Question

According to a Field Poll conducted February 8 – 17, 2005, 79% of California adults (actual results are 400 out of 506 surveyed) feel that "education and our schools" is one of the top issues facing California. We wish to construct a 90% confidence interval for the true proportion of California adults who feel that education and the schools is one of the top issues facing California.
Find a 90% confidence interval for the population proportion. (Round your answers to three decimal places.)

_______to_______

Answer 1

Answer: (0.760, 0.820)

Step-by-step explanation:

Let p be the population proportion of all adults feel that "education and our schools" is one of the top issues facing California.

Given : 79% (actual results are 400 out of 506 surveyed) of California adults feel that "education and our schools" is one of the top issues facing California.

i.e.Sample size : n= 506

Sample proportion : $\hat{p}=0.79$

Critical value for 90% confidence interval ( Using z-value table) :

$z=1.645$

Now, the 90% confidence interval for the population proportion will be :

$\hat{p}\pm z\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

i.e. $0.79\pm (1.645)\sqrt{\dfrac{0.79(1-0.79)}{506}}$

i.e. $0.79\pm (1.645)(0.0181)$

$\approx0.79\pm 0.030=(0.79-0.030,\ 0.79+0.030)\\\\=(0.760,\ 0.820)$

Hence, the 90% confidence interval for the population proportion= (0.760, 0.820)