A doctor released the results of clinical trials for a vaccine to prevent a particular disease. In these clinical trials, 400,000 children were randomly divided in two groups. The subjects in group 1 (the experimental group) were given the vaccine, while the subjects in group 2 (the control group) were given a placebo. Of the 200,000 children in the experimental group, 44 developed the disease. Of the 200,000 children in the control group, 121 developed the disease. Complete parts (a) through (f) below.
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- Find the test statistic for this hypothesis
- Determine the P value
- Interpert the p value
- State the conclusion for this hypthesis test
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- A researcher conducted a medical study to investigate whether taking a low-dose aspirin reduces the chance of developing colon cancer. As part of the study, 1,000 adult volunteers were randomly assigned to one of two groups. Half of the volunteers were assigned to the experimental group that took a low-dose aspirin each day, and the other half were assigned to the control group that took a placebo each day. At the end of six years, 15 of the people who took the low-dose aspirin had developed colon cancer and 26 of the people who took the placebo had developed colon cancer. At the significance level α = 0.05, do the data provide convincing statistical evidence that taking a low-dose aspirin each day would reduce the chance of developing colon cancer among all people similar to the volunteers?A study claims that the combination of vitamin C with vitamin E is more effective for preventing the flu than vitamin C alone. In the study, a group of researchers studied 1132 subjects. The subjects were randomly assigned to out of three study groups. The first group of 310 subjects were instructed to take 600 mg daily of vitamin C. The second group of 418 subjects were instructed to take 600 mg of vitamin C and 200 IU of vitamin E daily. The third group of 404 subjects were instructed to take 600 mg of vitamin C and 400 IU of vitamin E daily. The incidence of flu over a season was recorded. All subjects knew which treatment they were receiving.This study is double-blind not an experiment blind, but not double-blind not blind. none of the aboveA study claims that the combination of vitamin C with vitamin E is more effective for preventing the flu than vitamin C alone. In the study, a group of researchers studied 1132 subjects. The subjects were randomly assigned to out of three study groups. The first group of 310 subjects were instructed to take 600 mg daily of vitamin C. The second group of 418 subjects were instructed to take 600 mg of vitamin C and 200 IU of vitamin E daily. The third group of 404 subjects were instructed to take 600 mg of vitamin C and 400 IU of vitamin E daily. The incidence of flu over a season was recorded. All subjects knew which treatment they were receiving.In this study, which is the control group? the second group of 418 taking 200 IU of vitamin E the third group of 404 taking 400 IU of vitamin E the 1132 subjects the first group of 310 taking vitamin C alone There is no control group.
- A study claims that the combination of vitamin C with vitamin E is more effective for preventing the flu than vitamin C alone. In the study, a group of researchers studied 1072 subjects. The subjects were randomly assigned to one of three study groups. The first group of 410 subjects were instructed to take 600 mg daily of vitamin C. The second group of 358 subjects were instructed to take 600 mg of vitamin C and 200 IU of vitamin E daily. The third group of 304 subjects were instructed to take 600 mg of vitamin C and 400 IU of vitamin E daily. The incidence of flu over a season was recorded. All subjects knew which treatment they were receiving. The study reported that 300 subjects in each group self-reported no flu for the season, so the vitamin E does not help fight the flu. This statement is false because:• not all groups have the same number of subjects.• not all the subjects took vitamin C.• the amount of vitamin C was different. It would be better to show the results…Based on advancements in drug therapy, a pharmaceutical company is developing Resithan, a new treatment for depression. A medical researcher for the company is studying the effectiveness of Resithan as compared to their existing drug, Exemor. A random sample of 461 depressed individuals is selected and treated with Resithan, and 172 find relief from their depression. A random sample of 419 depressed individuals is independently selected from the first sample and treated with Exemor, and 121 find relief from their depression. Based on the medical researcher's study can we conclude, at the 0.01 level of significance, that the proportion p1 of all depressed individuals taking Resithan who find relief from depression is greater than the proportion p2 of all depressed individuals taking Exemor who find relief from depression? Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places and round your answers as…In the journal Mental Retardation, an article reported the results of a peer tutoring program to help mildly mentally retarded children learn to read. In the experiment, the mildly retarded children were randomly divided into two groups: the experimental group received peer tutoring along with regular instruction, and the control group received regular instruction with no peer tutoring. There were n1 = n2 = 30 children in each group. The Gates-MacGintie Reading Test was given to both groups before instruction began. For the experimental group, the mean score on the vocabulary portion of the test was x1 = 344.5, with sample standard deviation s1 = 49.1. For the control group, the mean score on the same test was x2 = 353.8, with sample standard deviation s2 = 50.5. Use a 5% level of significance to test the hypothesis that there was no difference in the vocabulary scores of the two groups before the instruction began. (a) What is the level of significance? (b) What is the value of the…
- In the journal Mental Retardation, an article reported the results of a peer tutoring program to help mildly mentally retarded children learn to read. In the experiment, the mildly retarded children were randomly divided into two groups: the experimental group received peer tutoring along with regular instruction, and the control group received regular instruction with no peer tutoring. There were n1 = n2 = 30 children in each group. The Gates-MacGintie Reading Test was given to both groups before instruction began. For the experimental group, the mean score on the vocabulary portion of the test was x1 = 344.5, with sample standard deviation s1 = 49.7. For the control group, the mean score on the same test was x2 = 354.0, with sample standard deviation s2 = 50.5. Use a 5% level of significance to test the hypothesis that there was no difference in the vocabulary scores of the two groups before the instruction began. (a) What is the level of significance? What is the value of the…A consumer products testing group is evaluating two competing brands of tires, Brand 1 and Brand 2. Tread wear can vary considerably depending on the type of car, and the group is trying to eliminate this effect by installing the two brands on the same random sample of 12 cars. In particular, each car has one tire of each brand on its front wheels, with half of the cars chosen at random to have Brand 1 on the left front wheel, and the rest to have Brand 2 there. After all of the cars are driven over the standard test course for 20,000 miles, the amount of tread wear (in inches) is recorded, as shown in Table 1. Car Brand 1 Brand 2 Difference(Brand 1 - Brand 2) 1 0.375 0.194 0.181 2 0.238 0.111 0.127 3 0.232 0.265 -0.033 4 0.201 0.184 0.017 5 0.366 0.203 0.163 6 0.206 0.106 0.100 7 0.254 0.311 -0.057 8 0.369 0.242 0.127 9 0.378 0.287 0.091 10 0.349 0.253 0.096 11 0.209 0.235 -0.026 12 0.352 0.188 0.164 Table 1 Based on these data, can the…A consumer products testing group is evaluating two competing brands of tires, Brand 1 and Brand 2. Tread wear can vary considerably depending on the type of car, and the group is trying to eliminate this effect by installing the two brands on the same random sample of 12 cars. In particular, each car has one tire of each brand on its front wheels, with half of the cars chosen at random to have Brand 1 on the left front wheel, and the rest to have Brand 2 there. After all of the cars are driven over the standard test course for 20,000 miles, the amount of tread wear (in inches) is recorded, as shown in the table below. Car 1 2 3 4 5 6 7 8 9 10 11 12 Brand 1 0.35 0.37 0.51 0.59 0.59 0.42 0.59 0.35 0.38 0.40 0.51 0.34 Brand 2 0.37 0.41 0.49 0.48 0.41 0.21 0.53 0.34 0.34 0.48 0.26 0.19 Difference(Brand 1 - Brand 2) −0.02 −0.04…