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Point) The number of men and women among professors in Math, Physics, Chemistry; Linguistics, and English departments from SRS of small colleges were counted, and t...

Question

Point) The number of men and women among professors in Math, Physics, Chemistry; Linguistics, and English departments from SRS of small colleges were counted, and the results are shown in the table below: Dept MMath Physics Chemistry Linguistics English Men 53 30 26 36 Women 10 16 22 Test the claim that the gender of a professor is independent of the department: Use the significance level & = 0.025(a) The test statistic is x2(c) Is there sufficient evidence to warrant the rejection of the c

point) The number of men and women among professors in Math, Physics, Chemistry; Linguistics, and English departments from SRS of small colleges were counted, and the results are shown in the table below: Dept MMath Physics Chemistry Linguistics English Men 53 30 26 36 Women 10 16 22 Test the claim that the gender of a professor is independent of the department: Use the significance level & = 0.025 (a) The test statistic is x2 (c) Is there sufficient evidence to warrant the rejection of the claim that the gender of a professor is independent of the department? A: Yes B. No



Answers

Test the given claim. Listed below are student evaluation scores of female professors and male professors from Data Set 17 "Course Evaluations" in Appendix B. Use a 0.05 significance level to test the claim that female professors and male professors have evaluation scores with the same variation. $$\begin{array}{l|l|l|l|l|l|l|l|l|l|l} \hline \text { Female } & 4.4 & 3.4 & 4.8 & 2.9 & 4.4 & 4.9 & 3.5 & 3.7 & 3.4 & 4.8 \\ \hline \text { Males } & 4.0 & 3.6 & 4.1 & 4.1 & 3.5 & 4.6 & 4.0 & 4.3 & 4.5 & 4.3 \\ \hline \end{array}$$

So we would be assuming that the ratings for the females is equal to the ratings for the males. And alternately we just want to know if there's a difference. So we would do and not able to and use a two tailed tests. So I always like to draw a picture. And so we're going to assume that the difference between males and females is zero and we're actually getting a difference that's negative, what we're doing to tell tassel we're getting a difference that's negative, but then we'll symmetrically go on the opposite side and these two together will be R. P. Value. Now I use the software to find how the degrees of freedom and it ends up, the degrees of freedom comes out to be 13.77 And I put all my data in the calculator and into two lists. And I found that the mean of the first group of the females was 4.2 and the mean of the males was 4.1. And then divided by, we had the standard deviation of the females being 0.7 to roughly one square divided by the sample size which was 10. And then the males was 0.35 28 squared divided by a sample size of again 10. And when I got that test statistic that test statistic came out to be negative 0.315 So it was very close, it was actually way up here and then symmetrically so that P value we can see is very large. And so two times the probability of getting a test statistic with 13.77 degrees of freedom being less than or equal to this value. That p value comes out to be about 75 76% 760.756 Excuse me 0.7576 And so we fail to reject the null. Mhm. Therefore it appears as though the ratings for our males and females here to be the same. Mhm.

So we're looking at the reading of professors and we would be assuming that the mean rating for females is equal to the mean rating for males and alternately that they're different. We just want to find out if there's a difference between so we will be doing a two tailed test and I always like to draw a little picture. But remember we're assuming that the difference between these two is zero, and we can also write hypotheses that way, if we so choose so our smallest sample sizes, let's see. Our smallest sample size is 40. Therefore we will use 39 degrees of freedom and we will find our test statistics by taking the difference in our means. And so we're going to take that 3.79 minus the 4.1 And then we will be dividing it by the first standard deviation is 0.51 squared divided by that sample size. And then the second standard deviation is 0.53 square divided by its sample size, which is 53. And when I do that calculation, I find out that that test statistic comes out to be negative 2.25 And now we want to find, because we're doing a two tailed tests, we have this as one of our test statistics and the other one would be up here for our two tailed tests. And these two areas together give us our p values. So we want to find the probability of getting a test statistic less than or equal to negative 2.25 And then we want to double it for the other table, other tail. And we could look up the critical value for this as well. But I'm not going to set that up this way, I'm going to go through and use my T. C. D. F. So my T. C. D. F. I'm gonna have my left bound be like negative uh 10,000 and then my upper will be this negative 2.25 And I will have my degrees freedom being 39. Again that's like this isn't a conservative way. And then we're going to multiply that probability times two. And when I do that I find out that I get 0.4975 By just a tad bit. This is less than 5%. So we do have evidence to reject the not we have sufficient evidence to reject. And now and claim that the ratings are different mm for males and females. Now let's find that confidence interval. And because we did a two tailed test we would be finding them with a 5% significance level. We would be finding a 95% confidence interval and our confidence interval with 39 degrees of freedom. We don't have it in our table. So I'm going to actually use my inverse t to find that. And so I use my inverse T. I can put in the area and I'm gonna put in that lower areas point oh 25 since it's a 2.5% at the bottom tail, 2.5% of the top tail. And then I'll put in the degrees of freedom as 39. Again, this is a conservative approach and then we'll get that with that T value is so our T. Star value will be negative 2.227 So now when we get our confidence interval we need the difference. Let's get what that difference is. So that difference that 4.1 Well, I should say it the other way around the 3.79 minus the 4.1 That comes out to a difference of negative 0.22 it's a negative 0.22 plus or minus. And then we have our value for our T. Star value and then we'll use that big standard deviation and uh let me go back and find what those standard deviations are. So 0.51 point 53 0.51 squared and the 0.53 squared. And let's go back for the sample size is 40 and 53. Yeah. Right. Money. Mhm. Well, I never noticed before how this was 0.53 and that was 53 interesting. Not that makes any difference. So let's find the margin of error, first of all. Yeah, so 2.227 times the square root of 0.51 squared, divided by the 40 plus 400.53 squared divided by 53. And that margin of air comes out to be oh goodness! Isn't this interesting about 0.22? So when we find this interval this bottom while the negative and then minus is going to be minus 0.44 and then when we add them, that's going to be zero. So we're right at that cut off point that uh those values would uh we would only have negative values, we don't have zero. So this would be just at that cut off point to say that we would have evidence to reject the null but notice how close we were to 5% here and that's why this is such a close and we're here to so then it talks about the comparison between another example that did this and had smaller sample sizes. And yes, there is a difference. When we use larger sample sizes, we can make differences. The same differences be significant with larger sample sizes. If these were smaller sample sizes, we would not have been able to say and got the same meaning, same standard deviation, we would not have been able to have sufficient evidence to reject them all.

