In the upcoming governor's election, the most recent poll, based on 900 respondents, predicts that the incumbent will be reelected with 55 percent of the votes. For the sake of argument, assume that 51 percent of the actual voters in the state support the incumbent governor (population parameter 0.51). The values of the population parameter and the standard error of the sample for the pool of 900 respondents. are respectively. O 051 and 0.0167 0055 and 0.51 0055 and 240 0240 and 0.0167
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- An organization published an article stating that in any one-year period, approximately 8.2 percent of adults in a country suffer from depression or a depressive illness. Suppose that in a survey of 100 people in a certain town, six of them suffered from depression or a depressive illness. Conduct a hypothesis test to determine if the true proportion of people in that town suffering from depression or a depressive illness is lower than the percent in the general adult population in the country. p' = Calculate ?x. Find the p-value.In the February 20th issue of the East Daily Times, it was reported that n = 418 insured patientswere surveyed on their opinions about insurance rates. Of the 418 surveyed, Y = 280 blamedrising insurance rates on large court settlements against doctors. With 95% confidence, estimate the parameter p, that is, the proportion of all insured patients who blame rising insurance rateson large court settlements against doctors.The National Institute of Mental Health published an article stating that in any one-year period, approximately 9.5 percent of American adults suffer from depression or a depressive illness. Suppose that in a survey of 100 people in a certain town, seven of them suffered from depression or a depressive illness. Conduct a hypothesis test to determine if the true proportion of people in that town suffering from depression or a depressive illness is lower than the percent in the general adult American population. find the p value
- A researcher is concerned that his new antihypertensive medication may be causing insomnia in some of his patients. Suppose he gathers an SRS of 65 patients treated with the study drug with a sample average of 6.6 hours of sleep and a σ=1.1. Assuming that insomnia can be quantified as an average of 4.5 hours of sleep, can we determine with 95% confidence that his drug avoids diagnosis of insomnia as a side-effect?An electrical engineer wishes to determine if, among two specific municipal buildings in town, Building “North” and Building “South”, whether the tensile strength of pipes (in psi) is not the same in each of these two buildings. A sample of pipes was chosen at random from both Building “North” and Building “South”, respectively. Using α = 0.05, which of the following statistical test, or parameter, would be best for determining whether tensile strength of pipes (in psi) is not the same in each of these two buildings? (Assume all statistical assumptions met.) a) Binomial Distribution b) Population Difference in Means (i.e., Unpaired Data) c) The Chi-Squared Test of Independence d) Population Mean Difference (i.e., Paired Data)According to a certain government agency for a large country, the proportion of fatal traffic accidents in the country in which the driver had a positive blood alcohol concentration (BAC) is 0.36. Suppose a random sample of 110 traffic fatalities in a certain region results in 49 that involved a positive BAC. Does the sample evidence suggest that the region has a higher proportion of traffic fatalities involving a positive BAC than the country at the α=0.05 level of significance?
- A major credit card company is interested in the proportion of individuals who use a competitor’s credit card. Their null hypothesis is H0: p=0.65H0: p=0.65, and based on a sample they find a sample proportion of 0.70 and a pp-value of 0.053. Is there convincing statistical evidence at the 0.05 level of significance that the true proportion of individuals who use the competitor’s card is actually greater than 0.65 ?A major credit card company is interested in the proportion of individuals who use a competitor’s credit card. Their null hypothesis is H0: p=0.65H0: p=0.65, and based on a sample they find a sample proportion of 0.70 and a pp-value of 0.053. Is there convincing statistical evidence at the 0.05 level of significance that the true proportion of individuals who use the competitor’s card is actually greater than 0.65 ? Yes, because the sample proportion 0.70 is greater than the hypothesized proportion 0.65. A Yes, because the pp-value 0.053 is greater than the significance level 0.05. B No, because the sample proportion 0.70 is greater than the hypothesized proportion 0.65. C No, since the sample proportion 0.70 is exactly 0.05 away from the hypothesized proportion 0.65. D No, because the pp-value 0.053 is greater than the significance level 0.05.A certain financial services company uses surveys of adults age 18 and older to determine if personal financial fitness is changing over time. A recent sample of 1,000 adults showed 410 indicating that their financial security was more than fair. Suppose that just a year before, a sample of 1,200 adults showed 420 indicating that their financial security was more than fair. (a) State the hypotheses that can be used to test for a significant difference between the population proportions for the two years. (Let p1 = population proportion most recently saying financial security more than fair and p2 = population proportion from the year before saying financial security more than fair. Enter != for ≠ as needed.) H0: p1−p2=0 Ha: p1−p2!=0 (b) Conduct the hypothesis test and compute the p-value. At a 0.05 level of significance, what is your conclusion? Find the value of the test statistic. (Use p1 − p2. Round your answer to two decimal places.) Find the p-value.…
- A certain financial services company uses surveys of adults age 18 and older to determine if personal financial fitness is changing over time. A recent sample of 1,000 adults showed 410 indicating that their financial security was more than fair. Suppose that just a year before, a sample of 1,200 adults showed 420 indicating that their financial security was more than fair. (a)State the hypotheses that can be used to test for a significant difference between the population proportions for the two years. (Let p1 = population proportion most recently saying financial security more than fair and p2 = population proportion from the year before saying financial security more than fair. Enter != for ≠ as needed.) H0: Ha: (b) Conduct the hypothesis test and compute the p-value. At a 0.05 level of significance, what is your conclusion? Find the value of the test statistic. (Use p1 − p2. Round your answer to two decimal places.) = Find the p-value. (Round your answer to…A researcher took a random sample of 200 new mothers in the United States (Group 1) and found that 16% of them experienced some form of postpartum depression. Another random sample of 200 new mothers in France (Group 2), where mothers receive 16 weeks of paid maternity leave, found that 11% experienced some form of postpartum depression. Can it be concluded that the percent of women in the United States experience more postpartum depression than the percent of women in France? Use α=0.05. Select the correct alternative hypothesis and decisionAccording to a certain government agency for a large country, the proportion of fatal traffic accidents in the country in which the driver had a positive blood alcohol concentration (BAC) is 0.33.Suppose a random sample of 109 traffic fatalities in a certain region results in 47that involved a positive BAC. Does the sample evidence suggest that the region has a higher proportion of traffic fatalities involving a positive BAC than the country at the α=0.05 level of significance?