Input the data as follows:x1 represents the number of cases of cancer in the sample fromMaine. x2 represents the number from New Hampshire. n1 andn2, represent the number of people in each sample (50,000).Because we are interested in whether the proportion of cases ofcancer is higher in Maine than in New Hampshire, choose >p2.Then select Calculate to produce the results shown at the right.2-ProeZTestx1:291n1:50000x2:251n2:50000P1:#P2 P2 >PZCalculate DrawIf the samples were truly random, the set of all possible differences between the sampleproportions would form a normal distribution. The informal description of a normal distribution is a"bell shaped curve". The results of the test show whether the difference observed in the data istypical of most of the differences or if the difference is in fact considerable.2-PropZTestP1>P2z=1.72282231P=.0424602946The screen shows a p-value of about 0.042. This value means that if there were no actualdifference between the cancer rates of the two states, the difference observed in these sampleswould happen by chance about 4.2% of the time. Many statisticians use 5% (0.05) as a maximumvalue to decide if an observation is rare, so they would conclude in this example that it is unlikely toget such results as these by chance. Statisticians use sampling to make predictions about thewhole population. In this example, it is likely that the actual cancer rate in Maine is higher than thatin New Hampshire, based on the sample data. For a p-value higher than 5%, statisticians wouldstate that there is not enough information to conclude that one rate was actually higher than theother. They would not conclude, however, that the there is no difference between the rates.P1=.00582P2=.00502P=.005422. Connecticut and Rhode Island are also neighboring states. Compare their cancer data using atwo-proportion test. Choose p2, as appropriate. What is the p-value for this test?Explain the meaning of this number in the context of this example. Do you have enoughinformation to conclude that one rate is larger than the other?O3. The sample data in this activity indicate that 268 out of 50,000 people in both the Connecticutand Massachusetts samples were diagnosed with cancer. Does this mean that the actual ratesof incidence of cancer in the total populations of the two states were the same? Compare theircancer data using a two-proportion test. Choose either p2. What is the p-value for thistest? Explain the meaning of this number in the context of this example.Ai4. What is the relationship between the p-value obtained for the p2 test and the valuefor the # p2 test?In*Snowing now6R4 In "Waste Not," the FBI discovers that the children in the area of a sinkhole seem to have anunusually high occurrence of cancer. FBI agent Megan Reeves believes that this might represent a"cancer cluster." Charlie warns her not to jump to this conclusion too quickly. The FBI may haveused a statistical technique called a two-proportion test to decide whether the rate of cancer seemsto be unusually high in this location. A two-proportion test is a statistical technique that can be usedto decide if the rate of cancer seems to be different in different areas of the country. This activitycontains an introduction to this test.Consider the following data on cases of cancer, based on a hypothetical random sample of50,000 people from each of six states in New England. A large sample is used here because therate of incidence of cancer is quite small.StateConnecticutMaineMassachusettsNew HampshireRhode IslandVermontCancer Cases268291268251281249Based on these data, does there seem to be an association between location and incidence ofcancer? One way of attempting to answer this question is to perform a statistical procedure calleda two-proportion test. This is one of many procedures for testing hypotheses that are studied instatistics classes.1. Maine and New Hampshire are neighboring states. What percent of the Maine sample hadcancer? What percent of the New Hampshire sample had cancer? Do you think that thedifference between these two percentages is considerable?EThe data shows that the proportion of cancer cases in the sample from Maine is higher than theproportion of cancer cases in the sample from New Hampshire. If the samples involved arerandomly chosen, a two-proportion test can determine if it is likely that the proportion of cases forthe entire state's population is higher in Maine than New Hampshire. A hypothesis test assumesthat there is no difference and inspects the data to see if they in fact do not support thisassumption. In this case, the hypothesis is that there is no difference in the cancer rates for thestates. The alternative hypothesis is that the cancer rate is higher in Maine than in NewHampshire. A two-proportion test can indicate that the initial hypothesis should be questioned.To perform this test on your calculator, press [STAT] and go to theTESTS menu. Then select 6:2-PropZTest....EDIT CALC TESTS1:2-Test...2: T-Test3:2-Samp2Test....4:2-SamPTTest....5:1-PropZTest.....582-ProPZTest...IZIZInterval***Snowing now

Given Information: Consider the given data: State Cancer Cases Connecticut 268 Maine 291 Massachusetts 268 New Hampshire 251 Rhode Island 281 Vermont 2491. Consider x1 as the number of cases of cancer in the sample from Maine and x2 as the number of cases of cancer in the sample from New Hampshire. x1=291, n1=50000x2=251, n2=50000 The percentage of Maine sample had cancer is, p^1=x1n1=29150000=0.00582=0.582% The percentage of the New Hampshire sample had cancer is, p^2=x2n2=25150000=0.00502=0.502% The difference between these two proportions: p^1-p^2=0.00582-0.00502=0.0008=0.08%2. Consider x1 as the number of cases of cancer in the sample from Connecticut and x2 as the number of cases of cancer in the sample from Rhode Island. The null and alternative hypotheses are: H0: There is no difference in the cancer rates of states.H1: The cancer rate is less in connecticut than in Rhode Island. The step-by-step procedure to perform the 2-prop Z test using TI-83 is shown below: Go to STAT> TESTS > 2PropZTest. Now enter the values of x1,n1,x2 and n2. Select the <p2 and then click on Calculate. The output is shown below: The p-value for this test is 0.2889. if there were no difference between the cancer rates of the two states, the difference observed in these samples would happen by chance about 28.9% of the time. Decision criterion: Reject the null hypothesis if p≤0.05Otherwise fail to reject the null hypothesis. Conclusion: Since the p-value is greater than the significance level. Therefore, the null hypothesis does not get rejected at a 5% level of significance. There is not enough information to conclude that one rate is bigger than the other rate.3. For the state of Connecticut: x1=268 , n1=50000 For the state of Massachusetts: x2=268 , n1=50000 Having the same cancer cases in the states of Connecticut and Massachusetts does not mean the actual rates of incidence of cancer in the total populations of the two states are the same. The step-by-step procedure to perform the 2-prop Z test using TI-83 is shown below: Go to STAT> TESTS > 2PropZTest. Now enter the values of x1,n1,x2 and n2. Select the <p2 and then click on Calculate. The output is shown below: The p-value will be 0.5 with either inequality. if there were no difference between the cancer rates of the two states, the difference observed in these samples would happen by chance about 50% of the time. 4. The p-value for the ≠p2 (two tailed ) test is twice the smaller of the two p-values from the other tests <p2 or >p2.