A random sample of 500 army recruits has a mean height of 68 inches with a standard deviation of 2.5 inches. If a 95% confidence interval is constructed, with all the conditions having been met, what is the margin of error?
68
6.02
0.22
184

0.22

Step-by-step explanation:

Sample given is 500, so use z-score for the critical value

Given 95% confidence interval;

∝=100-95 =5% =0.05

∝/2 = 0.05/2 =0.025 because you are interested with one tail area

1-0.025= 0.975 area to the left

proceed to z-table to read 0.975 = 1.96 as the critical value

standard deviation from the question is 2.5 but because this is a sample then;

standard error for the mean is= standard deviation/√sample= 2.5/√500

=0.1118

Margin of error=1.96*0.1118 =0.2191

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step-by-step explanation: