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4market researcher an automoblle company suspwcl dlfferences preldrred colar butreen male and (emale buyers: Adverllsaments - targeted different Aroups should take ...

Question

4market researcher an automoblle company suspwcl dlfferences preldrred colar butreen male and (emale buyers: Adverllsaments - targeted different Aroups should take such d fferences into account; if they exlst The researcher examines the most recont salas Information of & particular cor that comes three calors. (You may Iind useful to reference the approprlote table: chl: square table table) mleAutoroblle Jaana ent FcuultColor Janlis BlckChoose the competing hypotheses determing whctne: colo

4 market researcher an automoblle company suspwcl dlfferences preldrred colar butreen male and (emale buyers: Adverllsaments - targeted different Aroups should take such d fferences into account; if they exlst The researcher examines the most recont salas Information of & particular cor that comes three calors. (You may Iind useful to reference the approprlote table: chl: square table table) mle Autoroblle Jaana ent Fcuult Color Janlis Blck Choose the competing hypotheses determing whctne: color preference depends on the autcmobile Duyer $ sex Ho Color preference Independcnt of Inc automobi buyer $ sex HA: Color preference dependent on the automobllc buyer $ se Ho Color preference dependent on tie automabile buyer'5 s0x; HA' Color preference Independent of the autonoblle buyer $ sex 6 1 Calculute the value oftha teSt statlstic (Round the decimal placee ) decima places and nta anstyer [0 3 calculation Elete 62 Find the pvaluc 0.025 cuau 0.05



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(a) identify the claim and state $H_{0}$ and $H_{a},(b)$ find the critical value and identify the rejection region, $(c)$ find the test statistic $F,(d)$ decide whether to reject or fail to reject the null hypothesis, and (e) interpret the decision in the context of the original claim. Assume the samples are random and independent, the populations are normally distributed, and the population variances are equal. If convenient, use technology. The table shows the salaries of a sample of individuals from six large metropolitan areas. At $\alpha=0.05,$ can you conclude that the mean salary is different in at least one of the areas? (Adapted from U.S. Bureau of Economic Analysis) $$\begin{array}{|l|l|l|l|c|l|}\hline \text { Chicago } & \text { Dallas } & \text { Miami } & \text { Denver } & \text { San Diego } & \text { Seattle } \\\hline 43,581 & 36,524 & 49,357 & 37,790 & 48,370 & 57,678 \\37,731 & 33,709 & 53,207 & 38,970 & 45,470 & 48,043 \\46,831 & 40,209 & 40,557 & 42,990 & 43,920 & 45,943 \\53,031 & 51,704 & 52,357 & 46,290 & 54,670 & 52,543 \\52,551 & 40,909 & 44,907 & 49,565 & 41,770 & 57,418 \\42,131 & 53,259 & 48,757 & 40,390 & & \\& 47,269 & 53,557 & & & \\\hline\end{array}$$

As we're looking at the mean price of agricultural books at the 10% significance level. The mean is supposed to be 8.45 So that's gonna be our hypothesis. $8.45 to be more specific. But that's it. And then we want to find out if there is a difference so that is a does not equal to tail test your hypothesis. And our green is gonna give us our calculated chi square statistics. So it's gonna be 28 minus one. So 27 divided by the deviation squared. So 8.45 mhm squared times 9.29 squared. And that is going to get us a value of put into my calculator because divided by 8.45 squared. And we get 32.63 Yeah. Yeah. And that's our chi square graph right? There does not equal means we chop it off here and here for a good dose. If we land in the shade of reach right here, we will rejecting all hypothesis at the 10% significance level. That means we're looking at half 5% on both ends. So we're gonna be looking at 5% for the right end and 95% for the left end at a degree of freedom of 27. So we're looking at 27 here and 5% is going to get us 40.113 Mhm. And for the left side, gonna get us at the 95% level. Yeah, and that's gonna be 29 degrees of freedom. So 17.7 all eight 32 is smack dab in the middle there. So we're good. We fail to reject the hypothesis and that is all that was required of us. Um mm The standard deviation is indeed 8045 cents.

Okay so the following is an a nova test uh for toothpaste and the mean cost per ounce For very good stain removal, good stain removal and fair stain removal. So the first step is to state your hypotheses and this is always the way it's gonna be. Is h not is that the means are equal and then that the alternative is at least one of them is different with just one of those means is different. The second step is you need to find the critical value. Now you can find the critical value either using a table or using a calculator. Now I've written a program in the calculator. Someone use a calculator but you can certainly use a table as well. I'm not going to go into detail on how to Create the program in the T. I. 84. You can search that online if you want. But the the uh there is a table for you to do this as well. So for a critical value F. Star we're going to call it uh you need the degrees of freedom for the numerator which in this case will be 10. And that's the number of columns we had. Very good, good and fair. So that's three columns minus the one gives us too. So that's the numerator. And then the degrees of freedom for the denominators 12. That's the total number of data values, which is 15 minus the number of columns, which is three. So that is enough and an alpha equals 0.5. So you also need to know your alpha. So alpha equals 0.5. So with these three pieces of information you should be able to find the F. Star in the book. But just to show you on the calculator. So I've written a program called inverse F. So this is going to give you my F. Star and the area is your alpha value. The degrees of freedom for the numerator was too and then for the Denominator was 12 and you should get about 3.89 or 3.885 will go to three decimal places. So that's your f. star. 3.885. So anything to the right or anything greater than 3.885 will reject the null hypothesis. So the next step is to find our critical value. Um Which is F. Now you can do that manually certainly but I'm going to go and use the calculator because it's so much easier. So I've actually already pre set this one up. You gotta step edit here your data values. So this was your very good column. The L one, the good column was L. Two and then the fair column was L. Three. Then if you go to stat and then air over two tests and it's your very last one in nova. And we go second L one comma second L two comma second L three. And then we close our parentheses and that gives us everything we need. So the F is about 4.8. We'll call it. So f is 4.8. Okay, Which is somewhere over here. And that means in step four we reject the null hypothesis, reject. H not now. You can also look at the P value. So this p value is 0.25 And remember that alpha value was point oh five. So um that's actually preferred to the P value whenever the p values less than alpha. That's whenever we reject the null hypothesis and the p value 0.2 is less than 0.5. But either way you reject you can use the traditional or the P value methods. So then the last step is where we just basically state what our conclusion is. So there is sufficient evidence. We'll just go and say there is sufficient evidence to suggest that at least one mean cost per ounce is different from the others.

