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ClipboardFontAlignmentMuMoTest ApplicationParametric TeststatisticvalueNonparametric Test statisticvalueOne-sample testsZ-test for population meanStd NormalSign tes...

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ClipboardFontAlignmentMuMoTest ApplicationParametric TeststatisticvalueNonparametric Test statisticvalueOne-sample testsZ-test for population meanStd NormalSign test for population medianXorStd Normal or Sign Testt-test for population meanr-tableSign test for a population modet-tableTwo-sample testsDependent samples t-test for the difference between meanscoffee-table Paired sample sign testX Or 2 Standard Normal Sign Test Wilcoxon rank sumWilcoxon signed- rank testDependent samples z-test for th

Clipboard Font Alignment MuMo Test Application Parametric Test statistic value Nonparametric Test statistic value One-sample tests Z-test for population mean Std Normal Sign test for population median Xor Std Normal or Sign Test t-test for population mean r-table Sign test for a population mode t-table Two-sample tests Dependent samples t-test for the difference between means coffee-table Paired sample sign test X Or 2 Standard Normal Sign Test Wilcoxon rank sum Wilcoxon signed- rank test Dependent samples z-test for the difference between means Std Normal Wilcoxon rank sum test Std Normal t-test for the difference between means r-table Tests for 23 samples Three-way ANOVA F-table Kruskal-Wallis test Chi-Square Correlation Pearson product moment correlation coefficient Pearson spearman rank correlation Ws spearman Randomness Binomial (none)l(none) Runs test for randomness Runs Test THERE ARE TEN MISTAKES IN THE TABLE ABOVE CIRCLE THEM Questions



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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 brief 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 $\sigma=8$.

The following is the solution for # seven, Number seven, which involves the archaeology or regional distribution of raw materials and comparing it to a current excavation site. And I did a little preliminary work here. First I um was we were given the absurd value so I just copy those down and then I found I had to find the expected values and I use those percents that were given. And I multiplied by the sample size to get the expected value. So I took the 61.3% or .613 and I multiplied by 1486. And I got that's where I got this 910.918 and that's what I did for all these percent. And I got these values here. So I needed that in order to find and what I like to do on these is kind of look at these numbers and compare them. And if they are really, really close something that these two distributions are are are going to be roughly the same. And in this case they look pretty darn close to the same, like 53 53 1 97 and 1 94 and so on. They're pretty close to the same. So, I'm thinking that we're gonna get a pretty high p value here, pretty small chi square value because they're so close together. Um but that's just a hunch. All right, so let's do it mathematically. So, first portion of this is finding the significance level that's given to you. It's always the alpha value. So, the significance level here is 1% of .1. The second part of this is writing the Nolan alternative hypotheses. And you can wear these however you want. But basically, the null hypothesis is saying that these two distributions are the same, and the alternative is saying that they're not the same. So in this case we can say for the knoll the regional distribution of raw materials fits the distribution at the current excavation site. And then I'll just shorthand the alternative is just saying that it's not so regional distribution of raw materials does not fit the distribution at the current site. Okay, So the next part, we need to find the chi square value and we're gonna find out with technology, so we're actually going to kind of come back to that. The also another part of this is finding the expected values, which we actually already did, and we just want to verify that all these expected values are greater than five and they are. So that's one of those conditions for inference that all the expected values need to be greater than five. And then it also asked us what distribution we're gonna use, we're gonna use a chi square distribution, obviously because it's a chi square goodness of fit test. But then we also have to say how many degrees of freedom In this case, there are 4° of freedom because there are five categories. So 5 -1 gives us the degrees of freedom of four. Okay, so from here we're going to be in the technology. So again, you can use I'm using the T I 84 but you know, you can certainly use Excel S. P. S. S sas are, you know, whatever statistical software package you want, I find T- 84 to be pretty easy for elementary stats. Okay, so I went ahead and took the liberty and L one, I put the observed in an L two I put expected and then if you go back to stat air over to test and it's it's close to the bottom. So I went up first and it's the chi square G. O. F. Test and that stands for goodness of fit test. And assuming you put observed in no one you're going to change that to L. One expected will be L. Two and then your degrees of freedom, like I said, we'll be four and then we're gonna calculate that thing and that gives us everything we need. So that chi square value, that's right up here, that's what we want. So it's about 40.198 So like I thought it's a pretty small chi square value, 0.198 You know, it's gonna be pretty pretty much smaller than anything we're going to get. And we are also given the P. Value and I don't know if you saw it but that's a pretty much as big as you can get. So 0.995 Very large P. Value. So 0.995 And that P value is obviously greater than alpha. It's definitely greater than that 0.1 that we were given. And any time the P values greater than alpha, we fail to reject H. nine And we failed pretty hard that time. Right? So we're not definitely not going to reject because that is a very very strong P value. Very very large P. Value. And the last part of this is party. It's um stating a formal conclusion using the word. So again there are different ways of wording this and I'm gonna say there's a double negative here. But this is the way the ap exam likes it. So it says there is not sufficient evidence to suggest that the regional that the regional distribution of raw materials is different than the current excavation site. Okay, so that's the formal conclusion would come up with. Whenever we fail to reject meaning, there is not sufficient evidence to say that these two distributions are different.

