In a survey of 3363 adults, 1464 say they have started paying bills online in the...

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In a survey of 3363 adults, 1464 say they have started paying bills online in the last year. Construct a 99% confidence interval for the population proportion. Interpret the results. A 99% confidence interval for the population proportion is (D). (Round to three decimal places as needed.) Interpret your results. Choose the correct answer below. OA. With 99% confidence, it can be said that the population proportion of adults who say they have started paying bills online in the last year is between the endpoints of the given confidence interval. OB. With 99% confidence, it can be said that the sample proportion of adults who say they have started paying bills online in the last year is between the endpoints of the given confidence interval. OC. The endpoints of the given confidence interval show that adults pay bills online 99% of the time. Construct the indicated confidence interval for the population mean using the t-distribution. Assume the population is normally distributed. c=0.90, x=14.4, s = 3.0, n=8 (Round to one decimal place as needed.) C Construct the confidence interval for the population mean . c=0.98, x=15.5, = 7.0, and n = 55 A 98% confidence interval for is (). (Round to one decimal place as needed.) The state test scores for 12 randomly selected high school seniors are shown on the right. Complete parts (a) through (c) below. 1430 1223 981 692 720 839 Assume the population is normally distributed. 724 750 545 620 1443 945 (a) Find the sample mean. x= (Round to one decimal place as needed.) (b) Find the sample standard deviation. S= (Round to one decimal place as needed.) (c) Construct a 90% confidence interval for the population mean . A 90% confidence interval for the population mean is ( ). (Round to one decimal place as needed.) Find the minimum sample size n needed to estimate for the given values of c, , and E. c=0.98, = 8.8, and E = 2 Assume that a preliminary sample has at least 30 members. n = (Round up to the nearest whole number.) In a survey of 3363 adults, 1464 say they have started paying bills online in the last year. Construct a 99% confidence interval for the population proportion. Interpret the results. A 99% confidence interval for the population proportion is (D). (Round to three decimal places as needed.) Interpret your results. Choose the correct answer below. OA. With 99% confidence, it can be said that the population proportion of adults who say they have started paying bills online in the last year is between the endpoints of the given confidence interval. OB. With 99% confidence, it can be said that the sample proportion of adults who say they have started paying bills online in the last year is between the endpoints of the given confidence interval. OC. The endpoints of the given confidence interval show that adults pay bills online 99% of the time. Construct the indicated confidence interval for the population mean using the t-distribution. Assume the population is normally distributed. c=0.90, x=14.4, s = 3.0, n=8 (Round to one decimal place as needed.) C Construct the confidence interval for the population mean . c=0.98, x=15.5, = 7.0, and n = 55 A 98% confidence interval for is (). (Round to one decimal place as needed.) The state test scores for 12 randomly selected high school seniors are shown on the right. Complete parts (a) through (c) below. 1430 1223 981 692 720 839 Assume the population is normally distributed. 724 750 545 620 1443 945 (a) Find the sample mean. x= (Round to one decimal place as needed.) (b) Find the sample standard deviation. S= (Round to one decimal place as needed.) (c) Construct a 90% confidence interval for the population mean . A 90% confidence interval for the population mean is ( ). (Round to one decimal place as needed.) Find the minimum sample size n needed to estimate for the given values of c, , and E. c=0.98, = 8.8, and E = 2 Assume that a preliminary sample has at least 30 members. n = (Round up to the nearest whole number.)