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Suppose the distribution of salaries among recent college graduates is approximately Normal with a standard deviation of $8,689

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Suppose the distribution of salaries among recent college graduates is approximately Normal with a standard deviation of $8,689. In a simple random sample of 150 college graduates who have taken a statistics course, starting salaries had a mean of $44,583. a. The margin of error in an 81% confidence interval for the average starting salarly among recent college graduates who have taken a stastics course, based on this sample, is: b. The 81% confidence interval is: i ' ' < [4 < i as; (round each value to the nearest whole dollar) ' (round to the nearest whole dollar).

(1 point) In a simple random sample of 15 gas stations in the metropolitan area, the price per gallon of regular gasoline had a mean of $2.51 and a (sample) standard deviation of $0.75. Use this sample data to find a 98% confidence interval for the mean price per gallon of regular gasoline for all gas stations the metropolitan area. Round all answers to the nearest cent (two decimal places). a. The margin of error for this con?dence interval is: $ . b. The confidence interval is $ '

(1 point) In 2018, the prices of new mid-size cars sold by a local dealership followed an approximately Normal distribution with a standard deviation of $2,474. In a simple random sample of 50 new mid-size cars sold in 2018 by this dealership, the average sale prices had a mean of $18,839. a. Using this sample, a 98% confidence interval for the mean price of new mid-size cars sold by this dealership has a Margin of Error of $ 1 (round to the nearest whole dollar). b. Using this sample, a 98% confidence interval for the mean price of new mid-size cars sold by this dealership is $ :[? < [4 < $ iii (round each value to the nearest whole dollar) If prices for all new mid-size cars sold by this dealership did not follow a Normal Distribution, could we expect the above results to be accurate? OA. Yes, because the sample size is less than 100. O B. No, because the population's standard deviation is known. © 0. Yes, because the sample size is greater than 30. O D. No, because the sample size is less than 100. O E. Yes, because the population's standard deviation is known. 0 F. No, because the sample size is greater than 30.

(1 point) Based on a simple random sample of size 90, an 85% confidence interval for the mean SAT MATH score in some large population is 506 < u < 518. Which of the following is/are correct interpretation(s) of this confidence interval? There may be more than one correct answer; you must check all correct answers in order to get credit for this problem. A. There is an 85% chance that another sample of size 90 from this population would have a mean SAT MATH score between 506 and 518. _ B. 85% of individuals in this sample have a mean SAT MATH score between 506 and 518. C. 85% of all samples of size 90 from this population would have a mean SAT MATH score between 506 and 518. D. 85% of individuals in this population have a SAT MATH score between 506 and 518. E. There is an 85% chance that the true mean SAT MATH score for the entire population is between 506 and 518. _ F. There is an 85% chance the interval 506 < [4 < 518 contains the true mean SAT MATH score for the entire population.