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Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. (Note: Answers in Appendix D include technology answers based on Formula 9 - 1 along with "Table" answers based on Table $A$ - 3 with df equal to the smaller of $\boldsymbol{n}_{I}-\boldsymbol{I}$ and $\boldsymbol{n}_{2}-\boldsymbol{I} .$ ) Blanking Out on Tests Many students have had the unpleasant experience of panicking on a test because the first question was exceptionally difficult. The arrangement of test items was studied for its effect on anxiety. The following scores are measures of "debilitating test anxiety." which most of us call panic or blanking out (based on data from "Item Arrangement, Cognitive Entry Characteristics, Sex and Test Anxiety as Predictors of Achievement in Examination Performance," by Klimko, Journal of Experimental Education, Vol. 52, No. 4.) Is there sufficient evidence to support the claim that the two populations of scores have different means? Is there sufficient evidence to support the claim that the arrangement of the test items has an effect on the score? Is the conclusion affected by whether the significance level is 0.05 or 0.01? $$\begin{array}{|c|c|c|c|c|} \hline {}{} {\text { Questions Arranged from Easy to Difficult }} \\ \hline 24.64 & 39.29 & 16.32 & 32.83 & 28.02 \\ \hline 33.31 & 20.60 & 21.13 & 26.69 & 28.90 \\ \hline 26.43 & 24.23 & 7.10 & 32.86 & 21.06 \\ \hline 28.89 & 28.71 & 31.73 & 30.02 & 21.96 \\ \hline 25.49 & 38.81 & 27.85 & 30.29 & 30.72 \\ \hline\end{array}$$ $$\begin{array}{|c|c|c|c|} \hline {}{} {\text { Questions Arranged from Difficult to Easy }} \\ \hline 33.62 & 34.02 & 26.63 & 30.26 \\ \hline 35.91 & 26.68 & 29.49 & 35.32 \\ \hline 27.24 & 32.34 & 29.34 & 33.53 \\ \hline 27.62 & 42.91 & 30.20 & 32.54 \\ \hline\end{array}$$

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

Number eight. A researcher believes a new diet should improve weight gain in laboratory mice. If 10 control mice on the old diet gained an average of four ounces for the standard deviation of 40.3 ounces paul's and write that down. This is the old we have 10 mice. They gain an average of four ounces with a standard deviation of 40.3. While the average gain for the tin mice on the new diet gained an average of 4.8 with the standard deviation of 0.2. What is the P value? Assume all conditions for infants are meant, all the answer choices are written out in terms of formula so that we're not gonna solve. We're just gonna set up and then see which one matches. Uh First thing I want you to realize is an option C. And E. They've set up a binomial distribution. We are not going to use Obama mail distribution in this case so we can eliminate options C. And E. Um Also, if you take a look at A and B notice that in the denominator the radical is spread out across both parts of this. While an option D, it is not like that option D. Does not have um the standard deviations turning into variances. And the only way to that you can add standard deviations is if you turn them into a variance, then add them together. So option D. Can't be the answer either. So now we're between A and B. Now, multiple choice to make this pretty simple, we have a sample of 10 mice using the old method, a sample of 10 mice using the new method notice that the only difference between A and B is T or Z. So this whole question turns into a matter of is this A. T. Or is he? So what we're looking at here is means we have a group of 10 mice in the old method, a group of 10 mice and the new method, and we have standard deviations that relate to those samples. So I'm gonna change my notation here, so it's more appropriate. Whenever I'm reading, mean standard deviation is quickly jot those down. What we really have here is a sample average for the old mice, a sample average for the new mice, a sample standard deviation for the old mice. We'll just write that is oh sample standard deviation for the new mice. So because we don't have the population standard deviation and we only have the sample standard deviation, this is gonna be a. T. So the answer to eight is a probability that t is less than four minus 4.8. Um No four minus 4.8 would be the difference between the two. Now. Um Normally there is a mountain Zero here, that is not included because there's no need to subtract zero develop bar now we have to combine these standard deviations. Well you can't just add standard deviations, so we're gonna have to take each one and square them first. And because we are doing averages, there's a sample size of 10 in each one. We have to divide by 10, then take the square to the entire thing. So the answer is a but there's plenty of ways to arrive at your answer. Besides setting up the problem, process elimination eliminates binomial and eliminates um and anything else that's not appropriate at the time.

Okay. So this problem we got insects and chocolate bars are average is going to be 14.4 and a 225 chocolate bar. 225g. And we're looking what is the probability that we find? zero insects in a 225 grand bar. So that's gonna be the probability that X. Is equal to zero. Yeah. And our average is gonna be 14.4 since we're looking at the whole bar And it's gonna be our -144. Uh Well it's hard to write. Mhm. Okay. And then we'll we'll divide that by one. Of course this is gonna be one. So this will be our answer. Oh but wait there's one thing I forgot here access supposed to be zero. So this part zero which makes this one. So our answer is just gonna be eating the power of negative 14.4. Next for B we have 1/5 of the bar tested. So that's gonna be your same answer from last time. Since we're looking at the probability of X equals zero. But at a different average. So it's gonna be e to the power of negative 14.4 times 0.2 for that 1/5. And that's gonna be your answer for beat. Now for seat were asked was a probability that we get at least one in a 28.35 grand bar. So first things first probability of X is at least one Is going to be equal to the probability of 1- the probability that there's zero. Yeah. Sure. I should write x equals zero. Mm. Mhm. So then we'll do that. It's gonna be one Either power of native 14 4 Times 20%. No it's not 20. It's a different percentage this time. And that's gonna be 28 0.35 Out of the 225g times are 14.4 average. And there it is. So we're going to have that be our answer On 1914.4 times 28.35 over 2 25. The power of eight. We're gonna subject one for that. And now for part D. We have. So the standard deviation is just gonna be the square roots that I mean, Which is gonna be the square root of 14.4, which is gonna be about three years old. And we're looking at more than twice. So that's gonna be like 28th and Between 14 and 28. There's not a difference of three. So, yes, this is unusual.


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