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Conduct tne hypothesis test and provide the test statistic; critical value and value and state the conclusion: person randomly selected 100 checks and recorded the ...

Question

Conduct tne hypothesis test and provide the test statistic; critical value and value and state the conclusion: person randomly selected 100 checks and recorded the cents portions of - tnose = checks Tne table below lists those cents portions categorized according to the indicated values. Use significarice level lo lest Ihe claim Ihal Ihe four calegories are equally likely: The person expecled Inial many checks for whole dollar arounts would resull in disproporlionalely high frequenc for tne iirs

Conduct tne hypothesis test and provide the test statistic; critical value and value and state the conclusion: person randomly selected 100 checks and recorded the cents portions of - tnose = checks Tne table below lists those cents portions categorized according to the indicated values. Use significarice level lo lest Ihe claim Ihal Ihe four calegories are equally likely: The person expecled Inial many checks for whole dollar arounts would resull in disproporlionalely high frequenc for tne iirst category; but do the results support that expectation? Cents portion of check 0-24 25-49 50-74 75-99 Number Click here t0 view the chi-square distibution table. The test slatistic i5 (Type an integor e decimal The critical value is (Round t0 three decimab places as needed ) The P-value 9.348 (Round to iour decimal places as needed: ) State the conclusion: There sufficient evidence to warrant rejection of tne claim that the four categories are equally Vikely The results the frequencv for the first cateqory is disproportionately hiqh. to support the expectation tnat



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

So for this problem we're told the significance of 0.01. The claim is that we pennies are manufactured so that the waves have a standard deviation equal 0.0-30 graphs. And for this problem we want to identify them. No alternative hypothesis testes is six p values, critical values and the make a claim based on what were the data we did based on our process. So, first there's an empire, the no hypothesis. Alright, highlight that this is a protesting for sanity, aviation and that this equals 0.0 230 g from the claim. Yeah. The alternative, we're still testing Sander deviation but there is no greater than or less than So we don't really care as long as it does not equal 0.023 g. So this is a two tailed test. So secondly, let's now find our testes cystic. Yeah. For the test statistics, since we're testing the standard deviation, this is gonna be a chi square distribution. And we can find it to be M -1 squared divided by sigma squared where n is the length of the data. And as squared you can find using this form from the data And if you plug this in, I got a testis cystic of three points 81 for all right. So, once you have that, you can find your p value, you can find the p value is one of two ways either through the table. Um but if you do it through the table method, you won't get an exact value of finding a range that your p value falls under. Or you can use technology like our our stack crunch, which will give you a more accurate p value. So the main function in our final p values where chi square distribution with bp Yeah. Hi, square. And then we're going to have our testes cystic Degrees of freedom and -1 and we're gonna have this lower dot tail equals something At one point on this is a two tailed test. So this p value around to multiply this p value by two and remember this P value is a probability. So if we say the p value is equal to false and we get greater than one then we might we might not be doing the right tail. So let's try P value is equal to true. And we get a p value of zero point two five three and we like this because the p value still less than one. If you were to try this with the lower tell equals of false, you should get a p value greater the one which is not which is no longer a probability. So remember we can make in conclusions. If the p value, it's greater than alpha, we will accept the null hypothesis. Yeah. And if the p value less than alpha, we will reject no of assists. And here we can see that 0.253. Well that's greater than our alpha which is 0.01. So we should accept on our policies process for the sake of completion. Find the critical values and compare it with our testes cystic. Yeah. So you can use the table again. Find the degrees of freedom and alpha. But if you you the degrees of freedom or we're going to be chi squared with one minus off over to degrees of freedom and chi squared Alpha over 2° of freedom. Yeah. And from the table you should get the numbers. You can also use technology like our the functions are gonna be the same will be q chi square and for the most part inputs are the same. You're gonna putting off over to Your N -1. The only thing that's going to change is this lower dot tail. So we're gonna have one that's lower dot l equals true. And this gives him one 34. Yeah. Another one. The whole q chi square alpha over to and -1. Lower dato equals two false. And this gave me 21 point 955. So how do you make conclusions based on the critical values and the testes cystic. So if we have a high score distribution We have these points. The 1.34 And the 21.96. These are giving us two different areas and these are gonna be our alphas over to And both of these are gonna be our rejection region. So if our tested falls within those areas, we reject ever if it falls within the middle, that is our acceptance region. So we won't we will accept our normal processes. So we go back to see what our test statistic was. This is 3.814. So that falls probably somewhere around here, or this was our testes statistic. So again, our testes, it falls within our acceptance region. So we will accept the null hypothesis was which was the same conclusion. We came with the P values in alpha. So This is not statistically significant alpha level of 0.01. So we will have to accept that the standard deviation is equal to 0.023.

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.

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.

Problem. 17 8. Each note is that new one is smaller than or equal commute. Each one is equal to me. One is bigger than you, so the degree of freedom is the minimum off. Anyone minus one into my storage is 18 and 14, so it's equal toe food correspondent to critical value with offer. April 2.451 tail so T is equal to 1.761 So the actual reason contain old values. Great around 0.761 So the test statistic is equal to x one bar minus X to bar. So for 97 year zero minus 4 to 000 over square. Note on 8800 square over 19 plus 51 Use your square over 15, which is ableto 3.19 point. So it's a very all the tests statistic in the rectory does not have pointy inject. Three point is smaller and it's bigger than 1.761 So we reject then the hypothesis


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