Question

assume that a sample is used to estimate a population proportion P. Find a 98% confidence interval for a sample of size 254 with 85% successes. Enter your answer as an open interval (ie., parentheses) using decimals( not percents)

can someone tell me how this can be done on my TI-84+.

Answer 1

Solution:

Given ,

n = 254

Let denotes the sample proportion.

     = 85% = 0.85

Our aim is to construct 98% confidence interval.

c = 0.98

= 1- c = 1- 0.98 = 0.02

  /2 = 0.02 2 = 0.01 and 1- /2 = 0.99

= 2.326 (use z table)

Now , the margin of error is given by

E = /2 *  

= 2.326 * [(0.85 * (1 - 0.85)/254]

= 0.05211322432

Now the confidence interval is given by

( - E)   ( + E)

(0.85 -  0.05211322432 )   (0.85 + 0.05211322432)

0.79788677567 0.90211322432

Interval is (0.79788677567, 0.90211322432 )

Using TI84 plus calculator

x = n * = 215.9

Steps:

1) Press STAT

2)Go to 1 PropZint (One proportion z interval)

3)1 - PropZint(215.9 , 254 , 0.98) Press Enter

We get required interval