QUESTION 02 (10 points) Sample Size Determination for ClI of Population Proporti

Question

QUESTION 02 (10 points) Sample Size Determination for ClI of Population Proportion We need to find the homeownership proportion in Southern California. The homeownership proportion is believed to be around 54%. How many householders would you survey for a 95% Cl where the maximum likely sampling error is no more than 1.2%? Make sure to say the result in words. In addition, make a sketch to illustrate how you find the z-value Hint: Round up the final result to a whole number. Refer to some Excel lookups: 0.995 0.985 2.17009 0.975 1.95996 0.955 1.69540 0.965 NORM.S.INV(av)2.57583 1.81191

Explanation / Answer

Formula for Sample size determination for CI of proportion

Required sample size = n = {p-bar * (1- p-bar) * z^2}/SE^2

Where,

Required sample size, n =?

Value of z at 95% confidence level = 1.645 (By using “=NORM.S.INV (0.95)”)

The homeownership proportion is believed (prior judgment), p-par = 54% or 0.54   

Sampling Error, SE = 1.2% or 0.012

Therefore

n = {0.54 * (1- 0.54) * 1.96^2}/0.012^2

= 4667.06 or 4667 (rounding off to nearest whole number)

Therefore required sample size is 4,667.

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