A newspaper article reported that 400 people in one state were surveyed and 75% were opposed to a recent court decision. The same article reported that a similar survey of 500 people in another state indicated opposition by only 45%. Construct a 95% confidence interval of the difference in population proportions based on the data.
Answer to relevant QuestionsThe Healthy Eating Index measures on a 100-point scale the adequacy of consumption of vegetables, fruits, grains, milk, meat and beans, and liquid oils. This scale is called HEI2005 (Guenther et al.2007). There are two ...Independent random sampling from two normally distributed populations gives the following results: nx = 81; x-bar = 140; σ2x = 25 ny = 100; y-bar = 120; σ2y = 14 Find a 95% confidence interval estimate of the difference ...Test the hypotheses H0: μ ≤ 100 H1: μ > 100 Using a random sample of n = 25, a probability of Type I error equal to 0.05, and the following sample statistics. a. x-bar = 106; s = 15 b. x-bar = 104; s = 10 c. x-bar = 95; ...On the basis of a random sample the null hypothesis H0: μ = μ0 is tested against the alternative H1: μ > μ0 and the null hypothesis is not rejected at the 5% significance level. a. Does this necessarily imply that μ0 is ...In a random sample of 361 owners of small businesses that had gone into bankruptcy, 105 reported conducting no marketing studies prior to opening the business. Test the hypothesis that at most 25% of all members of this ...
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