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The sample size needed to estimate the difference between two population proportions to within a margin of error m with a significance level of α can be found as follows. In the expression m=z∗p1(1−p1)n1+p2(1−p2)n2−−−−−−−−−−−−−−−−−−−−√ we replace both n1 and n2 by n (assuming that both samples have the same size) and replace each of p1, and p2, by 0.5 (because their values are not known). Then we solve for n, and get n=(z∗)22E2. Finally, increase the value of n to...
all answers please!! 1. A company that produces brendis concerned about the distribution of the amount of sodium in its bread. The company takes a simple random sample of 100 slices of bread and compute the sample mean to be 103 milligrams of sodium per slice. Assume that the population standard deviation is 10 milligrams a) Construct a 95% confidence interval estimate for the mean sedium level. b) Construct a 99% confidence intervalestime for the mean sodium level. 2. Fill...
1. In an August 2012 Gallup survey of 1,012 randomly selected U.S. adults (age 18 and over), 53% said that they were dissatisfied with the quality of education students receive in kindergarten through grade 12. The bootstrap distribution (based on 5,000 samples) is provided. a). Would it be appropriate to use the normal distribution to construct the confidence interval in this situation? Explain briefly. b). The standard error from the bootstrap distribution is SE = 0.016. Use the normal distribution...
Find the indicated critical z value. the value of zan that corresponds to a confidence level of 97 80% Use the given degree of confidence 2) Of 88 selected and sample data to construct a confidence interval for the population proportion p adults selected randomly from one town 69 have health insurance. Find a 90% confidence interval for the true proportion of all adults in the town who have health insurance the given degree of confidence and sample data to...
4 Chapter 7 Test B 16. [Objective: Calculate and interpret confidence intervals for a proportion) A random sample of 950 adult television viewers showed that 48% planned to watch sporting event X. The margin of error is 4 percentage points with a 95% confidence level. Does the confidence interval support the claim that the majority of adult television viewers plan to watch sporting event X? No; the confidence interval means that we are 95% confident that the population proportion of...
Among a simple random sample of 326 American adults who do not have a four-year college degree and are not currently enrolled in school, 48% said they decided not to go to college because they could not afford school. Suppose an earlier hypothesis test determined that the data do not provide strong evidence that less than half of American adults who decide not to go to college make this decision because they cannot afford college. (a) Calculate a 90% confidence...
all answers please 3. The teachers union is concerned about the amount of time teachers spend each week doing schoolwork at home. A simple random sample of 56 teachers had a mean of 8.0 hours per week working at home after school. Assume that the population standard deviation is 1.5 hours per week. a) Construct a 95% confidence interval estimate for the mean number of hours per week a teacher spends working at home. b) Does the confidence interval support...
Hours after a governor was indicted on several counts of fraud and extortion, two reputable research firms polled citizens from the region asking whether they thought the governor should resign. The first poll revealed that 60% people in the region felt the governor should resign; it had a margin of error of plus or minus 4.8 percentage points. The other poll reported that 55% people in the region felt the governor should resign; it had a margin of error of...
1. Use the given degree of confidence and sample data to construct a confidence interval for the point) population proportion p. A survey of 865 voters in one state reveals that 408 favor approval of an issue before the legislature. Construct the 95% confidence interval for the true proportion of all voters in the state who favor approval. 0 0.438<p0.505 0 0.444 p0.500 0 0.435<p<0.508 O 0.471 p0.472 2. Use the given data to find the minimum sample size required...
A company surveyed 2400 men where 123 of the men identified themselves as the primary grocery shopper in the household a) Estimate the percentage of all males who identify themselves as the primary grocery shopper Use a 96% confidence interval Check the conditions fint bj Agrocery store owner believed that only 45% of men are the primary grocery shopper for their family, and targets his advertising cardingly. He whes to conduct a hypothesis est to see the fraction is in...