Question

10. 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 95​% confident that his estimate is in error by no more than three 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.

N=_____

​(Round up to the nearest​ integer.)

​b) Assume that a recent survey suggests that about 89​% of computers use a new operating system.

N=____

​(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.No, using the additional survey information from part​ (b) does not change the sample size.

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

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

D.No, using the additional survey information from part​ (b) only slightly increases the sample size.

9. A genetic experiment with peas resulted in one sample of offspring that consisted of 408 green peas and 160 yellow peas.

a. Construct a 95​% confidence interval to estimate of the percentage of yellow peas.

b. It was expected that​ 25% of the offspring peas would be yellow. Given that the percentage of offspring yellow peas is not​ 25%, do the results contradict​ expectations?

a. a. Construct a 95​% confidence interval. Express the percentages in decimal form.

____<p<____

​(Round to three decimal places as​ needed.)

Given that the percentage of offspring yellow peas is not​ 25%, do the results contradict​ expectations?

​No, the confidence interval includes​ 0.25, so the true percentage could easily equal​ 25%

​Yes, the confidence interval does not include​ 0.25, so the true percentage could not equal​ 25%

Answer 1

10)

Sample size = Z² $\alpha$ /2 * p ( 1 - p) / E²

a)

When prior estimate for proportion is not known then p = 0.50

N = 1.96² * 0.50 * 0.50 / 0.03²

= 1067.11

N = 1068 (Rounded up to nearest integer)

b)

Given, p = 0.89

N = 1.96² * 0.89 * 0.11 / 0.03²

= 417.88

N = 418 (Rounded up to nearest integer)

c)

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