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4Award 10.00 polntsGiven the following contingency table; conduct test for independence at the 10% significance level (You may find

Question:

4 Award 10.00 polnts Given the following contingency table; conduct test for independence at the 10% significance level (You may find useful to reference the ppropriate table: chi-square table or table) Variable Varable Choose the null ad alternative hypotheses_ Ho: The tvo variables are indepondent ; HA: The two variables are dependent Ho: The tvo variables are cependent ; HA: The twa variables are independeni; Calculate the value of the test statistic. (Round intermediate calculations to at least decimal places and final answer to decimal places_ Test statistic Find the p-value oniea-d 0.10 0,05 Dvalue 01'0 0.025 p-valv 0.05 0,01 p-value 0,025 p-valuc 0.01 What vour concluslon? Do not reject Ho; we Camdi conclude that Ine [wO categcries are dependent Reject Hc; cannat canclude Inat the twvo categories are dependent; Do not reject Ha; we can ccnclude that the two categories are dependent: Reject Ha; we can conclude that the two categores are dependenz rev; 06 2019 CS-170121 Hints Hint# References Difficulty: Medium Workshoct Learning Objective: 12-02 Conduct test tor Independence


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Incomes: Lawyers and Architects How do the average weekly incomes of lawyers and architects compare? A random sample of 18 regions in the United States gave the following information about average weekly incomes (in dollars) (Reference: U.S. Department of Labor, Bureau of Labor Statistics). $$ \begin{array}{l|ccccccccc} \hline \text { Region } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline \text { Lawyers } & 709 & 898 & 848 & 1041 & 1326 & 1165 & 1127 & 866 & 1033 \\ \hline \text { Architects } & 859 & 936 & 887 & 1100 & 1378 & 1295 & 1039 & 888 & 1012 \\ \hline \\ \hline \text { Region } & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 \\ \hline \text { Lawyers } & 718 & 835 & 1192 & 992 & 1138 & 920 & 1397 & 872 & 1142 \\ \hline \text { Architects } & 794 & 900 & 1150 & 1038 & 1197 & 939 & 1124 & 911 & 1171 \\ \hline \end{array} $$

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Yeah. We have the following matched pairs and we want to conduct a scientist and matched pairs testing the plane P less than 0.5 at alpha equals 0.16 Presents this question assessing our understanding of how to conduct a non parametric test. A scientist and matched pairs for population data. So we have to proceed through such a through D to solve first and a wee state alpha. And our hypotheses this is alpha equals 0.1 H P equals 0.5 Or an H A p is less than 0.5 are clean.

Now we can get the test and B we have the signs or rather the change from population would be as follows. There are 16 pluses minuses. So X is 13/16. The number of pluses over total plus minus. Thus we have zero equals excellent 00.5 over 0.25 and routed or 2.5 years in our next identify the P value from a normal pair of this is P equals PZ greater than actually magazine equals 0.62 Thus we conclude indeed, that we reject because P is less than equal to alpha.

We have evidence to support our hypothesis are cheering hypothesis. We have evidence to support the less than 0.5..

So we're going to assume that the two variables are independent and our two variables were the type of crash and type of vehicle. And alternately that the two variables are dependent. So our degrees of freedom. We take one less than the row total and one less than the column totals. We have two degrees of freedom.

Were using a five significance level. So it's a one tailed test on at the right tail and we put all .05 there. and with 2° of freedom we have that that Critical value is 5.99. So any time we have a value of a test statistic that's greater than 5.99. That's going to cause us to reject The null and using software.

I get that the the test statistic comes out to be extremely large. 139.03. So I can't even represent that on there. It's way down. We are definitely going to reject the null.

So we can conclude that there is evidence. There is evidence mhm. That there is a relationship between the type of crash. I am the type of vehicle, so vehicles don't have crashes in the same way, and you could continue to do some proportion test to see where those differences lie..

Okay. So we're going to calculate the sum totals of our first table. So in the first round we have 4:13 and 15. Yeah In the 2nd row we have 5 11 18. Okay.

So the some of that first one is going to be four plus 13 plus 15 is 32 16 Plus 18 is going to be 34. That adds up to be 66. And then down here we have a total of nine 24. Mhm. And 33.

All right, so now that we've got all these ones down we can get our expected table. So are expected value for this first one right here um is going to be the road times the column divided by the sample. So that's gonna be 32 and nine Divided by 66. Okay. Okay.

And we get 4.36 which is less than five. So we'll keep that in mind as we calculate the other values. So the next one we're gonna do the same thing. 32 and 24 Calculate this one right here. 32 times 24 divided by 66.

So we've got 11.6. And then in the next one we're going to have 32 times 33 Divided by 66, It's gonna be 16. It's bottom value right here. nine and 34 nine times 34 divided by 66. So we have 4.6.

And right here we're gonna call it quits because we got two values that are less than five. Therefore we cannot use the chi squared independence test. So that's part A cannot use chi squared now for part B. Um slightly different story. All right.

So now and we plug these values into Excel with our new data sheets We got over here is our original the top one. Okay. All our values right there, sum total of 66. Nice. All right.

Important part the expected values. None of them Our five are less than five. So that's great. That means we can use archive squared independence. Mhm She's very good.

And then we can take the difference of these two values. The expected minus the original. So it's going to be this one minus this one squared divided by the expected so that one again. Mhm. And then we get these respective values here.

Yeah. Yeah. So then we get our totals and then the sum this is our chi squared right here we could look it up with degrees of freedom of He Rose -1 columns -1. Nothing Israel degrees of freedom are equal. The Rose Rose -1 times columns -1.

But I just wanted to calculate the p value instead since I had an excel. And right here is our p value Of 0.12. So at the 1% significance level, we failed to reject the null hypothesis. And we can say that there is no, well, we say that there is not sufficient evidence to say that there is, so that an association exists between social class and frequency of..

All right, so we're gonna be comparing the ratings between Siskel and Ebert. Um No hypothesis is that there is no association between their ratings and alternative hypothesis is that there is an association between their ratings. So I've gone ahead and copied out these values into Excel and here we are plopped them right there, hold up to go fetch it. So here it is. Um what we're looking at here is our originals on the left here, this right here is our original And then this is our expected right here.

So we're expecting 17 0.78 So it's going to be about 18, 18 Down Mixed and Up Reviews. So right there and then we take the expected minus D. Observed squared and then divided by the expected to get us these values right here and then we take the sum of all of these values in there to get 146 for our chi square, which results in a p value of basically zero, it's super super small. So we reject the null hypothesis in favor of the alternate.

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