From a survey of coworkers you find that 50% of 184 have already received this y

Question

From a survey of coworkers you find that 50% of 184 have already received this year's flu vaccine. An approximate 98% confidence interval is (0.414, 0.586). How would the confidence interval change if the sample size had been 2944 instead of 184? How would the confidence interval change if the confidence level had been 95% instead of 98%? How would the confidence interval change if the confidence level had been 99% instead of 98%? Select the correct choice below and fill In the answer box within your choice. The new confidence interval would be narrower. The standard error would be divided by a factor of. The new confidence interval would be wider. The standard error would be multiplied by a factor of .

Explanation / Answer

Lets findout xbar and margin of error

xbar = .414+.586 / 2 = .5

Margin of errror = .086

So, at 98% CI the Z is 2.35

a. For sample size 2944 instead of 184?

ThAT' 16 times the sample size , or 1/4 times the MOE

So, MOE = .086/4 = .0215

Confidence interval becomes .5+/- .0215 instead of .5+/- .086

To answer in terms of question given:

The new CI would be narrower. The SE would be divided by factor of .25

b.

MOE of .086 will change to (.086/2.35)*1.96= .0602 ( adjusted for Z)

The new CI would be narrower. The SE would be divided by factor of 1.2

c.

MOE of .086 will change to (.086/2.35)*2.58= .0944 or .086 *(2.58/2.35) = .086*1.098 ( adjusted for Z)

The new CI would be wider . The SE would be divided by factor of 1.098

c.

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