The following is a solution to number 10 and we're going to see if the days and I think it's january and Maui, there's a town in Maui or city in Maui and we're seeing if it follows a normal distribution, the average daily temperatures In January and there are 20 years worth. So it's like 600 20 or something days. They they looked at and there are two parts to this. The first part Um says, Okay, explain what these percentages mean, and essentially it just means that it's the empirical rule. Okay, so the, the reason why they did 34% and 34% is that makes up 68% and empirical rule says that 68% of the data lie within one standard deviation of the mean. So I'm kind of like looking at the middle of that chart right now and um that's where we get the minus sigma plus sigma. So we have a mean, or this is the at least the hypothesis, the mean was 68. I'm saying that in a race ago and so 68 and the standard deviations for that means 68% of the data is between 64 and 72°.. All right, if you do 68 minus 4, 64 68 plus 4, 72 then 95% lies within two standard deviations in the means. That's where we get the 13.5 and 13.5. If you add 13.5 plus 34 plus 34 plus 13.5, that equals 95%. So two standard deviations, if you take 68 minus two times four, that gives you 60 and then 68 plus two times four, that's 76. So 95% of the data is but then is between 60 and 76 degrees. And then finally, likewise, the 2.35 on either end, If you add all that together, you get basically 100%. So 99.7%, you know anything outside of that is definitely an outlier. So that means in this case Um essentially all the data will be between 56° and 80°.. Okay, so that's the first part, that's kind of the meaning behind that. Now let's look at the actual data. So we have observed days, so 14 data values were 14 days were, you know, within three standard deviations on the low side And then 86 within two standard deviations on the low side, and then 207 within one standard deviation and so on, and so forth. And then to 15, so that's all given to you. And then the expected, I just like we've been doing I did the 2.35 took the percent times the sample size, and that gives me the expected data values. Okay, So now let's take a look at The questions that says, what's the significance level? Well, that's given to you, it's the 1% significance level, and that's gonna be your office, your alphas .1. And then we need to write down our hypothesis. So the null hypothesis is that these are matching, these two distributions are matching. So, in other words, the average daily temp the average daily temperature in january for whatever this town is, I think something somewhere in mali follows a normal distribution with I mean 68° and then standard deviation for all right. And then the alternative is just essentially saying not so, you can pretty that up if you want, but the average daily temperature in january does not follow a normal distribution. Alright. Part B says, what's the test statistic? What's the chi square value? We're gonna use technology to find that. But the second part of that says um have the conditions of inference been met and we look to see to make sure that the expected value is greater are are all greater than five and they are. So if you look at these expected values, they're all greater than five, that's good. And then the last part of this, what type of distribution we're gonna use? We're gonna use the chi square distribution obviously because the chi square goodness of fit test, But we need to say how many degrees of freedom and we're gonna use 5° of freedom since there are 6 um observed values. So there are six different buckets I guess you could think of And -1 is five. Okay, so now let's look at some technology so I'm gonna use the IT4 but you can certainly use different form of software. You can certainly you can even use the the formula if you want. Although that might take a while. So I took the liberty to go ahead and type all this stuff in. If you had a cal can edit or Staten edit At L1 is where my observed values are and then L2, these are my expected values and it should be a mirror image. These data values should be the same once you reach the peak. Okay? So then you go to stat once you type those in and then tests and it's one of the last ones. So I went up first since the chi square G O F. That stands for goodness of fit test. L one is your observed L two is you're expected at least. It is in my case. Now, if you have those switched or if you have them in different columns, you'll need to change those. And then the degrees of freedom in this case is five, so then calculate and that gives us everything we need. So that first part, that's the chi square value and that is point 256 and 7.256, which is a super small one. And if you look at if you look at that p value, I mean, that's one of the biggest p values ever seen. So p values 10.998 I pretty much can't get much better than that. That is a very, very big P value and it is definitely greater than alpha, which means we fail to reject. That's two words, by the way, I fail to reject. H Not All right, So, this does follow, you know, almost identical. And you can kind of see that these data values are really close to the same thing. So, um yeah, so, this does follow a normal distribution and concluding it again, you don't need to write it like this, you can deviate from this, but I would say there is not sufficient evidence to suggest that the average daily temperature in january does not follow a normal distribution With means 68 and standard deviation four Kind of a double negative there. So, another way you could say that is um the average daily temperature in January does in fact follow a normal distribution with mean 68 and standard deviation for

Theme. Each note is that the one is smaller than or equal to Muto, and each one in that New York is bigger than you. So the degree of freedom, which is in one lost in two miles to I mean, which is equal to fit. So the critical, very correspondent is toe degree freedom, equal to 50 and off, April toe open or five one tail. So using Table five is equal to 1.6 seventies. So the fool the standard deviation, which is equal to square root off and one minus 1 20 to minus one times this one square 0.89 square well in the Tu minus one. So 13 minus one times is to square as 2.7 mine squared over and 1% from minus, which is 22 plus 30 minus two, which is ableto open a 30. So the distance statistic so T is equal to x one bar one x two so fine by forcing one that's 5.12 Oh, which is open 833 square root off one over and one yes, one over and to which is equal to 1.454 So if the value of this this is is in the rejection reasons and now have process rejected, So as 1.454 is smaller than 1.676 Okay, so we fail to reject the null hypothesis you

Any time we're doing a chi square test, it's always a one sided test in the right tail, the police mhm.


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