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QuesTion 23A police chief wants to determine crime rates are different for four different areas of the city (East(1) , West(2) , North(3); and South(4) sides); and ...

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QuesTion 23A police chief wants to determine crime rates are different for four different areas of the city (East(1) , West(2) , North(3); and South(4) sides); and obtains data on the number of crimes per day in each area: The one-way ANOVA table shown belowS0uL Vanaton 3elmean uIcups20.20Wmcin CrcunsAt the 1% significance level, the critical value is 2.383.103 86

QuesTion 23 A police chief wants to determine crime rates are different for four different areas of the city (East(1) , West(2) , North(3); and South(4) sides); and obtains data on the number of crimes per day in each area: The one-way ANOVA table shown below S0uL Vanaton 3elmean uIcups 20.20 Wmcin Crcuns At the 1% significance level, the critical value is 2.38 3.10 3 86



Answers

Conduct the hypothesis test and provide the test statistic and the P-value and/or critical value, and state the conclusion. The police department in Madison, Connecticut, released the following numbers of calls for the different days of the week during a February that had 28 days: Monday (114): Tuesday (152): Wednesday (160); Thursday (164): Friday (179); Saturday (196): Sunday (130). Use a 0.01 significance level to test the claim that the different days of the weck have the same frequencies of police calls. Is there anything notable about the observed frequencies?

For this problem, we're testing to see if there's a statistically significant difference in the mean property values between two cities. For purposes of calculating, tax exempt properties were given two samples one for city A of sample size, uh, 16 and city be also sample size 16 without going into the calculations directly. We have an X bar of property values for City A of 24.75 and this is in ah, millions of dollars. And for city be we have a sample mean of 42.94 we're not given population various, So we have to calculate variance standard deviation from the sample For City A. That's standard deviation is 25.97 And for city be that standard deviation this 72.74. All right now we're calculating We're looking for differences between the underlying population means and we're assuming that they're normally distributed. So even though our sample size is less than 30 we're going to proceed on the basis that the underlying population is normally distributed and there before we can use either normal or T distribution. Statistics are no hypothesis. Is that the underlying means are equal. And if we disprove that, then we'll be looking at an alternative hypothesis that they are not equal. So we are not making any presumption about if they are unequal, in which direction which city we're not presuming or assuming which city has a higher mean value. Therefore, this is going to be a two tailed distribution, a two tailed test. Now our sample statistic is going to have the standard form of the difference in the sample means minus the difference in the expected values according to the null hypothesis and standardized by dividing by a standard error in this case, there we go now, since we are hype are no hypothesis is that the two means are equal. This term will simply turn into zero and our sample statistical simplify to this x x bar a minus expert be over our standard error term. We are asked to make this test with a significance level of Alfa equals 5% or 50.5 Also, because we are using the sample standard deviation rather than the population standard deviation, we're gonna have to use a student T distribution. This statistic is distributed according to student's T distribution, which requires that we determined degrees of freedom. All right, degrees of freedom equals the lesser of either an A minus one or N B minus one. Now, in this case, they both equals 16. Therefore, our degrees of freedom is 16. Minus one equals 15. No. So if we have a test statistic that is distributed T with degrees of freedom equals 15 and an Alfa 0.5 we can look this up and determined that are critical. Value is plus or minus 2.1315 and that can be looked up. If we were to visualize this Oh, that would suggest that any values of the test statistic falling within this range would lead us to not reject the null hypothesis. And in fact, if we do the calculations, we in fact come up with a T statistic of negative 0.9419 which is about there. And so this leads us to not reject the null hypothesis. Which is to say we do not have reason to assume that there is a statistically significant difference in the property values between the two cities

In problem 19. We're going to be getting the critical regions and critical valleys for three different cases. So in the first case, he we given the non hypotheses me one equals mu to be called me. Three equals before and that gives us, uh a full different factors. So we can say C equals four seized. A number of factors were also given that and equals 18 and they never off significance. Lunder equals zero 0.5 So to get the critical region and critical value for this case scenario, we need to get the degrees of freedom for the numerator on for the denominator. So the degrees of freedom for the factors will be given by C minus one and the degrees of freedom. But the era will be given by and minus C. So this bets This occurs when into the numerator degrees of freedom and the denominator degrees of freedom. And in this case, the degrees of freedom will be equal to for minus one, which is three and 18 minus four, which is 14. So in this case, we need to get the F money that corresponds to three degrees of freedom for the new maritime, 14 degrees ing freedom for the denominator with the level of significance of Europe in here. If I So in this case the value will be three 14 and 0.5 So this will be the critical Bundy for F end from the Tibbles. It corresponds to the tree point 34 So the region the critical region, it's going to be elected using a bag room. This zero, this is one and this is 3.34 So we can share this area. Is there critical region for the first case? In the second case be we're tool that the non hyper this is his me one equals me to equal. Smith three equals beautiful equals mu five that queues this five doctors were also given that n equals 15 and the level of significance London zero 0.1 So, in the same way, we want to get the degrees of freedom. Um, and we'll have to get the degrees of freedom that's follows F A C minus one en may not see and our for which 0.1 in this case, it's going to be f C five minus one 15 minus Phi and zero point the Roban. So this would be if four degrees of freedom for the numerator and 10 degrees of freedom for the genome, Eaton and Europe in Terrebonne level of significance and from the table, this value equals five point name name. So this is the critical body. And so the critical region will be obtained a Swallow's zero. And this is one on this lease. Life won't name any. So we can should the critical region for bad case and lastly k c. We have They're not help with issues hitch not as me one equals mewtwo because mitri and that gives us three factors. We're also told that the sample size is 25 and equals 2 to 5. On the level of significance, he's 0.5 So we're going to follow the same procedure to get the critical value, and then we can determine the critical region. So in this case, the degrees of freedom for the new Meritor would be three minus one that sense that he's seem in this one. Then for the dinner militate Enman C, which is 25 minus three and then the level of significant 0.5 So this would be F 22 a 0.5 level of significance. And when you fine, you look for this value of the table, he will look teen 3.44 So the critical value is 3.44 And for the critical region, we have to do the same thing. How zero there and one here and 3.44 is mocked at So I should add Region is critical region for that case.

Alright for this problem we're going to verify that we can use the chi squared independence tests before we stayed out our hypotheses. So we got lawyers making bank um 250,000, to 500,000. More than 500,000 in the government, judicial, private and salaried sectors. So this is our original table right here. All good. And then uh I just copied it from the last table since we had the same dimensions and adjusted it that way. So this is our expected table right here. Um So what we're looking at right here are 12 three values that are less than five, which violates our second term per second assumption being that we can't have expected frequencies that are less than five. So with that said we cannot apply the chi squared test. Okay?


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