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Question (20 marks)(a) Let %z be the systematic sample mean We know that %, is an unbiased estimator off the NJ population mean. Show that Var(;) = "7152, (all...

Question

Question (20 marks)(a) Let %z be the systematic sample mean We know that %, is an unbiased estimator off the NJ population mean. Show that Var(;) = "7152, (all symbols have their usual meaning) According to POPD Ltd; there were 118 Covid-19 cases reported in March 2020 in Kenya . The number of cases in each reporting region is listed below with the regions numbered from North to South and identified by numbers only:Region Number Number of CasesSuppose that public heath researcher had decide

Question (20 marks) (a) Let %z be the systematic sample mean We know that %, is an unbiased estimator off the NJ population mean. Show that Var(;) = "7152, (all symbols have their usual meaning) According to POPD Ltd; there were 118 Covid-19 cases reported in March 2020 in Kenya . The number of cases in each reporting region is listed below with the regions numbered from North to South and identified by numbers only: Region Number Number of Cases Suppose that public heath researcher had decided to estimate the total number of Covid-19 cases in Kenya in March 2020 not by contacting each ofthe N-l2 reporting regions; but by selecting One in-three systematic sample from the 12 regions. Selecting starting region number on the list at random from the numbers 1, 2 3, the sample includes that region and every third region thereafter on the list. (0) List every possible systematic sample that could be obtained with the design marks) To estimate the total number of cases XT (the true value is 118) the researcher multiplies the total number of cases in the sample by 3. Is this estimate unbiased for the population total (Xr) Justify your answer_ marks)(111) (jii What is the true variance of the researchers estimate? (3marks) For an estimate of the variance in pait (ii1) . the researcher used 12(12-4)52/4 where s" is the



Answers

Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. (Note: Answers in Appendix D include technology answers based on Formula 9 - 1 along with "Table" answers based on Table $A$ - 3 with df equal to the smaller of $\boldsymbol{n}_{I}-\boldsymbol{I}$ and $\boldsymbol{n}_{2}-\boldsymbol{I} .$ ) Blanking Out on Tests Many students have had the unpleasant experience of panicking on a test because the first question was exceptionally difficult. The arrangement of test items was studied for its effect on anxiety. The following scores are measures of "debilitating test anxiety." which most of us call panic or blanking out (based on data from "Item Arrangement, Cognitive Entry Characteristics, Sex and Test Anxiety as Predictors of Achievement in Examination Performance," by Klimko, Journal of Experimental Education, Vol. 52, No. 4.) Is there sufficient evidence to support the claim that the two populations of scores have different means? Is there sufficient evidence to support the claim that the arrangement of the test items has an effect on the score? Is the conclusion affected by whether the significance level is 0.05 or 0.01? $$\begin{array}{|c|c|c|c|c|} \hline {}{} {\text { Questions Arranged from Easy to Difficult }} \\ \hline 24.64 & 39.29 & 16.32 & 32.83 & 28.02 \\ \hline 33.31 & 20.60 & 21.13 & 26.69 & 28.90 \\ \hline 26.43 & 24.23 & 7.10 & 32.86 & 21.06 \\ \hline 28.89 & 28.71 & 31.73 & 30.02 & 21.96 \\ \hline 25.49 & 38.81 & 27.85 & 30.29 & 30.72 \\ \hline\end{array}$$ $$\begin{array}{|c|c|c|c|} \hline {}{} {\text { Questions Arranged from Difficult to Easy }} \\ \hline 33.62 & 34.02 & 26.63 & 30.26 \\ \hline 35.91 & 26.68 & 29.49 & 35.32 \\ \hline 27.24 & 32.34 & 29.34 & 33.53 \\ \hline 27.62 & 42.91 & 30.20 & 32.54 \\ \hline\end{array}$$

So Chevy Jebbie Shelves rule is 100%. Are 100 times 1 -1 of her case squared percent of the data values is within K standard deviations where K is greater than one. So our solution is the first time the Z score, which is x minus mu divided by sigma. Ah this is equal to 17.28 -14, divided by 0.4. This equals 8.2. So using Chevy chiefs inequality or sorry, the rule, we'll just plug that in 100 times one minus 1/8 10.2 squared and we get approximately mhm 98 51%. With the other weights are at least the 17 oz.

We have a sample with N equals 83. And of that 83 members of the sample 64 satisfy a certain condition. Want to meet our equal 64. Now we want to use the sample data to test the claim that p does not equal 2.75 for the population at a confidence level of 5% or alpha equals 0.5 Now that we've identified the confidence level, we can go to the following procedural steps to conduct this hypothesis test first. Is the normal distribution appropriate to use? Yes, it is. Because both N. P and Q. P. R greater than five. B. One of the hypotheses were testing we're testing whether or not H not P equals 10.75 or H a p does not equal 0.75 Ak We're conducting a two tailed test. Next compute P hot. And the test statistic P hot is all over. N equals 20.77 Z is given by the formula on the right, taking in as input P hat P Q n N producing down to Z equals 0.42 Next we compute the P value. We use the table to see the area outside Z equals plus and minus 0.42 as we shaded in yellow and the graph on the right. This P value corresponds to 0.6745 Next we reject H not, no, we do not. P is greater than alpha and we interpret this signing to suggest that we lack evidence that P does not equal .75.

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.

Soon. Number 28. What's your name? The level of significance is mentioned is all for equal 0.1 on our hypothesis, which state that the population mean new is equal to the value mentioned in the claim, so it would two new equal 4.55 g. The alternative hypothesis stated the opposite off the novel hypothesis according to the clean, thus using lesson. So each one to mu is less than 4.55. Dreams is alternative hype with his old listen. Then the test is left. 15 is the alternative. Hypotheses uses Biggers in, so the desk is right field. The alternative hypothesis uses non equal, so the test is toe field, so the answer will be left field question. Number be given export equal 3.75 being equal 4.55 Zita equal 0.7 and unequal six. The sampling distribution off the sample mean export is normal because the population distribution X is assumed to be normal. The sampling distribution off the sample mean has mean new and stirred and standard deviation. Zita over square root and does it? Value is a sample mean decreased boy. The population mean divided by the standard division so that equal explore minus new over Zita over square root in equal 3.75 minus 4.55. Mhm 0.7 over square root six equal. Minus 2.8. Question number C Riddle Sport A and B. It snowed too. Mu equal 4.55 g each one to mu is less than 4.55 g and did equal minus 2.8. The B value is the population off obtaining a value more extreme or equal to the standard test static Zet that remind the probability using tepidly. So, probability equal probability off export is less than 3.75 Equal probability off. That is less than minus 2.8. Equal 0.26 Question number We? Yeah, given Alpha giving Goma equal 0.1 Result port. See, P equal points here is you 26 If the value is smaller than significance level Alfa, then then I'll Hypothesis is rejected. Mm. Is listening 0.1 So I reject h No. If we rejected them on hypothesis the data is the data easy, statically significant at level Alfa Question number e result, 40 each node to new equal, 4.55 g and each one to mu is listening four point 55 graham result for the project. Each node. There is sufficient evidence to support the claim that the mean, the mean weight off these periods in three sports off the Grand Canyon is less than 4.55 g.


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