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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 of independence? (e) Interpret your conclusion in the context of the application. Use the expected values $E$ to the hundredths place. Sociology: Movie Preference Mr. Acosta, a sociologist, is doing a study to see if there is a relationship between the age of a young adult $(18$ to 35 years old) and the type of movie preferred. A random sample of 93 young adults revealed the following data. Test whether age and type of movie preferred are independent at the $0.05$ level. $$ \begin{array}{l|c|c|c|c} \hline & \multicolumn{4}{c} {\text { Person's Age }} & \\ \cline { 2 - 4 } \text { Movie } & \multicolumn{1}{c} {18-23 \mathrm{yr}} & \multicolumn{1}{c} {24-29 \mathrm{yr}} & \multicolumn{1}{c} {30-35 \mathrm{yr}} & \text { Row Total } \\ \hline \text { Drama } & 8 & 15 & 11 & 34 \\ \hline \text { Science fiction } & 12 & 10 & 8 & 30 \\ \hline \text { Comedy } & 9 & 8 & 12 & 29 \\ \hline \text { Column Total } & 29 & 33 & 31 & 93 \\ \hline \end{array} $$

It's a kind of an interesting question if we we would be assuming that the anxiety level if we are going from easy to difficult is equal to the anxiety level. If the questions were difficult to easy and alternately, we just want to know if there's a difference and easy to difficult and not equal to difficult to easy. And so we're going to assume that that difference is actually zero so that the, I'll just write E G minus D. E. That that difference is zero. And when I put this into my calculator, um we have the lists have different sample sizes. We have that first group of the E. T. D. I have, there are 25 numbers versus the D two E. There are 16 numbers and we could use the degrees of freedom of 15, but I'm going to use the degrees of freedom from the formula. And that degrees of freedom ends up being uh 30 almost 39. So 38 point well, I'm just going to call it 39 it's approximately 39 degrees of freedom. And that test statistic that we're going to get, we need to take the mean, which was 27.1152 minus the mean of the other group, which was 31.7 to 8125 And then divided by the square root of. And the first standard deviation all around that a bit is 6.857 one square, divided by the sample size, which was 25. And then the second standard deviation was 4.26 square divided by the sample size of 16. And when I got that test statistic, the test statistic came out to be, I'll just read it up here negative 2.6 566 And so it came out down here. And since we're doing a two tailed test, we also use this one. And so what's the likelihood of getting a test statistic in this distribution that is less than or equal to that negative 2.6566 That gives us this tale and then we want the other tales to double it. And that p value comes out to be a 0.114 So at a 5% significant level, this is smaller than 5%. So we would have sufficient evidence to reject now. Yeah. And say that the mean anxieties are different so that they're different. The means are different. However, at a 1% significance level we would fail to reject the novel. Mhm. And we'd have to say here that they're actually not. They appeared to be the same. So it does depend on our significance level. How did pick you want to be. But it is an interesting idea in all my years of teaching, I never thought about the arrangement of having them all go easy, too difficult or difficult to easy. So what interesting concept.

The following is a nova test based on the mean salaries for different metropolitan areas. So the alternative or the null hypothesis is that all the means are the same. So there are six metropolitan areas, I think it goes Chicago, Dallas Miami, Denver san Diego and Seattle. Uh So the null hypothesis is that all the means are the same. And then the alternative is that at least one of them is different. The second step is to find the critical value and you can do that using either software or a table, But they're essentially three things you need. The first thing is your alpha value, your significance level and that's usually given to you the problem and that's .05. Then you need the degrees of freedom for the numerator and the degrees of freedom for the denominator. And the way you find that Is the degrees of freedom for the numerator is the number of categories -1. So there were six cities that we looked at our metropolitan areas, so 6 -1° of freedom would be five for the numerator. And then for the denominators, the total number of data values minus the number of categories. So there were 36 data values minus the six metropolitan areas. So 30 is your degrees of freedom for the denominator. So that should be enough to use a table. But I use a calculator and I wrote a program in here called inverse. F. I'm not going to show you how to how to write the program. You can youtube it if you wish. Um But this is what I do. So um I put in my area which is my alpha value, my degrees of freedom is five and then my degrees of freedom for the denominator is 30 and That gives me my critical value. About 2.534 2534 is my critical value. I call f. star. So 2.534. Okay so anything greater than 2.534. We reject the annual hypothesis that all the means are the same And anything less than 2.534. We failed to reject meaning the h not is true. Okay so the second step is to find the F statistic and there's a formula but it's a bit of a mess. I always use software you know technology is a great thing. So if you go to stat and you can type in your data values. So these are the mean salaries um So again L1 I think was Chicago and then this is the mean salary for Dallas Miami Denver San Diego and Seattle. So there are six categories. And if you go to stat tests and then we're gonna go to the Unova test and then you just type in your columns separated by commas remember there were six columns, six data columns that we used and we need to make sure that all of them are in there and last one and then also you know make sure you separate those by commons, otherwise it's going to read it wrong. So then um that gives us everything we need. So the F. Is the F statistic, that's the third step. So we're looking at this it's about 2.281 as our F. Value. So two point 281 is our f statistic Which is actually barely in the non rejection region 2.281. So that means we fail to reject. Okay and also we can verify that with this p value here. So the p values 0.7 which is a pretty small p value, but it's still in this case greater than the alpha value. So the alpha value remembers point oh five, so it's barely greater than the alpha value. And whenever it's greater than the alpha value, uh we failed to reject, I should probably put H not there, so we failed to reject H not whenever the P values greater than the alpha. Okay. So then the last step is to summarize everything with actual words. So what does this all mean? It just means that there is not sufficient evidence, there is not sufficient. I guess you could say statistical evidence to suggest that the mean salaries from the different metropolitan areas are different. Okay. And that's the five step process for an Innova one way and over test

