A study of two hundred teens found that the number of hours they spend on social networking sites each week is normally distributed with a mean of 12 hours. the population standard deviation is 4 hours. what is the margin of error for a 98% confidence interval?
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The confidence interval formula is as follows :
C.I = X + /- (α/√n) Zβ/2
The β = 1 - 0.98 = 0.02
0.02/2 = 0.01
Z0. 01 = 2.57
α = 4
n = 200
X = 12
√200 = 14.14
Doing the substitution :
α / √n = 4/14.14 = 0.2829
0.2829 × 2.57 = 0.7271
The interval is :
Lower limit : 12 - 0.7271 = 11.2729
Upper limit : 12 + 0.7271 = 12.7271
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