| 17 | The lower end of a 95% interval estimate of the mean annual salary of persons with an MBA degree five years after graduation is reported to be $81,200. The interval is based on a sample of 310 persons. The sample standard deviation is $23,356. What is the point estimate of the salary? (Note: The z-score was used in building the interval) | |||||||
| a | $85,100 | |||||||
| b | $84,900 | |||||||
| c | $84,100 | |||||||
| d | $83,800 | |||||||
| 18 | In the previous question, what is the upper end of the interval? | |||||||
| a | $86,400 | |||||||
| b | $86,700 | |||||||
| c | $87,500 | |||||||
| d | $87,700 | |||||||
I'm doing something wrong because every answer I get is slightly off.
Lower Bound = 81200
std.dev = 23356 , n = 310
LOwer Bound = mean -z *(s/sqrt(n))
81200 = mean - 1.96 *(23356/sqrt(310))
mean = 2600 + 81200
mean = 83800
18)
Upper Bound = 83800 + 1.96 *(23356/sqrt(310))
Upper Bound = 86400
17 The lower end of a 95% interval estimate of the mean annual salary of persons...
The lower and upper end of a 95% interval estimate of the population proportion are, respectively, 0.582 and 0.658. 19 The point estimate to build this interval is, a 0.634 b 0.630 c 0.628 d 0.620 20 The sample size to build this interval estimate is, a 675 b 649 c 627 d 615