a sample of recent graduate salaries taken form a normally distributed populatio
ID: 3245522 • Letter: A
Question
a sample of recent graduate salaries taken form a normally distributed population was reported as follows (in thousands of dollars) 42, 38, 30, 29, 47, 51, 33, 31, 34, 45, 44, 35, 40, 39, 37, 36, 41, 38, 52, 43 a. what is the point estimate of the mean? b. make a 99% confidence interval for . Be sure to report the standard deviation and any other relevant statistics that helped you create this interval. c. what is the maximum error of the estimate reported for part b? d. suppose the confidence interval obtain in b is two wide. Describe all the possible ways the width of this interval may be reduced. Which of these is the best alternative? e. Suppose another same of 20 s drawn such that the mean is 42 and the standard deviation is 5. Can you prove the two samples are statistically different?
mean: 38.75
SD: 7.19557613
Explanation / Answer
From the given data we have
(a) point mean estimate = 39.25
(b)
Hence 99% CI = (35.1946, 43.3054)
(c)
Maximum error of estimate = Margin of error (ME) = 4.0554
(d)
In order to reduce the width of CI
1) Decrease the confidence interval, it is 99% now. this can be reduced to 95% or 90% which will decrease the value of t which results into decrease in the value of ME.
2) Increase the size of sample(n), right now it is 20. Increasing the value of n will reduce the margin of error this will reduce the width of CI
Mean 39.25 std. dev. 6.3394 n 20 std. error = std.dev./sqrt(n) 1.4175 t (two tail for 0.01) 2.8609 margin of error = t*SE 4.0554Related Questions
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