Here is a 95% confidence interval estimate of the proportion of female medical school students: 0.449<p < 0.511 (based on data from the Journal of the American Medical Association ). What is the best point estimate of the proportion of females in the population of medical school students?
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- A pharmaceutical company is running tests to see how well its new drug lowers cholesterol. Fourteen adults volunteer to participate in the study. The total cholesterol level of each participant (in mg/dL) is recorded once at the start of the study and then again after three months of taking the drug. The results are given in the following table. Construct a 90% confidence interval for the true mean difference between the cholesterol levels for people who take the new drug. Let Population 1 be the initial cholesterol level and Population 2 be the cholesterol level after three months. Round the endpoints of the interval to one decimal place, if necessary. Total Cholesterol Levels (in mg/dL) Initial Level Level after Three Months 190 192 230 207 206 182 229 199 224 181 195 209 226 214 190 189 224 196 219 206 185 195 208 204 194 198 225 216 Please highlight the answerFor a confidence level of 94%, find the critical value for a confidence interval on a one sample proportion.Here are summary statistics for randomly selected weights of newborn girls: n = 167, x= 30.6 hg, s= 6.2 hg. Construct a confidence interval estimate of the mean. Use a 99% confidence level. Are these results very different from the confidence interval 29.2 hg <<32.2 hg with only 13 sample values, x= 30.7 hg, and s= 1.8 hg? What is the confidence interval for the population mean µ? hgHere are summary statistics for randomly selected weights of newborn girls: n=217, x 26.5 hg, s = 6.1 hg. Construct a confidence interval estimate of the mean. Use a 99% confidence level. Are these results very different from the confidence interval 24.7 hgIndentify the t-score for a 95% confidence interval if the sample size is 50 t=A study of 420,098 cell phone users found that 0.0324% of them developed cancer of the brain or nervous system. Prior to this study of cell phone use, the rate of such cancer was found to be 0.0335% for those not using cell phones. Complete parts (a) and (b). a. Use the sample data to construct a 95% confidence interval estimate of the percentage of cell phone users who develop cancer of the brain or nervous system Do not round until the final answer. Then round to three decimal places as needed.] RThe distribution of study times for first-year students follow a Normal distribution with a standard deviation σ = 78 minutes. Suppose a sample of 431 students found that the mean study time for these students was x¯= 195 minutes per week. Use this to find a 95% confidence interval to estimate the mean study time for all students. We are 95% confident that the mean study time for first-year students is between (Answer) and minutes ( Answer) minutes.A researcher is attempting to calculate a 95% confidence interval for the mean average wingspan of Statsian Swallows in a particular rainforest. They take a sample and find that of the 16 swallows surveyed, the mean average was 3.14 inches and the standard deviation was 0.2 inches. What is the 95% confidence interval for this sample?Here are summary statistics for randomly selected weights of newborn girls: n = 204, x = 28.4 hg, s =7.7 hg. Construct a confidence interval estimate of the mean. Use a 98% confidence level. Are these results very different from the confidence interval 26.6 hg < u < 31.2 hg with only 14 sample values, x = 28.9 hg, and s = 3.3 hg? What is the confidence interval for the population mean u? hgA Gallup poll of 1487 adults showed that 43% of the respondents have Facebook pages. Construct a 95% confidence interval for the proportion of adults who have a Facebook page.A poll reported that only 1316 out of a total of 1746 adults in a particular region said they had a "great deal of confidence" or "quite a lot of confidence" in the military. Assume the conditions for using the CLT are met. Complete part (a) (a) Find a 95% confidence interval for the proportion that express a great deal of confidence or quite a lot of confidence in the military, and interpret this interval. The 95% confidence interval for the proportion that express a great deal of confidence or quite a lot of confidence in the military is: ( ? , ? )Here are summary statistics for randomly selected weights of newborn girls: n= 190, x 32.9 hg, s= 6.2 hg. Construct a confidence interval estimate of the mean. Use a 95% confidence level. Are these results very different from the confidence interval 31.4 hgSEE MORE QUESTIONSRecommended textbooks for youMATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons Inc
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