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put this solution on YOUR website!Q1. if Z is a standard normal variable, find the probability. P(-0.73
Comment: Something must be missing in this posting.
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Q2. express the confidence interval in the form of +/-E, -0.052
Comment: Something is missing here.
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Q3. find the critical value Z(alpha/2) that corresponds to a degree of confidence of 91%
Half of 91% is 45.5%
The z-score that corresponds to 4.5% and to 95.5% is -1.6954 and +1.6954
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Q4. find the minimum sample size you should use to assure that your estimate of P will be within the required margin of error around the population P. Margin of error:0.04 confidence level:94% P and q unknown.
E=z*sqrt(pq/n)
sqrt(n)=[z*sqrt(pq)]/E
sqrt(n) = [1.88079 sqrt(.5*.5)]/0.04
sqrt(n) = [0.9403968049]/0.04
sqrt(n) = 23.5099
Square both sides to get:
n=552 (rounded down to the whole number)
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Cheers,
Stan H.
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