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Suppose we have a multinomial population with four categories: $\mathrm{A}, \mathrm{B}, \mathrm{C},$ and $\mathrm{D} .$ The nullhypothesis is that the proportion of items is the same in every category. The null hypothesis is$H_{0} : p_{\mathrm{A}}=p_{\mathrm{B}}=p_{\mathrm{C}}=p_{\mathrm{D}}=.25$A sample of size 300 yielded the following results.$\begin{array}{lllll}{\mathrm{A} : 85} & {\mathrm{B} : 95} & {\mathrm{C} : 50} & {\mathrm{D} :70}\end{array}$Use $a=.05$ to determine whether $H_{0}$ should be rejected. What is the $p$ -value?

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

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Lectures

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so this question were given the following hypotheses and have to figure out you know what is true. We have to conduct a hypothesis test and ah are no hypothesis is that all the proportions for each of our treatment groups are equal, and our alternate hypothesis is that not all of the proportions for our treatments are equal. So I created this table over here. That's a pretty neat way of organizing our information. So the first column asked us to describe the expected proportion. And because our no hypothesis is that all the proportions are equal, that means this would be equal to one divided by the number of treatment groups that we have, which is for so this is equal to 10.25 So each of our expected proportions are 0.25 And now we have our observed frequency, which is just the number off, um, of, ah, values that we have over or get so hum in our first group A. This is equal to 85. Our second group. This is 95 our third group. This is 50 and our fourth group is 70 and our expected frequency is the proportion or are expected proportion times our total sample size and our sample size and equals 300. So this would this over here would be equal to 0.25 times, 300. So this would be 75 it would be the same for all of our treatment groups. And now the difference would be so This is basically the given information and now we have to conduct a hypothesis test. So we are going to do a chi squared goodness of fit test and this is the formula for that. What we're going to calculate is the actual frequency minus the expected frequency squared over the expected frequency and take the sum for each treatment group, or take the sum of all of these values for each treatment groups so we can collect this information this way. So our difference is gonna be the reserve frequency minus the expected frequency which is 85 minus 75 which is 10 and then 95 minus 75 which is 20 50 minus 75 negative, 25 70 minus 75 negative five and now we're going to square our differences. So that will be 104 100 6 25 and 25. And now we have to compute the, um, this square difference over the expected um frequency. So that will be equal to 1.3. Repeating for Group A 5.3 Repeating for Group B 68.3 Repeating for groups C and 0.3 repeating for Group D And now we have to come up with a chi squared value so our chi square will be the sum of our last column. So if we take this, some archives squared is equal to 15.3 repeating. So this over here would be equal to 15.3 repeating. And now we have to discover our degrees of freedom in order to come up with a P value and our degrees of freedom and acai squared is equal to the number of treatment groups that we have minus one and we have four categories on treatment. I'm sorry. We have four categories, so four minus one is equal to three. So we have three degrees of freedom. So our key value would be our ah chi squared value with degrees of freedom of three. So with that, we get that our P value must be less than 0.5 So our P value is less than 0.0 five from our chi square table. And since we are where we are comparing against an Alfa of zero point 05 Um, yeah, our alpha 0.5 If you want to make sure so are Alfa is your point 05 And because 0.5 is obviously less than 0.5 we reject the knoll. So what does that mean for us? That means that there is sufficient evidence to reject the claim that each distribution over the proportion are the same.

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