Question

1.)

In a recent poll, 100 people were asked if they liked cats, and 81% said they did. Based on this, construct a 99% confidence interval for the true population proportion of people who like cats.

Use the following approximate critical values (z-scores) to perform the calculations by hand:
--Use z = 1.645 for a 90% confidence interval
--Use z = 2 for a 95% confidence interval
--Use z = 2.576 for a 99% confidence interval.


Give your answers as decimals, to 4 decimal places.

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2.)

If n=480 and ˆpp^ (p-hat) = 0.88, construct a 99% confidence interval.

Calculate this interval using the 1PropZInt program in your calculator. Give your answers to three decimals.

< p <

Answer 1

1)

Solution:

Confidence Interval for population proportion(p)

Given,

n = 100 ....... Sample size

Let denotes the sample proportion.

     = 81% = 0.81

Our aim is to construct 99% confidence interval.

c = 0.99

= 1- c = 1- 0.99 = 0.01

  /2 = 0.01 2 = 0.005 and 1- /2 = 0.995

Search the probability 0.995 in the Z table and see corresponding z value

= 2.576 is the critical value.

Now , confidence interval for population proportion(p) is given by:

     

0.81 - 2.576*       0.81 + 2.576*

0.81 - 0.1010567131 <     < 0.81 + 0.1010567131

0.7089 <     < 0.9111

is the required 99% confidence interval for the population proportion....

2)given ,

n = 480

= 0.88

using calculator the required 99% CI for the population proportion is

0.841 <     < 0.917