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You wish to test the following claim (lat signilicance level ol & 0,001 For the context of this problem, /44 141 where the first data set represents pre-test an...

Question

You wish to test the following claim (lat signilicance level ol & 0,001 For the context of this problem, /44 141 where the first data set represents pre-test and the secord data set represents post-test.Il;4 Hla:| < 0You believe the population of difference scores Is normally distributed, but you do rot know the standard deviation: You obtain pre-test and post-test samples for n 38 subjects_ The average difference (post pre) Is ( 10 with standard deviation of the differences of 94/ 44,5.W

You wish to test the following claim (lat signilicance level ol & 0,001 For the context of this problem, /44 141 where the first data set represents pre-test and the secord data set represents post-test. Il;4 Hla:| < 0 You believe the population of difference scores Is normally distributed, but you do rot know the standard deviation: You obtain pre-test and post-test samples for n 38 subjects_ The average difference (post pre) Is ( 10 with standard deviation of the differences of 94/ 44,5. What is the test statistic for this sample? (Report answer accurate t0 three decimnal places. test stalistic What Is the p-value for this sample? (Report answer accurate t0 four decimal places. p-value The p-value I5. = less than (or equal t0) ( grcater than ( Thl: test scatistic Iead? dechsion tO. reject tie null accept te nUll fajl t0 reject the null



Answers

Testing the Difference Between Two Means (a) identify the claim and state $H_{0}$ and $H_{a}$, (b) find the critical value(s) and identify the rejection region(s), (c) find the standardized test statistic t, (d) decide whether to reject or fail to reject the null hypothesis, and (e) interpret the decision in the context of the original claim. Assume the samples are random and independent, and the populations are normally distributed. If convenient, use technology. Annual Income A personnel director from Pennsylvania claims that the mean household income is greater in Allegheny County than it is in Erie County. In Allegheny County, a sample of 19 residents has a mean household income of $\$ 49,700$ and a standard deviation of $\$ 8800 .$ In Erie County, a sample of 15 residents has a mean household income of $\$ 42,000$ and a standard deviation of $\$ 5100 .$ At $\alpha=0.05,$ can you support the personnel director's claim? Assume the population variances are not equal. (Adapted from U.S. Census Bureau )

The following is a nova test based on the mean salaries for different metropolitan areas. So the alternative or the null hypothesis is that all the means are the same. So there are six metropolitan areas, I think it goes Chicago, Dallas Miami, Denver san Diego and Seattle. Uh So the null hypothesis is that all the means are the same. And then the alternative is that at least one of them is different. The second step is to find the critical value and you can do that using either software or a table, But they're essentially three things you need. The first thing is your alpha value, your significance level and that's usually given to you the problem and that's .05. Then you need the degrees of freedom for the numerator and the degrees of freedom for the denominator. And the way you find that Is the degrees of freedom for the numerator is the number of categories -1. So there were six cities that we looked at our metropolitan areas, so 6 -1° of freedom would be five for the numerator. And then for the denominators, the total number of data values minus the number of categories. So there were 36 data values minus the six metropolitan areas. So 30 is your degrees of freedom for the denominator. So that should be enough to use a table. But I use a calculator and I wrote a program in here called inverse. F. I'm not going to show you how to how to write the program. You can youtube it if you wish. Um But this is what I do. So um I put in my area which is my alpha value, my degrees of freedom is five and then my degrees of freedom for the denominator is 30 and That gives me my critical value. About 2.534 2534 is my critical value. I call f. star. So 2.534. Okay so anything greater than 2.534. We reject the annual hypothesis that all the means are the same And anything less than 2.534. We failed to reject meaning the h not is true. Okay so the second step is to find the F statistic and there's a formula but it's a bit of a mess. I always use software you know technology is a great thing. So if you go to stat and you can type in your data values. So these are the mean salaries um So again L1 I think was Chicago and then this is the mean salary for Dallas Miami Denver San Diego and Seattle. So there are six categories. And if you go to stat tests and then we're gonna go to the Unova test and then you just type in your columns separated by commas remember there were six columns, six data columns that we used and we need to make sure that all of them are in there and last one and then also you know make sure you separate those by commons, otherwise it's going to read it wrong. So then um that gives us everything we need. So the F. Is the F statistic, that's the third step. So we're looking at this it's about 2.281 as our F. Value. So two point 281 is our f statistic Which is actually barely in the non rejection region 2.281. So that means we fail to reject. Okay and also we can verify that with this p value here. So the p values 0.7 which is a pretty small p value, but it's still in this case greater than the alpha value. So the alpha value remembers point oh five, so it's barely greater than the alpha value. And whenever it's greater than the alpha value, uh we failed to reject, I should probably put H not there, so we failed to reject H not whenever the P values greater than the alpha. Okay. So then the last step is to summarize everything with actual words. So what does this all mean? It just means that there is not sufficient evidence, there is not sufficient. I guess you could say statistical evidence to suggest that the mean salaries from the different metropolitan areas are different. Okay. And that's the five step process for an Innova one way and over test

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.

18. So it's no if that new one is equal to mean to, and each one is that anyone is not equal commute. So the degree of freedom, the minimum off in one way, this one, which is 17 and 18 to minus one, which is 19th. So the video is 17, using all for April toe 0, 00.11 teen. So for table five, the critical, very equal toe postive or negative 1.74 So the rectory, then they contain all value smaller than negative 1.74 and all various larger than 1.74 So that is the statistic, if that makes one minus x two bar. So it's 56900 minus five 7800 minus me one minus me, too over square root off this one squared over in one, which is one toe. Once you lose your square over and once with the 18 lost, it's too square. So it's 8000 square over into approximately negative 4.267 So it's a very off statistics. Even in the final hypotheses rejected so and this value realize between the booze values. So we failed to reject the null hypothesis

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.


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