$ distribution has a mean $\mu$ of approximately $75^{\circ} \mathrm{F}$ and standard deviation $\sigma$ of approximately $8^{\circ} \mathrm{F}$. A 20 -year study $(620$ July days) gave the entries in the rightmost column of the following table. (i) Remember that $\mu=75$ and $\sigma=8 .$ Examine Figure $6-5$ in Chapter $6 .$ Write a bricf explanation for Columns I, II, and III in the context of this problem. (ii) Use a $1 \%$ level of significance to test the claim that the average daily July temperature follows a normal distribution with $\mu=75$ and $\sigma=8$

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.

We have the following two data sets A and B, A and B. As follows. We want to test the claim. There is no difference in the underlying distributions that output equals 20.1 significance. This question is testing our understanding of how to implement non parametric statistical tests, particularly the ranks, some tests for which we proceeded through. That's a three D below to complete and a wee state alpha and hypotheses. Office 30.1 A stated agent is distributions are the same. Age distributions are different and be we can be the test at we see the sampling distribution which is normal and check the requirements which are met. So first we can be the ranks as follows. Thus we have N one and 2. You are sigma are as defined here. Next we have our equal some of rings from a 1 54.5. So Z is our medics Newark over signal article 1.38 Thus we compute the pR value from a normal distribution as two PZ grid and the non equals 20.16 as shown here. Thus, we conclude that we fail to reject nonsense P is greater than alpha, which means we lack evidence to support the alternative hypothesis.

