Answers

2015-11-12T22:08:19+05:30
The proportion of nutritionally deficient boys is
p̂=21/170 = 0.1235.
The standard error of proportion is
SE = √p̂(1 − p̂)n
= √0.1235(1 − 0.1235)/170 = 0.0252.
A 95% confidence interval for the proportion of nutritionally deficient boys is
CI = (p̂− z∗SE; p̂+ SE) = (0.1235 − 1.96 ∙ 0.0252; 0.1235 + 1.96 ∙ 0.0252) = (0.0741; 0.1729).
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