Homework: Ch9 - Estimating Parameters Score: 0 of 1 pt 9.2.17 Save 16 of 25 (21

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Homework: Ch9 - Estimating Parameters Score: 0 of 1 pt 9.2.17 Save 16 of 25 (21 complete) HW Score: 83.27%, 20.82 of 25 p Question Helpi A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x, is found to be 106, and the sample standard deviation, s, is found to be 10 (a) Construct a 98% confidence interval about if the sample size, n, is 11 (b) Construct a 98% confidence interval about if the sample size, n, is 20. (c) Construct a 99% confidence interval about if the sample size, n, is 11 (d) Could we have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed? Click the icon to view the table of areas under the t-distribution. (a) Construct a 98% confidence interval about if the sample size, n, is 11. Lower bound: upper bound: (Use ascending order. Round to one decimal place as needed.)

Explanation / Answer

solution:-
given that mean = 106 , standard deviation = 10
confidence interval formula = x +/- t * s / sqrt(n)
(a) given df = 10 , 98% confidence value is t = 2.764
=> 106 +/- 2.764 * 10 / sqrt(11)
=> (97.7 , 114.3)

(b) df = 19 , 98% confidence t = 2.539
=> 106 +/- 2.539 * 10/sqrt(20)
=> (100.3 , 111.7)

(c) df = 10 , 99% confidence t = 3.169
=> 106 +/- 3.169 * 10/sqrt(11)
=> (96.4 , 115.6)

(d) No, the population needs to be normally distributed , and the sample size is too small

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