In the planning stage, a sample proportion is estimated as 2formula204.mml = 40/

Question

In the planning stage, a sample proportion is estimated as 2formula204.mml = 40/50 = 0.80. Use this information to compute the minimum sample size n required to estimate p with 99% confidence if the desired margin of error D = 0.12. 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.) Confidence Level n

99% confidence level is 74
90% ?

Explanation / Answer

In the planning stage, a sample proportion is estimated as 2formula204.mml = 40/50 = 0.80. Use this information to compute the minimum sample size n required to estimate p with 99% confidence if the desired margin of error D = 0.12. 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.) Confidence Level n

99% confidence level is 74

P=0.8

For 99%, z=2.58

d=0.12

Sample size = (z2*p*(1-p))/d2

= (2.582*0.2*0.8)/0.122

=73.96

The sample size required=74

90% ?

For 90%, z=1.64 ( two decimals)

n= (1.642*0.2*0.8)/0.122

=29.88

The sample size required=30

( if we take z value more than 2 decimals, we get sample size of 31).

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