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- A newspaper story claims that more houses are purchased by singles now than singles 5 years ago. To test this claim, two studies were conducted on the buying habits of singles over the past 5 years. In the first study, 500 house purchases in the current year were randomly selected and 200 of those were made by singles. In the second study, again 500 house purchases were randomly selected from 5 years ago and 167 of those were made by single people. Test the newspaper's claim using a 0.05 level of significance. Is there sufficient evidence to support the newspaper's claim? Let singles now be Population 1 and let singles 5 years ago be Population 2. Step 2 of 3: Compute the value of the test statistic. Round your answer to two decimal places.A bag of candy contains 5 different types of colored candies; red, green, blue, yellow, and orange. According to the manufacturer, bags should contain an equal number of each color. Students in a statistics class decided to use a chi-square procedure to test the manufacturer’s claim. They opened a bag of candy and recorded the number of candies of each color. The results are shown in the following table. Red Green Blue Yellow Orange 17 24 20 25 14 Which color contributes most to the chi-square test statistic? Red A Green B Blue C Yellow D OrangeSuppose that 20% of the adult population in Palestine smoke cigarettes only, 15% smoke Argeeleh only, and 10% smoke both cigarettes and Argeeleh. Suppose that 15% of those who smoke cigarettes only have respiratory problems, 10% of those who smoke Agreeleh only have respiratory problems, 20% of those who smoke both cigarettes and Argeeleh have respiratory problems, and 5% of non-smokers have respiratory problems. If an adult person is randomly selected from the population: What is the probability that he/she has respiratory problems? If the selected individual has respiratory problems, what is the probability that he/she is non-smoker? If the selected individual has respiratory problems, what is the probability that he/she smokes both cigarettes and Argeeleh?
- A poll released this week found that in a random sample of registered voters, 60% indicated that they think a fenmale "will run" for the presidency, 30% said a female “will not run," and 10% had “no opinion." When asked their opinions on whether or not a female could be elected, 66% of those who said a female "will run" thought a female could be elected; 25% of those who thought a female "will not run" thought a female could be elected; whereas, 20% of those who had no opinion said 4. that a female could be elected. а. [2] What percentage of registered voters (in this sample) thought that a female could be elected? [2] Given that a person thought that a female could be elected, what is the probability that this person said a female "will not run" for the presidency? b. [3] Given that a person thought that a female could be elected, which is more likely: that this person said a female “will not run" for the presidency or that this person said a female “will run" for the presidency? C.An anger-management course claims that, after completing its seminar, participants will lose their tempers less often. Always a skeptic, you decide to test this claim. A random sample of 12 seminar participants is chosen, and these participants are asked to record the number of times that they lost their tempers in the two weeks prior to the course. After the course is over, the same participants are asked to record the number of times that they lost their tempers in the next two weeks. The following table lists the results of the survey. Using these data, test the claim at the 0.01 level of significance assuming that the population distribution of the paired differences is approximately normal. Let participants before completing the anger-management course be Population 1 and let participants after completing the anger-management course be Population 2. Number of Times Temper Was Lost during a Two-Week Period Before 10 6 8 10 6 3 10 3 8 After 6 7 4 4 8 6 4 9 3 7 Copy Data Step 2 of 3:…A social psychologist hypothesized that a white person is more likely to cooperate with another person than a black person in a task requiring cooperation (this would be considerate an alternative hypothesis). To test this idea he conducted an experiment in which white subjects either worked on a task with a white person or black person (both of these “others” were confederates of the experimenter). Thirty-one subjects were randomly assigned to the “white-other” treatment (say this is treatment 1) and 31 were randomly assigned to the “black-other” treatment (this is treatment 2). If the mean cooperation scores, on a scale from 0 to 20, were 14 for those subjects in the “white-other” treatment and 12 for those in the “black-other” treatment, run an appropriate test to test the social psychologist’s hypothesis. Let us further say that the variance for the cooperation measure for the “white-other” treatment was = 2.48 and the variance for the “black-other” treatment was = 2.52. Use =0.05…
