Find the critical value z /2 that corresponds to the given confidence level. 89% Za/2 = (Round to two decimal places as needed.)Find the critical value z /2 that corresponds to the given confidence level. 89% Za/2 = (Round to two decimal places as needed.)Use the sample data and condence level given below to complete parts (a) through (d). In a study of cell phone use and brain hemispheric dominance, an Internet survey was emailed to 2551 subjects randomly selected from an online group involved with ears. 1143 surveys were returned. Construct a 99% confidence interval for the proportion of returned surveys. @ Click the icon to view a table of z scores. a) Find the best point estimate of the population proportion p. (Round to three decimal places as needed.) A certain drug is used to treat pain, but it is dangerous because it is addictive and can be lethal. In clinical trials, 218 subjects were treated with the drug and 53 of them developed nausea. Complete parts (a) and (b) below. 3. Construct a 99% (Wald) condence interval estimate of the percentage of users of the drug who develop nausea. % a. 08 1 .2 >1. 2 6 X-Values Select the correct choice below and, if necessary, ll in the answer box to complete your choice. '33:? A- 201/2 = (Round to two decimal places as needed.) '33:? 3- tel/2 = (Round to two decimal places as needed.) {:1- C. Neitherthe normal northe t distribution applies. Assume that we want to construct a condence interval. Do one ofthe following, as appropriate. (a) Find the critical value 201,2, (b) find the critical value tan, or (c) state that neither the normal nor the t distribution applies. Confidence level is 95%, o = 26.4, and the histogram of right-eye vision measurements from a random sample of 67 adult males is shown to the right. Frequency 0 50 100150 200 Vision in Right Eye Select the correct choice below and, if necessary, ll in the answer box to complete your choice. A- 20\Here are summary statistics for randomly selected weights of newborn girls: n = 204, x = 28.2 hg, s = 6.1 hg. Construct a confidence interval estimate of the mean. Use a 98% confidence level. Are these results very different from the confidence interval 27.0 hg