Question

In a pool of n = 1000 randomly selected teenagers, 450 indicated that they like Netflix.

a) Construct a 95% confidence interval for the proportion of teenagers who like Netflix and interpret it.

b) Why the result in a) is approximately valid.

c) If you wish to estimate the proportion of teenagers who like Netflix correct to within 0.025 with 95% confidence, how large should the sample size n be?

Answer 1

Solution :

Given that,

n = 1000

x = 450

Point estimate = sample proportion = = x / n = 450/1000=0.45

1 - = 1-0.45=0.55

At 95% confidence level

= 1 - 95%  

= 1 - 0.95 =0.05

/2 = 0.025

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

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

= 1.96 (((0.45*0.55) /1000 )

E = 0.0308

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.45 -0.0308 < p < 0.45+0.0308

0.4192< p < 0.4808

The 95% confidence interval for the population proportion p is : 0.4192,0.4808

(c)

Solution :

Given that,

= 0.5 ( estimate is not given than use 0.5)

1 - = 1 - 0.5= 0.5

margin of error = E = 0.025

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 ( see the 0.025 value in standard    normal (z) table corresponding z value is 1.96 )

Sample size = n = (Z/2 / E)2 * * (1 - )

= (1.96 / 0.025)2 * 0.5* 0.5

= 1536.64

Sample size = 1537