Question

In: Statistics and Probability

A retirement planning firm conducted a survey of 80 working Americans and found that only 40%...

A retirement planning firm conducted a survey of 80 working Americans and found that only 40% of them had adequate retirement savings for their age group. Create the 99% confidence interval for the proportion of working Americans with adequate retirement savings.

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Expert Solution

Solution :

Given that,

n = 80

Point estimate = sample proportion = = 0.40

1 -   = 1- 0.40 =0.60

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576 ( Using z table )

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

= 2.576*((0.40*0.60) /80 )

=0.1411

A99% confidence interval for population proportion p is ,

- E < p < + E

0.40-0.1411 < p < 0.40+0.1411

0.2589< p < 0.5411

The 909% confidence interval for the population proportion p is : 0.2589, 0.5411

orchestra answered 2 years ago
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