A web based soltware compaiy is inlerested in estirnaling the proportion ol indi

Question

A web based soltware compaiy is inlerested in estirnaling the proportion ol individuals who use the Firelox browser. I a sample of 200 ofdividuals, 31 users slated thal they used Firefox. Using this dala, construct a 99% confidence interval for the proportion of all individuals that use Firefox. a) What is the lower limit on the 99% confidence interval? Give your answer to three decimal places. b) What is the upper limit on the 99% confidence interval? Give your answer to three decimal places. c) Google states that the proportion of all individuals that use Firefox is 0.2. Based on the interval above, does the Google claim seem reasonable Yes because 0.2 is inside the interval No because 0.2 is inside the interval Yes bese 0.2 is not inside the interval No because 0.2 is not inside the interval. d If you hadn't already taken the sample listed above, what sample size would be required so that the width of the 99% confidence interval would be at most 0.02 units wide? Be as conservative as possible with your answer!

Explanation / Answer

p = 31/200 = 0.155

a) At 99% confidence interval the critical value is z0.005 = 2.58

The 99% lower confidence interval is

p - z0.005 * sqrt(p * (1 - p)/n)

= 0.155 - 2.58 * sqrt(0.155 * (1 - 0.155)/200)

= 0.155 - 0.066

= 0.089

b)The 99% upper confidence interval is

p + z0.005 * sqrt(p * (1 - p)/n)

= 0.155 + 2.58 * sqrt(0.155 * (1 - 0.155)/200)

= 0.155 + 0.066

= 0.221

c) Yes, because 0.2 is inside the interval.

d) Margin of error = 0.01

or, z0.005 * sqrt(p * (1 - p)/n) = 0.01

or, 2.58 * sqrt(0.155 * (1 - 0.155)/n) = 0.01

or, sqrt(n) = 2.58 * sqrt(0.155 * (1 - 0.155))/0.01

or, sqrt(n) = 93.37

or, n = 8718

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