- 7. A new drug study was conducted by a drug company. In the study, 10 people were chosen at random to take Vitamin X for 3 months and then have their cholesterol levels checked. In addition, 10 different people were randomly chosen to take Vitamin Y for 3 months and then have their cholesterol levels checked. All 20 people had cholesterol levels between 8 and 10 before taking one of the vitamins. The drug company wanted to see which of the 2 vitamins had the greatest impact on lowering people's cholesterol. The following data was collected: Vitamin X 7.27.5 5.2 6.5 7.7 10 6.47.6 7.77.8 Vitamin Y 4.8 4.44.5 5.1 6.5 8.03.1 4.6 5.2 6.1 (a) Draw a box-and-whisker plot for both sets of data on the same number line. (b) Use the double box-and-whisker plots to compare the 2 vitamins and provide a conclusion for the drug company. (e) State the skewness of both distributions and justify your answer.A school psychologist is interested in determining if children with attention deficit hyperactivity disorder (ADHD) learn better if English literature is read to them rather than having them read the material alone by themselves. A random sample of 10 sixth graders with ADHD is selected and divided into two groups of n=9. One of the groups has a story read to them (Listening Group) and the other reads the story alone by themselves (Reading Group). A test on the story is given after each group has finished reading or hearing the story. The following scores were obtained with 20 being a perfect score. what is the cohen’s d ?Xu and Garcia (2008)conducted a research study demonstrating that 8-month-old infants appear to recognize which samples are likely to be obtained from a population and which are not. In the study, the infants watched as a sample of n = 5 ping-pong balls was selected from a large box. In one condition, the sample consisted of 1 red ball and 4 white balls. After the sample was selected, the front panel of the box was removed to reveal the contents. In the expected condition, the box contained primarily white balls like the sample, and the infants looked at it for an average of M = 7.5 seconds. In the unexpected condition, the box had primarily red balls, unlike the sample, and the infants looked at it for M = 9.9. The researchers interpreted the results as demonstrating that the infants found the unexpected result surprising and, therefore, more interesting than the expected result. Assuming that the standard error for both means is σM = 1 second, draw a bar graph showing the two sample…
- According to the CCD, 1 in 25 adults have mutation A in their blood. Test TT for that mutation will give a positive result 95% of the time when the adult actually has the mutation and a negative result the other 5% of the time. If the adult does not have the mutation, then the TT test gives a positive result 1% of the time and a negative result the other 99% of the time. Suppose an adult is randomly selected and given the TT test. Need help with e. a.What is the probability that the selected adult has mutation A? b. What is the probability that the selected adult has mutation A and test TT gives a positive result? c. What is the probability that test TT gives a positive result for the selected adult? d. What is the probability that test TT gives a negative result for the selected adult? e. What is the probability that the selected adult has the mutation given that test TT gives a positive result? f.If 1000 random adults are selected, about how many would be expected to give a positive…In a series of studies, Sebastian Deri and his colleagues (2017) asked multiple groups about the number of friend groups they had and how often they attended different social events. They then asked participants how they thought their own numbers compared to other people's: Did participants see themselves as less social, the same as, or more social than others? In the first study, the researchers reported that: "Three hundred four participants from Amazon's Mechanical Turk (150 women; Mage = 37.1; 78% White, 10% Black, and 6% Asian) completed a survey in exchange for modest payment" (p. 860). What do the researchers mean by "Amazon's Mechanical Turk" with respect to their sample of participants? O A. An archival data set that already exists and from which researchers can use data O B. An online tool that researchers can use to recruit people to participate in research in exchange for a small payment O C. An e-book sold by Amazon that includes contact information for possible…2. Suppose we want to test whether Canadiansand American viewers have the same reaction to acertain documentary film. A random sample of 6Canadians and 7 American viewers werecollected. Each are asked to watch and then ratethe film, from 1 (most negatively), to 10 (mostpositively). The data are shown on theaccompanying table:Canadians 2 10 1 4 6 1Americans 4 7 7 5 5 3 8a. Explain why we cannot test for the equality of means. That is, we cannot test Ho: Average rating among Canadians is equal to average rating among Americans. b. Use Wilcoxon Rank Sum procedure to test whether Canadians and Americans tend to rate the film similarly. Include the test statistic. State your conclusion in the context of the problem. Use the 10% significance level. c. Use Normal distribution to approximate the p-value. Would you recommend using it to carry out the test instead of the table?