A simple random sample of birth weights in the United Stateshas a mean of 3433g.

Question

A simple random sample of birth weights in the United Stateshas a mean of 3433g. The standard deviation of all birth weights is495g. a.) Using a sample of 60, construct a 98% confidence intervalestimate of the mean birth weight in the United States. b.) Using a sample size of 60,000 construct a 98% confidenceinterval estimate of the mean birth weight in the UnitedStates. c.) Which of the preceding confidence intervals is wider?Why? A simple random sample of birth weights in the United Stateshas a mean of 3433g. The standard deviation of all birth weights is495g. a.) Using a sample of 60, construct a 98% confidence intervalestimate of the mean birth weight in the United States. b.) Using a sample size of 60,000 construct a 98% confidenceinterval estimate of the mean birth weight in the UnitedStates. c.) Which of the preceding confidence intervals is wider?Why?

Explanation / Answer

mean, x = 3433
standard deviation, s = 495

a.) sample size, n = 60
hence standard error, e = s/ n = 495/60 = 63.9
for 98% confidence, z value = 2.33
confidence interval estimate of the mean birth weight in the UnitedStates = x ± ze
= 3433 ± (2.33*63.9) = (3284.113 , 3581.887 )


b.) sample size, n = 60,000
hence standard error, e = s/ n = 495/60000 = 2.02
for 98% confidence, z value = 2.33
confidence interval estimate of the mean birth weight in the UnitedStates = x ± ze
= 3433 ± (2.33*2.02) = (3428.2934 , 3437.7066 )


c.) confidence interval in case (a) is wider because itssample size is small. in case of (b) since the sample size islarge, hence the standard error becomes less which makes theconfidence interval less wider

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