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Q2. Stef works for an aquarium and has noticed that the number of aquarium visitors per week may depending on the duy of the week: Based on her experience with the ...

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Q2. Stef works for an aquarium and has noticed that the number of aquarium visitors per week may depending on the duy of the week: Based on her experience with the aquarium, Stef expects that Monday through Thursday should have the same numbers of visitors, Friday and Sunday should each have twice the numher of visitors as O One of the lirst Iour days; and Saturday is expected to have three tirnes the attendance as on one of the first four days:In a randomly selected week; 2500 people visited th

Q2. Stef works for an aquarium and has noticed that the number of aquarium visitors per week may depending on the duy of the week: Based on her experience with the aquarium, Stef expects that Monday through Thursday should have the same numbers of visitors, Friday and Sunday should each have twice the numher of visitors as O One of the lirst Iour days; and Saturday is expected to have three tirnes the attendance as on one of the first four days: In a randomly selected week; 2500 people visited the aquarium: the results are shown below. Test whether the data distribution differs from Stef" expected distribution_ Day Visitors Monday 173 Tuesday 220 Wednesday 256 Thursday 274 Friday 357 Saturday 732 Sunday 488 Provide the null and alternalive hypotheses lwr the chi-square lest: B. Calculate the expected counts for each weekday. Create similar table as above for the x" components and calculate the /" test statistic_ D. Would you reject or fail to reject Ho at the 0.05 signilicance level? Justily Four ansicr



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. Ecology: Fish The Fish and Game Department stocked Lake Lulu with fish in the following proportions: $30 \%$ catfish, $15 \%$ bass, $40 \%$ bluegill, and $15 \%$ pike. Five years later it sampled the lake to see if the distribution of fish had changed. It found that the 500 fish in the sample were distributed as follows. $\begin{array}{cccc}\text { Catfish } & \text { Bass } & \text { Bluegill } & \text { Pike } \\ 120 & 85 & 220 & 75\end{array}$ In the 5 -year interval, did the distribution of fish change at the 0.05 level?

We have data sets A and B listed here. We want to test the claim that there is no difference in the underlying distributions for populations A and B at the alpha equals 0.1 significance level. We proceed through steps A through D listed below this question is testing an understanding of non parametric tests in particular how to implement the ranks on test. So first we stayed alpha and hypotheses, this is alpha equals 0.1 has stated H. And distribution of the same H. A distributions are different and be we can take the test that we also see the sampling distribution which is normal and check the requirements which are met. So the ranks are as follows. We compute N one and two as 12 and 12. That's when you are is 1 50 single. Rs 17.32 are this summer brings from A is 1 66. Z is not unless you're over signal articles 660.92 Thus from a normal distribution we compete the P value as two PZ greater than absolute values equals 20.28 Thus, we can conclude, indeed, that we fail to reject nation on, since P is greater than alpha, which means we lack evidence to support the alternative hypothesis H. A.

What we want to conduct A p D. T. A pair differences tests at the alpha equals 1% confidence level. Testing the claim that the population means A bar next pr are not equal. We have the data for A and B. Given below assuming amount shapes, not your distribution on the right. I've already calculated D. Bar noted that an equal seven and calculated SD as 70.47 So we proceeded to do the five steps listed below to solve this first. We evaluate the requirements and hypotheses. So the requirements to use the students distribution have been met because the distribution shape we have degree of freedom and minus one equals six. Are null hypothesis is mute equals zero. Or alternative is beauty does not equal zero. And we're testing at alpha equals 00.1 confidence Next will compute the test statistic and P value. So the statistic is T equals D. Bar divided by SD over. Uh huh. 2.083 from a tea table. This puts p between .1.05. So we can conclude that P is greater than alpha, which means we fail to reject the null hypothesis, which means that we lack evidence that beauty does not equal to about.

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.


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