Okay. What do we have with us? We have X as a random variable. That represents the average daily temperature in Para Night in January. For the town off Hannah, Now X has a mean off approximately 68 F. Okay, so we can see that the mean is 68 all right? And the standard deviation sigma is approximately 4 F. All right, Now, there is a 20 year study that is one of 620 January days, and we have their observations with us in the table. All right, so it is Just look at the table. Okay, So the first column is reason under. Of course. Okay, after that, we have the temperature. Then we have expected person from the normal. Then we have the expected person from normal curve. Okay. After that, we have the observed values. After that, we have the observed values. We have the observed values. All right. Now, the region under the curve is the first category Is the reason between three standard deviation and to standard deviation towards the left. Okay, so this is going to be new, minus three sigma. And let me just write this anti. Call him. It'll just take a minute. Then we have new minus sigma. Then we have new. We just have milk. Then we have new plus sigma. And then we have mute lis. All right, X, then we have new minus two sigma. Mean minus two Sigma new minus Sigma men, New plus sigma. New plus two Sigma New plus three sigma. All right. Now, what are the exact boundaries this is given in this column? We have 56. We have 60 than 64. 68 72 76. All right, these are all less than equal to science. Okay, 80 76 70 to 68 64 60. All right. Now, if the normal distribution actually fix the, you know, actually fix the observations that we have, then what should be the expected percentage from the normal Come? Like if I draw this normal distribution? Let's say that this is the mean right? This is my one standard deviation of it. I'm to standard deviation of I'm 300 deviation of. Okay, so this one is mu minus three sigma. And this one is mu minus two. Signal what is actually the area what percent off values lie in this region. If this actually is a normal distribution, the expected present attribute 2.35%. Right? So these are the values over here in this column. So this is 2.35%. And since it is symmetrical, the last one will also be 2.35%. Then it is 13.5% 13.5%. And since this is symmetrical, this is also 13.5%. This is 64% and this is also 64%. Now, what is exactly the observed values that we have with us? Okay, The observed values are 14 86 207 Then we have 215 Then we have 83 then we have 15. All right, so now let's just look at the questions. What exactly are they asking us? They're saying impact one that mu and sigma this and this. Now what? Our columns 12 and three. In the context of this problem, well, call them one. What has actually happened happened over here is they have divided the entire distribution into various categories. So column one is going to tell us that this particular reason the first region is the reason between three Sigma and two Sigma towards the left like and this value is the difference between two sigma and one Sigma towards the left and so on. Okay, so this is column one column six Give Sorry. This column to gives us the exact boundaries and column three is nothing but the expected value. If this really were to follow normal distribution, what is the percentage of values that we expect to find between new minus three Sigma Nu minus two Sigma that is in this region, right? What is expected amount? It is 2.35 What is expected amount in this region, that is between 23 sig mind to sigma it is 13.5%. Right. And then over here we have the observed values, the values that have actually been observed. All right, now, what is this question to question two says we have to use a 1% level of significance, which means what is our Alfa R Alfa for this question is 0.1 all right to test the claim that the average daily January temperature follows a normal distribution with Mu 68 sigma for All right. So in orderto do this, in order to perform this analysis, we're going to use the chi Square statistic. And the first time of the Chi Square statistic is to find the expected values to find the expected values to find the expected values. And the formula for that is the total sample sizes. The total sample size multiplied by the probability multiplied by the probability multiplied by the probability off each category off each category. Okay, off each category. All right, so what exactly is going to be my sample size? If I add all of these up, this should be 6. 20 as it is given to us, right? 6, 20 days. All right, so let us just use that formula way. This is the expected values for all the categories expected values. All right, so this comes over here, so let's put the formula into action. Okay? So I'm going to use my calculator for this. I want to 0.35% off 6. 20. So this is 0.235 multiplied by 6. 20. This is 14.57 expected values. 14.57 Then I have 13.5% off. 6. 20 0.135 in 26 20. This is 83.7, 83.7. Then we have 64% or 0164 into 6. 20. This is 3 96 3 96 point eight. All right. Just a moment. This is 0.6. For how can this be 0.64 This has to be 34. My mistake. I apologize for this. This should be 34 right? So this is 34. 34%. All right, so this is 0.34 into 6. 20. This is 210.8. So if I just erase this, this thing is 210 pointed Now, since this is symmetrical, this is a normal distribution. So this will also be 210.8. This will be 83.7 and this will be 14.57 These will be the expected values. Now what now? In order to calculate the Chi Square statistic for all the categories we're going to apply the formula observed value minus the expected value. Whole square divided by the expected value. And in the end, I'm going to sum them Allah, and it will give me the overall chi square statistic for my question. All right, so let's just look at this inaction. These will be the individual chi square value, the individual chi square values for all the categories. We're going to send them all up in the end. So this is the difference between 14 point 57 and 14 and we square this and this. I'm getting us 0.32 and I divide this by the expected value, that is 14.57 So this is 0.2 0.2 Similarly, this is 86 minus 83.7. I square this and I divide this by 83.7, 83 0.7. This is 0.6 0.6 Then we have the difference between 210.8 on 207 I square this and I divide this by 210.8. So this is 0.685 Recognized this as zero point 0.7 All right, then we have difference between two and five and 210.8. We square this and divide this by 210.80 point 80.8 Okay, then The difference between these two is 20.7, and we square this divide this by 83.7. So this is 0.0 0.55 Texas. Right. This is six. And then the difference between 15 and 14.57 We square this and divided by 14.57 So this is 0.1 0101 Now we, some all of these up. So this is 0.22 plus 0.6 plus 0.7 plus 0.8 plus 0.6 plus 0.1 And this is 0.248 So I can say that my guys question to stick for this problem is 0.248 All right, now that I have Mike ice Question District, what else do I need in order to get my p values? I need the degrees of freedom DF I need d f. And how do I find d f? This is given by the formula Number of categories. Number off categories minus one. What is the number of categories that we have? We have six different categories, Right? 123456 So this is going to be six minus one six minus one. Or I can write this as five. So my degrees airfield I was five. Now I have my kaist question the steak. I have my degrees of freedom. So you can either use a chi square table and it will give you an approximate P value. Or what you can do is you can use an online software or statistical tool so that you get your exact P value. My guys question is thickest 0.248 This is 0.248 and my degrees of freedom is five, as we just saw are significant level is 0.1 and we can see that a P value is 0.9985 My P value is 0.9985 What was my Alfa? My Alfa was 0.1 hence I can say that as my P value is greater than Alfa I failed to reject. I fail to reject my null hypothesis. H not Okay now I think we also forgot to write the hypothesis initially. Okay, so let's just look at it briefly. What is going to be my null hypothesis for this question? What was my null hypothesis? My hypothesis waas that normal distribution normal distribution fits the distribution off days Just a moment for the distribution of the average daily January temperature fits the distribution fits the distribution off the average daily January temperature, January temperature and what would be my alternative hypothesis? My alternative hypothesis would be that the are not similar, right? That the normal distribution does not fit the average daily January temperature. So they are not similar. Let me just write it like this. They are not similar. All right? So I cannot reject my null hypothesis. So what is going to be my answer? This line is very important. I will say that I don't have I don't have enough statistical evidence. Enough statistical evidence to suggest I do not have enough statistical evidence to suggest that the normal distribution dozen doesn't fate The January temperatures. The distribution doesn't fit the distribution off average daily January temperatures, and this is how we go about doing this question.

