In a sample of 50 women, it is determined that 53% have their HDL cholesterol level checked on a regular basis. Find the margin of error of a 95% confidence interval for the proportion of all such females

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Jun 22nd, 2015

E=z_{α/2}/(2√ n)

95% level of confidence. We want to look up thez-scorez*for which the area between -z* and z* is 0.95. From the table we see that this critical value is 1.96.

since α = 1 - 0.95 = 0.05, we see that α/2 = 0.025

We now search the table to find thez-score with an area of 0.025 to its right. We would end up with the same critical value of 1.96.

E = 1.96 / 2√ 50

E = 6.93

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