Following is the solution in number 14 at one way Innova test. Uh and this is about the mean sales prices for three cities. And the null hypothesis here is that the mean sale prices are the same for these certain houses. And then the alternative is that at least one of them is different. The second step is to find the critical value and you need three pieces of information to find the critical value. One is your alpha, Your significance level, that significance level. In this case, that's usually given to you is 10. They also needed the degrees of freedom for the numerator, which is the number of categories in this case, the number of cities minus one. So there were three cities that we looked at minus one is two, so degrees of freedom for the numerator is two degrees of freedom for the denominator is the total number of data values minus the number of categories. So in this case, if you counted up those data values, there were 31 -3 cities that we looked at. So 31 -3 is 28. So that's what we need. So from there you can use a table or you can use software. I'm gonna use software. So I wrote a program and I called it inverse. F. I'm not going to show you how to write this program. You can youtube it if you wish. But um it makes Makes it easier for me. So the area is the alpha value. So we'll put in .10 for that. Degrees of freedom from the numerator was too. And then degrees of freedom for the denominator was 28. And that's going to spit out my f. star my alpha value and my uh critical value which is about 2.503. Let's call it 2.503 is my F. stars 2.503. So anything greater than 2.503. We're going to reject h not anything less than 2.503. And we're gonna fail to reject the null. The next step is to find the f statistic and you can do that manually but especially with bigger data sets that can be really time consuming. So I went ahead and punch this into stat. Edit. And these are my data values. So these are the I think these are in thousands of dollars but these are the mean sale prices and if you go back to stat and then tests and the very last one in nova And you put in your columns just make sure you separate them. This is on the T. 84 by the way but make sure you separate them by commas otherwise it's not gonna read it right so nova for those three columns and that's gonna give us everything we need to. The f statistic is about 0.966 Let's go and write that down. So 0.9 66 which is somewhere over here. So that lands in the non rejection region. So that's actually gonna tell us why our fourth step which is the decision and we're going to fail to reject H not since the F statistic is less than the critical value. Now you can also use the P value method that's what this second piece of information is good for it. Now this other stuff doesn't really matter. Um You can just kind of ignore it because this is really what we need. We need the F statistic and we need the p value and the p values pretty large. It's about 0.39 And what you do is you explicitly compare the P value with your alpha value. So the p value in this case is greater than your alpha value. 0.39 is bigger than 0.10. And any time you're P values greater than alpha, you failed to reject H nine. If it's less than alpha then you you reject. And then the final step is to conclude this, you know, with actual words and bring it back to the question at hand. And so what we're going to say is that there is not enough evidence or there is not sufficient statistical evidence. So there's not sufficient evidence to suggest that the mean sales price prices of houses In the three cities are different. Okay that's the five step in Nova process.

In this problem, We're going to be looking at the study that was conducted to investigate the association between cell phone use and the hemispheric brain dominance. So we have two samples. The first sample has, uh, 216 subjects who preferred to use their left year on bond, and and yet 166 are right 100. So the first proportion off right 100 people is 166 out off the 216 people who preferred to use their left ear. The second sample was coming from 452 subjects would prefer to use their right here. So this is from the left here. And this is for those who prefer to use their right ear on. For those who prefer to use your there, Right here are 400 36 were right handed out off 452. So we're going to be conducting tests, uh, to test the claim that the writ off right handedness for those who prefer to use their left here, it's less than the rich off right handedness for those who prefer to use their right here. In other words, the fast proportion is much greater than the second proportion. Okay, so let's compare and fast to do a hypothesis test, we will need to right there. Now hypothesis on the alternative hypothesis. So they're not. Hypothesis is p one he calls p two on the second hypothesis, the alternative hypotheses p one iss less than p two. Which is to say that this fraction this proportion is much smaller. That when you're and you're right, when you prefer to use your left here, then fewer people will be, uh, right handed. No for the zed, uh, calculate the value of that. We need to go to substitute the values into the formula. And this gives us the value of that off negative. 7.933 Now we need to get a critical value for this being a one tailed test and the level of significance being 0.1 then the critical value offset would be equal to negative 2.33 Because this time we're saying that p one IHS less than p two. So therefore it's going to be negative and not positive. Now we can compare the two by during the critical region and as you can see this would be negative 2.33 and the the left Month will be the critical region. And with that, we would have to notice that the calculated Valley offset is within the critical region. And for that reason we reject the null hypothesis. Rejecting the null hypotheses means that there is sufficient evidence to support the claim that the rate off right handedness for those who profound to use their left here is less than the rate off right handedness for those who prefer to use your right ear for cell phones. So it's much easier for you to use your your right hand if you're right handed. In other words. So we tested clean using, uh ah confidence interval. And for that, we need to complete the value off the margin of error has shown in this formula, and when you substitute the values you obtained e us zero point 0698 Next you substitute the values into the confidence interval expression and you find the limit will be negative 0.26 six aunt, the other limit his negative 0.1 to 16. So notice that all these values are negative, meaning zero is not contained within the confidence limits. And that means that there is a significant difference between the two proportions. And because the interval consists of negative numbers only it appears that they claim is supported. Therefore, we can see that the difference between the populations does appear to have practical significance.


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