Question

A sociologist wants to construct a 96% confidence interval for
the proportion of children aged 5 – 11 living in New York who own
cell phones.

a) A survey by Mediamark Research and Intelligence estimated the
nationwide proportion to be 0.20. Using this estimate, what sample
size is needed so that the confidence interval will have a margin
of error of 0.02?

b) Estimate the sample size needed if no estimate of the nationwide
proportion is given.

Answer #1

Solution:

Given ,

confidence level c = 96% = 0.96

= 1 - c = 1 - 0.96 = 004

/2 = 0.02

Using Z table ,

= 2.054

a) Here , p = 0.20

1 - p = 1 - 0.20 = 0.80

E = 0.02

The sample size for estimating the proportion is given by

n =

= (2.054)^{2} * 0.20 * 0.80 / (0.02^{2})

= 1682.6404

= 1683 ..(round to the next whole number)

Answer : Required Sample size is n = **1683**

b)

Here p is not known

Take p = 0.5

1 - p = 1 - 0.5 = 0.5

The sample size for estimating the proportion is given by

n =

= (2.054)^{2} * 0.5 * 0.5 / (0.02^{2})

= 2636.8225

= 2637 ..(round to the next whole number)

Answer : Required Sample size is n = **2637**

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