Question

Determine the margin of error for a 98% confidence interval to estimate the population proportion with a sample proportion equal to 0.90 for the following sample sizes n=125, n=220, n=250

Answer #1

Solution:

Formula for margin of error for the population proportion is given as below:

Margin of error = E = Z* sqrt(P*(1 – P)/n)

Where, P is the sample proportion, Z is critical value, and n is sample size.

We are given

P = 0.90

Confidence level = 98%

Z = 2.3263

(by using z-table)

For n = 125

Margin of error = Z* sqrt(P*(1 – P)/n)

Margin of error = 2.3263*sqrt(0.90*(1 - 0.90)/125)

**Margin of error =** **0.062421**

For n = 220

Margin of error = 2.3263*sqrt(0.90*(1 - 0.90)/220)

**Margin of error =** **0.047052**

For n = 250

Margin of error = 2.3263*sqrt(0.90*(1 - 0.90)/250)

**Margin of error =** **0.044138**

It is observed that, as the sample size increases, the margin of error decreases.

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