I am having trouble with A-D steps this is an example by my professor just need help with 1-3 and interpreting the process he used
Given into Given 1 = 420105 x = 137 confidence level. 95 % = = 1- 95 %- 5% Want Confidence literal of the proportion sample Then the sample proportion 137 statistic P = = 42005 - D.0003 262 critical The critical value is value Zo = Inv Norm ( 1 - 2 ) = In Norm ( 1 - 2-2 ) = Inv Norm ( 0975 ) - 1.91 Heme the margin margin of 0 vol 3 262 ( 1- 0 0801262 ) E = 23 = 196 n 420 00 5 - 191 0.000 $262 . 079 96738 420 005 - 1.96 . 0 00 002786 calculation Then the confidence interval is ( a , b ) = ( 2027 16% 0. 38 8 % ) the confidant where a = p - E = 0 010 } 262 - 0.0600 5 461 - 0 00 02716 = 0.027 16 % 6 = P + E = 0 0 0 3252 + 0.0#1541 1 - 0 080 3808 = 0.03808 1 * Interpretation Therefore . we are 95 % confident that the interval from 0.02716 % to 0. 03808 % actually contains the True population proportion of the cell phone users who developed came of the brain or nervous system4: A study of 420, 005 cell phone users found that 137 of them developed cancer of the brain or nervous system. Prior to this stud)r of cell phone use. the rate of such cancer was found to be 0.0346% for those not using cell phones. Construct a 95% condence interval estimate of the proportion of cell phone users who developed cancer of the brain or nervous system. Round your answer to ve decimal places as needed. 1) In a simple random survey of 89 teachers who taught college statistics, 73 said that it was the most satisfying, most enjoyable course they had ever taught. In order to construct a 98% confidence interval estimate of the proportion of all community college statistics teachers who feel this way, let's following the steps: a) What is the critical value? b) What is the margin of error? c) Construct a 98% confidence interval. d) Interpret the above confidence interval. (You must follow the format from the example of instructor's video. Other way of interpretation will result in little credit or no credit.)2) To determine the mean cost of groceries in a certain city, an identical grocery basket of food is purchased at each store in a random sample of ten stores. If the average cost is $47.52 with a standard deviation of $1.59. To find a 98% confidence interval estimate for the cost of these groceries in the city. (Assume that the population is normally distributed), let's complete the following steps: a) What is the critical value? b) What is the margin of error? c) Construct a 98% confidence interval. d) Interpret the above confidence interval. (You must follow the format from the example of instructor's video. Other way of interpretation will result in little credit or no credit.)3) The values listed below are waiting times (in minutes) of customers ate a certain bank, where customers enter a single waiting line that feeds three teller windows: 6.5 6.6 6.7 6.8 7.1 7.3 7.4 7.7 7.7 7.7 To construct a 95% confidence interval for the population standard deviation, we will answer the following questions: (Assume that the population is normally distributed). a) What is the critical values (There are two of them)? b) Construct a 98% confidence interval. c) Interpret the above confidence interval. (You must follow the format from the example of instructor's video. Other way of interpretation will result in little credit or no credit.)
