Question

The mean weight of 50% newborn babies selected from 60 randomly at a local hospital. Assume the weight of newborn babies has approximately normal distribution. a. Find the Critical Value 99% for the mean weight of all newborn babies at this hospital. b. Find the Margin of Error for the 99% confidence interval all newborn babies at this hospital. c. Find a 99% Confidence Interval for the mean weight of all newborn babies at this hospital.

Answer 1

Solution :

Given that,

n = 60

Point estimate = sample proportion = = 0.50

1 - = 0.50

a)

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

b)

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 2.576 * (((0.50 * 0.50) / 60)

= 0.166

c)

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.50 - 0.166 < p < 0.50 + 0.166

0.334 < p < 0.666