Suppose that x = 11 observations possess the characteristic of interest from n = - 32 observations. Let p be the population proportion possessing the characteristic. (a) Compute a two-sided 99% confidence interval for p. (b) Compute a two-sided 95% confidence interval for p.
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- Suppose that the height of Asian males aged 35-44 in the United States is normally distributed with a standard deviation of 3 in. Jeremy takes a simple random sample of 16 such men and notices that none of them are unusually tall or unusually short. He then calculates the average of the sample to be 67 in. and is planning on using this to find a 90% z-confidence interval for the true mean height. Can Jeremy use this information to find the z-confidence interval? Complete the following sentences. The population distribution is and the population standard deviation is Therefore, the sample to find a 90% z-confidence interval.The ages of registered voters in Smith County are normally distributed with a population standard deviation of 3 years and an unknown population mean. A random sample of 18 voters is taken and results in a sample mean of 55 years. Find the margin of error for a 95% confidence interval for the population mean. z0.10z0.10 z0.05z0.05 z0.025z0.025 z0.01z0.01 z0.005z0.005 1.282 1.645 1.960 2.326 2.576 You may use a calculator or the common z values above. Round the final answer to two decimal places.Pedro was working on a report on Swiss and Austrian psychiatrists. He decided to find a 90 percent confidence interval for the difference in mean age at the time of significant psychiatric discoveries for Swiss versus Austrian psychiatrists. He found the ages at the time of psychological discovery of all the members of both groups and found the 90 percent confidence interval based on a t-distribution using a calculator. The procedure he used is not appropriate in this context because O The sample sizes for the two groups are not equal. O The entire population is measured in both cases, so the actual difference in means can be computed and a confidence interval should not be used. O Age at the time of psychological discovery occurs at different intervals in the two countries, so the distribution of ages cannot be the same. O Age at the time of psychological discovery is likely to be skewed rather than bell shaped, so the assumptions for using this confidence interval formula are not…
- Suppose X1,..., Xn is a random sample from an exponential distribution with mean e. If X = 17.9 with n = 50, find (a) a one-sided 95% confidence interval for 0, and (b) a two-sided 95% confidence interval for 0.Suppose we know that a confidence interval for a population proportion is (0.105,0.355), with a sample proportion of p̂=0.23. What is the margin of error?6. You buy a package of 122 Smarties and 19 of them are red. What is a 95% confidence interval for the true proportion of red Smarties? (0.092, 0.220) (0.103, 0.230) (0.085, 0.199) (0.109.0.210) A. B. C. D.
- 7. What is the 95% confidence interval for the difference in the population means, µ1 - µ2, given the data from the two independent samples: n1 =56, x1= 58.2, s1= 4.3, n2=45, x2=60.2, s2=5.5 B. [-3.28 , -0.72] C. [-3.60, -0.40] D. [-3.65 , -0.35] A. [-3.96 , -0.04] 8. What is the 98% confidence interval for the difference in the population means, µ1 - µ2, given: n1= 22, x1= 41.4, s1= 5.9, n2= 19, x2= 43.2, s2= 6.5 D. [-3.48 , -0.12] B. [-3.07 , -0.53] A. [-3.12, -0.48] C. [-3.18, -0.42]A sample of 1100 observations taken from a population produced a sample proportion of 0.38. Make a 90 % confidence interval for Round your answers to three decimal places. = ( ) to ( )Credit card ownership varies across age groups. Suppose that the estimated percentage of people who own at least one credit card is 64% in the 18–24 age group, 84% in the 25–34 age group, 75% in the 35–49 age group, and 77% in the 50+ age group. Suppose these estimates are based on 465 randomly selected people from each age group. (a) Construct a 95% confidence interval for the proportion of people in the 18–24 age group who own at least one credit card. (Round your answers to four decimal places.) to Construct a 95% confidence interval for the proportion of people in the 25–34 age group who own at least one credit card. (Round your answers to four decimal places.) to Construct a 95% confidence interval for the proportion of people in the 35–49 age group who own at least one credit card. (Round your answers to four decimal places.) to Construct a 95% confidence interval for the proportion of people in the 50+ age group who own at least one credit card. (Round your answers to four…
- Suppose a new standardized test is given to 108 randomly selected third-grade students in New Jersey. The sample average score Y on the test is 62 points, and the sample standard deviation, sy, is 10 points. The authors plan to administer the test to all third-grade students in New Jersey. The 95% confidence interval for the mean score of all New Jersey third graders is (.). (Round your responses to two decimal places.) Suppose the same test is given to 216 randomly selected third graders from lowa, producing a sample average of 66 points and sample standard deviation of 13 points. The 90% confidence interval for the difference in mean scores between lowa and New Jersey is -6.22.). (Round your responses to two decimal places.) The p-value of the test of no difference in means versus some difference is 0.0024. (Round your response to four decimal places.) Can you conclude with a high degree of confidence that the population means for lowa and New Jersey students are different? OA.…A researcher takes a random sample of 100 Americans that have never had a pet and a separate random sample of 100 Americans that are current or previous pet owners. She is interested in determining whether there is a difference in the proportion of individuals who believe in the afterlife among these two groups. She observes that 76 of the "never pet owners" and 84 of the "ever pet owners" believe in the afterlife. Determine the margin of error for a confidence interval for the difference in the two appropriate proportions at a 95% confidence level. a. The assumptions necessary to perform the calculation are not satisfied. b. 0.0837 c. 0.0719 d. 0.1103 e. 1.96Suppose that your 95% confidence interval for p is (0.68, 0.74). Based on the confidence interval what would your decision be for the hypothesis test H0: p=0.8 versus H1:p≠0.8 at the α=0.05 level?
