Question

# 13. In the planning stage, a sample proportion is estimated as p^ = 36/60 = 0.60....

13. In the planning stage, a sample proportion is estimated as p^ = 36/60 = 0.60. Use this information to compute the minimum sample size n required to estimate p with 95% confidence if the desired margin of error E = 0.09. What happens to n if you decide to estimate p with 90% confidence?

(You may find it useful to reference the z table. Round intermediate calculations to at least 4 decimal places and "z" value to 3 decimal places. Round up your answers to the nearest whole number.)

 Confidence Level n 95% ? 90% ?

Please do not answer if you are not sure, THANK YOU! :)

13)

Solution :

Given that,

= 0.60

1 - = 0.40

margin of error = E = 0.09

At 95% confidence

Z/2 = Z0.025 = 1.96

sample size = n = (Z / 2 / E )2 * * (1 - )

= (1.96 / 0.09)2 * 0.60 * 0.40

= 114

sample size = n = 114

At 90% confidence

Z/2 = Z0.05 = 1.645

sample size = n = (Z / 2 / E )2 * * (1 - )

= (1.645 / 0.09)2 * 0.60 * 0.40

= 81

sample size = n = 81

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