Question

In a survey of 2003 adults in a recent​ year, 706 made a New​ Year's resolution to eat healthier. Construct​ 90% and​ 95% confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals.

The​ 90% confidence interval for the population proportion p is

l

Answer 1

Solution :

Given that,

n = 2003

x = 706

Point estimate = sample proportion = = x / n = 706/2003=0.352

1 -   = 1- 0.352 =0.648

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

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

= 1.645 *((0.352*0.648) /2003 )

E = 0.0176

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.352-0.0176 < p <0.352+ 0.0176

0.3344< p < 0.3696

The 90% confidence interval for the population proportion p is : 0.3344, 0.3696

(B)

Solution :

Given that,

n = 2003

x = 706

Point estimate = sample proportion = = x / n = 706/2003=0.352

1 -   = 1- 0.352 =0.648

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96   ( Using z table )

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

= 1.96 *((0.352*0.648) /2003 )

E = 0.0209

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.352-0.0209 < p <0.352+0.0209

0.3311< p < 0.3729

The 95% confidence interval for the population proportion p is : 0.3311, 0.3729