In the planning stage, a sample proportion is estimated as = 56/70 = 0.80. Use t

Question

In the planning stage, a sample proportion is estimated as = 56/70 = 0.80. Use this information to compute the minimum sample size n required to estimate p with 95% confidence if the desired margin of error D = 0.09. What happens to n if you decide to estimate p with 90% confidence? Use Table 1. (Round your intermediate calculations to 4 decimal places and "z" value to 2 decimal places. Round up your answers to the nearest whole number.)

In the planning stage, a sample proportion is estimated as p-hat = 56/70 = 0.80. Use this information to compute the minimum sample size n required to estimate p with 95% confidence if the desired margin of error D = 0.09. What happens to n if you decide to estimate p with 90% confidence? Use Table 1. (Round your intermediate calculations to 4 decimal places and ''z'' value to 2 decimal places. Round up your answers to the nearest whole number.)

Explanation / Answer

Here phat = 56/70 = 0.80 , Margin of error D = 0.09

confidence level = 95% so Z0.95 = 1.96

Minimum Sample Size n = ( Z0.95 / D)2 phat (1 - phat)

= ( 1.96 / 0.09)2 * 0.80 *(1-0.80)

= 75.88

= 76 (rounding to nearest whole number)

confidence level = 90% so Z0.90 = 1.64

Minimum Sample Size n = ( Z0.90 / D)2 phat (1 - phat)

= ( 1.64 / 0.09)2 * 0.80 *(1-0.80)

= 53.13

   = 53 (rounding to nearest whole number)

So we observed that value of n decreases from 76 to 53 as the value of confidence level decreases from 95% to 90%.

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