For this exercise. We are given the following information. So we have these hypotheses and we're told that the true parameter P is p prime and P prime is less than the null hypothesis proportion. So therefore the alternative hypothesis is actually true now. For part, they were asked to show the following. Where's Ed is the test statistic for a one proportion test. So the one proportion test statistic is given as follows. Now, if we calculate the expectation on Zed, everything inside the brackets is constant except for the sample proportion. So the hypothesized proportion and and are both constant. So the expected value can be rewritten like this. This is because the expected value for a constant is that constant. Now, we also know from Chapter two that the expected value for a sample proportion is the true value. So this becomes and this is what we're trying to show, at least at least with respect to the expected value. So now, for the variants now, in our test statistic again, everything is a constant except for this sample proportion. So that is the only varying parameter. Now I will rewrite this like this, so I'm taking this constant here outside of the variance brackets. So therefore I squared. So that's why these square root symbol has disappeared. And we also know from Chapter two that the variance for a sample proportion is given by the following. So therefore we have the following result and this is what else we were trying to show in part A. So that completes part A. Now in part B, we want to show at the power of the lower tailed test is that is we want to find the probability of getting a test statistic lesson or equal to negative of the critical value when the alternative hypothesis is actually true. Now, if our test statistic is that and we have just found its expected value and it's variants, we would expect that this value is normally distributed as standard normally distributed because we are taking a parameter, we're subtracting its mean from it and then normalizing by its standard deviation. So to find the power of the test, we're looking for the probability that the standard normal distributed parameter is less than negative, the critical value and now just plugging in our values for the standard deviation, which is the square root of the variance of the test statistic and the expected value for the test statistic. And if you look at what's asked for in question be when you were asked to show this right here now for Part C were given this information. So we have some hypotheses and were given the true proportion as 0.8 and we're testing at a significance level of 0.1 This means that are critical. Value is negative, 2.3 to 6, and we are also given n equals 225. And so we know that we know that the alternative hypothesis is actually true and were asked, What is the probability that our test will detect this? What is the probability that our test will result in us rejecting the null hypothesis? In other words, what is the probability of us getting a test statistic less than minus 2.3 to 6? So all we have to do here is plug in all the numbers into this formula, and this gives us a value of 0.978 So the power of this test, the probability of rejecting the null hypothesis when the alternative hypothesis is true, is 0.978

Following is a solution video to number 24 and this is where we compare to means uh for the soil, water content for field a compared to feel b. And the first part is just to verify that the mean and the standard deviations are in fact these numbers, So the mean for field day is 12.53 and the standard deviation is 2.39 And then the mean for field B was 10.77 with the standard deviation of 2.4, and it says use a calculator something use this T I 84 I went ahead and typed in the means, or sorry, the data values L one and L two, so L one represents fuel day and then L two represents, you'll be and if you go to stat falcon and it's one of our stats and we're going to change that to L one and calculate and that gives us everything we need. So the X bar, is that 12.53 So that's verified. And then we're looking at the s the standard of the sample standard deviation is about 2.39 So for the field A. That is correct. And then let's just go and double check field be. So we're gonna change it to L. Two now because that's where field B. Is, and then that verifies its 10.77 for the mean, then about 2.40 for the standard deviation. So the first part is done, that's verified. And then the second part it says conduct not formally, but we're basically just going to conduct a uh uh two sample T. Test with an alpha value of point oh five. And we're gonna go back to the calculator because doing it by hand can be pretty cumbersome. So if you go to stat and then test now, since we already have, we're gonna go to two sample t. Test since we already have the data in there. I'm just gonna use the data instead of the summary stats. So list one is L. One that's the field A. List to is L. Two. That's field B. And then these frequencies just keep them the same and then we have the the alternative hypothesis and you kinda have to use your context clues here. But this one actually explicitly states is field A. Is the soil content or water content higher or greater than. So I'm going to change this to greater than So μ one is greater than you to warm you. A. is greater than YouTube pulled is usually zero or no unless they tell you otherwise. So we can go ahead and calculate and that's gonna give us everything we need. So the T. Value if you want you can put it down there. But really it's just this P. Value that I want to see. So 00.27 is the p value. So P value equals 0.27 And then we compare that with the alpha value and it's less than the alpha value. So any time it's less than the alpha value. That means we reject the no hypothesis. So we're rejecting the statement that these means are the same. And we are I guess you could say accepting the fact that these two means are in fact the water content and field A. Is greater than the water content and field be on average. So we can type this out as there is enough evidence to suggest that field A. Has on average a higher soil content soil. Sorry, water content. Banfield beat. Okay so this field A. Has on average a higher soil water content on field because since we are rejecting that null hypothesis and accepting the alternative hypothesis.


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