Question

Question: A researcher wishes to estimate the proportion of adults who have high-speed Internet access. What size sample should be obtained if she wishes the estimate to be within 0.030.03 with 9090 % confidence if (a) she uses a previous estimate of 0.580.58 ? (b) she does not use any prior estimates?

A researcher wishes to estimate the proportion of adults who have high-speed Internet access. What size sample should be obtained if she wishes the estimate to be within

0.030.03

with

9090 %

confidence if

(a) she uses a previous estimate of

0.580.58 ?

(b) she does not use any prior estimates?

Answer 1

To estimate the sample size needed to estimate the proportion of adults with high-speed internet access, we can use the formula: n = (z^2 * p * (1-p)) / E^2. If a previous estimate of 0.58 is used with a 90% confidence level and a margin of error of 0.03, the sample size is approximately 615. If no prior estimate is used and a conservative estimate of p = 0.5 is assumed, the sample size is approximately 1068.

Step-by-step explanation:

To estimate the sample size needed to estimate the proportion of adults with high-speed internet access, we can use the formula:

n = (z^2 * p * (1-p)) / E^2

Where:

• n is the sample size
• z is the z-score corresponding to the desired confidence level
• p is the estimated proportion of adults with high-speed internet access
• E is the desired margin of error

a) If the researcher uses a previous estimate of 0.58 and wants a 90% confidence level with a margin of error of 0.03, we can plug in the values into the formula:

n = (1.645^2 * 0.58 * (1-0.58)) / 0.03^2

Solving this equation, we get a sample size of approximately 615.

b) If the researcher does not use any prior estimates, we can use a conservative estimate of p = 0.5 and repeat the calculation:

n = (1.645^2 * 0.5 * (1-0.5)) / 0.03^2

This gives us a sample size of approximately 1068.