Following a solution number 13 goodness of fit test. And this compares the U. S. Census data with a town in our city in California as a sample. And we're just seeing if the to agree. So we have observed data values I think yeah the sample is 1215 and they observed data values are here 127 for these different ethnicities. So I think 127 maybe maybe it was black and then 40 was I can't remember 480 I think was maybe Anglo and then Latino 502 and so on. Maybe asian. So these are all ethnicities. And then here we have expected Percents uh we take the percent times the sample size and that gives the expected value. So I went ahead and did that work before making this video and then we can answer these pretty quickly. So the alpha value is one. So it gives you that in the actual problem this is 1% um is the significance level. And then we also need to state what are alternative are null and alternative hypotheses are, So the knoll is always that these two distributions are about the same. So the census population agrees with this particular study. And then the alternative is that they don't they don't match, they're not the same. So the knoll let's go and say the census distribution and the sample distribution agree. And then for the alternative, let's just say not. So the census distribution and the sample distribution do not agree. Okay part B. We're asked to find the chi square value. We're going to use technology to find that. But you can certainly use the formula. Um If you so wish I'm gonna use the T. I. 84. I think it's the easiest to use with the smallest data set. But you can use Excel, you can use argue sas S. P. S. S you know whatever program you like mini tab. But I use T. I-84, especially for small data sets. And then we also need to check a couple of things. The main thing is that the expected values are all greater than five and they are. So that's one of those conditions for inference, the expected values need to be greater than five. So that allows us to use the chi square distribution and we can say how many degrees of freedom there are. We have to say that actually there are five degrees of freedom. And the reason why we have five degrees of freedom is the number of categories minus one. So there are six Ethnic categories Here. six. Right here and then we subtract one and that's where we get five. Okay, so from here we can use our technology. So if you go to stat and then edit, you can see already punch those in 127 44 80 and so on and so forth. So L one is my observed values. L two are my expected values, that's where I put mine. But you can put them wherever you just may need to change a little bit later on, but I would just do that and then tests. It's the D option. So I went up first because it's so far down the chi square geo F. That stands for goodness of fit test and the observed is L. One for me and the expected L two for me. Now if you put it in different categories you'll need to change those. But L one and L two if you copied off of that and then the degrees of freedom in this case would be five. Remember that was six minus one. And then we calculate and that gives us everything we need. So the chi square Is this top number here? We'll go and say 13.683. Alright so whoops So chi square value is 13.683. And that p value is also given, I don't know if you saw it or not but this p value here it's about .018. Yeah .018. Okay so then we explicitly compare that p value with the alpha value and it's just big enough That alpha values one this is 18 which is slightly bigger than the alpha. So whenever it's bigger than the alpha than we fail to reject. H not so we're accepting h not to be a true statement. So these are the U. S. Census census distribution and the sample distribution do in fact agree. So the way we're going to write this as a conclusion will say there is this is one way to write it. Now. You don't have to write it like this, but there is not sufficient evidence to suggest that the census distribution and the sample distribution do not agree. Now I'm not a big fan of that because there's a double negative, but you could also say there um uh there is enough evidence or there should be that there there is enough evidence to suggest that these two distributions do. In fact, agree.


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