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Compute the F-ratio test for the significance of Factor A?Use for #9-#I0. study was conducied the eflectiveness supermarket sales strategies- At one supermarket the...

Question

Compute the F-ratio test for the significance of Factor A?Use for #9-#I0. study was conducied the eflectiveness supermarket sales strategies- At one supermarket the price level (regular: reduced price. and cosi supermarket) and display level (normal display Fpice, Hai display enace plus chd of aisle. and twice the display space) Wcrc tested determine ifthey had any efiect 0n the Weekly sales of a particular supemarket proxluct Each of the combinations of price leve and display level were put in

Compute the F-ratio test for the significance of Factor A? Use for #9-#I0. study was conducied the eflectiveness supermarket sales strategies- At one supermarket the price level (regular: reduced price. and cosi supermarket) and display level (normal display Fpice, Hai display enace plus chd of aisle. and twice the display space) Wcrc tested determine ifthey had any efiect 0n the Weekly sales of a particular supemarket proxluct Each of the combinations of price leve and display level were put in place for rndomly Tected wcek and the weekly sales of the product wcre recorded. Each combination was used three times over the COuTSC of the experiment The results of the study are shown pelon Ktrulir Furrr Faict Krutru Tutital [Tatela Aoamal Flu: End ANOVA Table Em MS [Ji-alay 14[54 Xt MR +IT DJi-alav"ParL 317U 137676 Lun ahhs 7 [dentily the treatment groups used this experiment; 4) The three price levels used by the supermarket b) The three display levels used by the supermarket The nine combinuone of price level display level used by the supemarket d) The weekly sales collected for each of the weeks 10. Find the test statistic for determining whether the factor of Display significant; 4) 0.2499 b) 495 257.93 d) 1708.48 Rattal



Answers

(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) $$\begin{array}{|l|l|l|l|c|l|}\hline \text { Chicago } & \text { Dallas } & \text { Miami } & \text { Denver } & \text { San Diego } & \text { Seattle } \\\hline 43,581 & 36,524 & 49,357 & 37,790 & 48,370 & 57,678 \\37,731 & 33,709 & 53,207 & 38,970 & 45,470 & 48,043 \\46,831 & 40,209 & 40,557 & 42,990 & 43,920 & 45,943 \\53,031 & 51,704 & 52,357 & 46,290 & 54,670 & 52,543 \\52,551 & 40,909 & 44,907 & 49,565 & 41,770 & 57,418 \\42,131 & 53,259 & 48,757 & 40,390 & & \\& 47,269 & 53,557 & & & \\\hline\end{array}$$

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.

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.

All right. We want to test the hypothesis. New equals 64.5 versus the null hypothesis or alternative hypothesis does not equal 64.5 for population standard deviation 0.6 given the following data at alpha equals 0.1 confidence. This question is testing your knowledge of how to conduct a hypothesis test for population renew. When the standard deviation sigma is known. We see through the five steps listed here to solve first, we verify normality. So from the normal probability bar on the left, in the box, on the right, we conclude that the data is normal without outliers. Thus we can proceed to solve so and be we calculate that sample size and X five simple meat. This is N equals 22 X 5 64.7 Thus you see we can calculate our test statistic, this is zero equals experiments were not over segment about 10, which gives negative 38.53 for alpha equals 0.1 We continue the critical value in step D. For two tailed tests using a Z. Table where the area H 2.5 are critical value is plus or minus 2.58 Thus we can conclude in step E the following first. Let's rewrite our critical Z score correctly. Since our Xena is to the left of our negative 2.58 we can conclude that zero is in the critical region. So we reject H. Not because you know it's a critical region. We also address the fact that ALPHA equals 00.1 is more reasonable than Alpha equals 0.1 because there are severe punishments for making the type point ever in this problem I. E. There's a severe punishment for making an area generally. So we want to keep the 0.0.1 stringent criteria.

Uh, h note in that new one is mother than or equal to me too. And each one is that new one is bigger than mutant. So the degree of freedom, which is the minimum off anyone Where? This one in two minus one. So is 16 and 13. So the minimum 13 and the critical value in the row Corresponding toe degree. Freedom is off. Ableto appoint one on table five. So the critical value is 1.35 So the distances is Is that x one bar minus X tow bar over. Squared off S one squared over N one plus is two squared over into. She is equal to 3.429 So at 3.429 is it's a bigger than 135 So we reject this offer is won t This value is from table five


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