Problem. Number number 52. Uh, question A. Explain why it is reasonable to assume that the scores for the two students our independence And the answer is, um, suit is oh, uh, selected a trend them and randomly selected incident can be assumed tohave independent score. That's why we can as you in rebellion and we've been best. I think so. This is good, Jimmy. And this is the explanation for question. Be question be, um we know that which would be what are they expected value and sort of deviation off the difference between the scores Female minus minus mail. We know that you or female is equal to 1 20 but and sigma for the female is equal to 28 and the mean for the main is equal to ah 15 and the standard deviation off the mean equal to it's totally fine. Okay, this is from and the given the property off the means using the property with the means, we can say that new female minus man is equal to ah, you off the female minus, um, you off the media so it can be and when 20 minus one five. So it is 15 and we can see it say the same thing for a live aliens, The variance squared, Ah, female minus man equal. And the variants the radiance where the mean female squared plus the mean off man man square. So we can't say that. All right, so this is the sun, that deviation. Look for me, of course. So this is a standard deviation off the female, plus the sender deviation or the variance. Sorry for the mail. So the variance for the minute is 28 28 squared close the variance for the mail. It is 35. It's queer. So the final answer and he is Ah, do so then a nine. So from that we can say that that's standard deviation, which is the square root of the variance. So the center of the Asian off Ah, female minus meal is equal to X squared off 2000 and nine. So is the phone that enter is 44 a 0.8 8 to 1. And this is for question, be for which is he for questions? See, um, the question here is from the information giving you can can you find the probability that the woman chosen scores higher than the man. If so, find this probability of not explained in Hawaii. And we way we say that this is not invested with so so see rancher is not possible. Huh um, To find the probability is not possible to find the probability because we don't know the distribution off the score. For example, we don't know it normally or uniform distribution, what at the time for distribution. That's why we say that it is not possible to calculate this probably.

A problem. Number 52 and a of the students where selected that trend them and randomly Ah, selected signals. And me, I assume we have in revenge Bent course it's was his explanation or point a six point B Ah, for fun. Be, um we know that the new off if it's 1 20 there's some will be the agent Ah, stand on the vida off Rash is 28 and then you, uh um ah, the new open is well working with five. Still, division is a 30 crime. Okay, So if x and y are independent so the new of X minus Y is able to the new vax minors, um, you'll want so badly Ah ah If, uh my and it's me old, uh, we're very rooms in capital letter. So it would be, ah, mean or graph minus the mean man. So this is 1 20 minus 25 which is, I mean volumes minus and is equal to, um, variance or Istanbul. Ah, these in Rex V squared? Uh, yes, it's under the ignition or end a square. This is a property and it's a property. If x and Y are independent s so we can say so it squared plus, uh, 35 Where? Certain paper toe percent of the end. But mine eso we can say that somebody division f my understand is equal to the square root there somewhere on nine, which is equal to 44 point 8 to 1 that one night. And this is the answer for for me. Ah, question. See? And you can see that not we'll see. Approved. Burkle finds on the probability. Uh ah. Do we really think make a wish? Way? Doughnuts? No, at the distribution. Oh, the scores, for example. A normal uniform. Well, we don't know the distribution. Um, we don't know the distribution for for the scores. What is its signature is not possible to find in the Caribbean. Obviously, the question for 3 50 to see.


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