1. Construct the indicated confidence interval for the population mean using a t-distribution. c = 0.80 , sample mean of distribution= 105 , s = 10 , n = 22 The confidence interval is (_ , _) (round to the nearest tenth as needed) 2. Find the margin of error for the given values of c, o, and n. c = 0.90 , standard deviation= 3.1 , n = 64 E = ( Round to three decimal places as needed 3. In a random sample of 26 people, the mean commute time to work was 34.1 minutes and the standard deviation was 7.2 minutes. Assume the population is normally distributed and use a t-distribution to construct a 95% confidence interval for the population mean . What is the margin of error of mean? Interpret the results . -The confidence interval for the population mean is ( _ , _ ) (round to one decimal place as needed) - The margin of error is _ (round to one decimal place as needed) A-With 95 % confidence , it can be said that the population mean commute time is between the bounds of the confidence interval . B-With 95 % confidence , it can be said that the commute time is between the bounds of the confidence interval . C-If a large sample of people are taken approximately 95 % of them will have commute times between the bounds of the confidence interval . D-It can be said that 95 % of people have a commute time between the bounds of the confidence interval . 4. Find the critical value zc necessary to form a confidence interval at the level of confidence shown below. c = 0.83. Zc= (Round to two decimal places as needed) 5. Let p be the population proportion for the following condition . Find the point estimates for p and q . A study of 4746 adults from country A found that 2750 were obese or overweight . The point estimate for p, p^ , is _ (Round to three decimal places as needed) The point estimate for q^ is _ (Round to three decimal places as needed) 6. In a survey of 614 males ages 18-64 , 390 say they have gone to the dentist in the past year . Construct 90 % and 95 % confidence intervals for the population proportion . Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals. -The 90 % confidence interval for the population proportion p is (_ ,_) ( Round to three decimal places as needed . ) The 95 % confidence interval for the population proportion p is (_ , _) ( Round to three decimal places as needed . ) -Interpret your results of both confidence intervals . A. With the given confidence , it can be said that the population proportion of males ages 18-64 who say they have gone to the dentist in the past year is between the endpoints of the given confidence interval . B. With the given confidence , it can be said that the population proportion of males ages 18-64 who say they have gone to the dentist in the past year is between the endpoints of the given confidence interval . C. With the given confidence , it can be said that the sample proportion of males ages 18-64 who say they have gone to the dentist in the past year is between the endpoints of the given confidence interval . Which interval is wider ? A-The 90 % confidence interval B- The 95 % confidence interval 7-Use the given confidence interval to find the margin of and the sample mean (12.5, 20.7) The sample mean is _ ( Type an integer or a decimal ) The margin of error is _ (Type an integer or a decimal .) 8-The grade point averages (GPA ) for 12 randomly selected college students are shown on the right . Complete parts (a) through (c ) below . Assume the population is normally distributed . 2.5 3.4 2.7 1.9 0.7 4.0 2.4 1.3 3.7 0.4 2.2 3.1 A) find the sample mean_ (round to decimal places as needed) B) find the sample standard deviation _ (round to decimal places as needed) C) Construct a 95% confidence interval for the population mean - A 95% confidence interval for the population mean is (_ ,_) (round to decimal places as needed) 9-Find the critical value for the confidence level c = 0.98 and sample size n = 27 tc= _ ( Round to the nearest thousandth as needed .) 10-Use the confidence interval to find the estimated margin of error . Then find the sample mean A store manager reports a confidence interval of (43.8, 81.6) when estimating the mean price (in dollars) for the population of textbooks -The estimated margin of error is _ (Type an integer or a decimal.) -The sample mean is _ ( Type an integer or a decimal .) 11-People were polled on how many books they read the previous year. How many subjects are needed to estimate the number of books read the previous within one book with 99% confidence? Initial survey results indicate that standard deviation= 14.7 books A 99% confidence level requires __ subjects . ( Round up to the nearest whole number as needed). 12-In a random sample of 55 refrigerators , the mean repair cost was $146.00 and the population standard deviation is $17.60 Construct a 90 % confidence interval for the population mean repair cost . Interpret the results . Construct a 90% confidence interval for the population mean repair cost. The 90% confidence interval is (_ ,_ ) (Round to two decimal places as needed.) Interpret you results . Choose the correct answer below . A)The confidence interval contains 90 % of the mean repair costs . B)With 90 % confidence , it can be said that the confidence interval contains the true mean repair cost . C) With 90 % confidence , it can be said that the confidence interval contains the sample mean repair cost. 13-Use the given confidence interval to find the margin of error and the sample proportion. (0.630, 0.658) E= _ (type an integer or as a decimal) hat(p)= _ (type an integer or as a decimal)