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Test the following hypoheses by using the $\chi^{2}$ goodness of fit test.$$ H_{0} : p_{A}=.40, p_{\mathrm{B}}=.40, \text { and } p_{\mathrm{C}}=.20 $$$$ \begin{array}{c}{H_{\mathrm{a}} : \text { The population proportions are not }} \\ {p_{A}=.40, p_{\mathrm{B}}=.40, \text { and } p_{\mathrm{C}}=.20}\end{array} $$A sample of size 200 yielded 60 in category $\mathrm{A}, 120$ in category $\mathrm{B},$ and 20 in category $\mathrm{C}$ .a. Use the $p$ -value approach.b. Repeat the test using the critical value approach.

a. There is sufficient evidence to reject the claim of the specific distribution.b. There is sufficient evidence to reject the claim of the specific distribution.

Intro Stats / AP Statistics

Chapter 11

Comparisons Involving Proportions and a Test of Independence

Descriptive Statistics

Confidence Intervals

The Chi-Square Distribution

Missouri State University

Cairn University

Oregon State University

University of St. Thomas

Lectures

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03:28

Test the following hypothe…

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Perform the test of hypoth…

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Perform the indicated test…

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Apply the chi-square goodn…

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Consider the following hyp…

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We'Ll be performing a high square goodness of fit test and we're, assuming that the proportion of a is .4, the proportion that are in b is .4, and the proportion that is in c is .2, and then that not all above are correct or you can put Down what they had but i'll just write that down, and so, if we look at our little chart, we have that for our categories of a b and c or number of observed, values is 6120 and 20 point, and i put these values into list 1 and We know that our proportions are the .4. These are the expected proportions as .4 .4.2 and these total up to 200 point, and so in my list 3. This is my list. 2. In my list 3, i am putting my list 2 times 200 point and i get my expected values and my expected values are 80 and 80 and 40 and notice that they add up to 200 point, and so our ki squared statistics, which will have 2 degrees Of freedom is the 60 minus the 80 squared divided by the 80 plus the 120 minus the 80 squared divided by 80 plus 20 minus 40 squared divided by 40, and that ki squared statistic. I'M also going to use my software and do my goodness of fit tests. So i go to stat and tess. I happen to have a t i 84 and i'm going to gomisafit, and i have my observed in list 1 and i have my expected in list 3 and then you do have to tell the calculator how many degrees of freedom- and we do have 2 degrees Of freedom, 1 less than the number of categories, and so that calculated value comes out to be 35 and for part a. We want to do this with the p value and we find out that the probability of ki squared with 2 degrees of freedom being greater than or equal to 35 is equal to a very small value. .7034567 and then a 25. And this is definitely smaller than the .01 significance level that we're using. So we have evidence to reject the null. So we have sufficient evidence to say that this model is not being held. True and if we do this with the critical value, the critical value for the ki squared distribution, if we want to look for ki, squared value, ki, squared value, critical value, that has 2 degrees of freedom and has .1 in the upper tail, that value in my Table is, is 9.21 and anything higher than that will be to reject the nal and notice that our test statistic is way larger than that 9.21. So again, we would make that decision to reject the null and say that the model is not holding true.

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