Question

A programmer plans to develop a new software system. In planning for the operating system that he will​ use, he needs to estimate the percentage of computers that use a new operating system. How many computers must be surveyed in order to be 90 ​% confident that his estimate is in error by no more than five percentage points question mark Complete parts​ (a) through​ (c) below. ​a) Assume that nothing is known about the percentage of computers with new operating systems. nequals nothing ​(Round up to the nearest​ integer.) ​b) Assume that a recent survey suggests that about 98 ​% of computers use a new operating system. nequals nothing ​(Round up to the nearest​ integer.) ​c) Does the additional survey information from part​ (b) have much of an effect on the sample size that is​ required? A. ​Yes, using the additional survey information from part​ (b) dramatically reduces the sample size. B. ​Yes, using the additional survey information from part​ (b) dramatically increases the sample size. C. ​No, using the additional survey information from part​ (b) does not change the sample size. D. ​No, using the additional survey information from part​ (b) only slightly increases the sample size. Click to select your answer(s).

Answer 1

(A) When prior estimate of proportion is not given

z score for 90% confidence interval is 1.645 {using z distribution table}

Margin of error (ME) = 5% = 5/100 = 0.05

Sample size

this implies

Rounding to nearest integer, we get sample size n = 271

(B) when proportion = 98% = 98/100 = 0.98

z score for 90% confidence interval is 1.645 {using z distribution table}

Margin of error (ME) = 5% = 5/100 = 0.05

Sample size

this implies

Rounding to nearest integer, we get sample size n = 21

(C) Sample size without prior estimation is bigger as compared to the sample size with prior estimation.

A. ​Yes, using the additional survey information from part​ (b) dramatically reduces the sample size

This is correct option