Question

Construct a 90% confidence interval for (P1-P2) in each of the following situations. a. n1-400, p1-0.67; n2-400, p2 = 0.56. b. n1 = 180; p1 = 0.31; nz" 250, p2 = 0.25. c. n1 = 100; p1 = 0.46; n2 = 120, pz" 0.61.

Answer 1

a)
`(0.67-0.56) - 1.64 \sqrt{(0.67(1-0.67))/(400) +(0.56(1-0.56))/(400)}=0.054`

`(0.67-0.56) + 1.64 \sqrt{(0.67(1-0.67))/(400) +(0.56(1-0.56))/(400)}=0.166`

And the 90% confidence interval would be given (0.054;0.166).

b)
`(0.31-0.25) - 1.64 \sqrt{(0.31(1-0.31))/(180) +(0.25(1-0.25))/(250)}=-0.0122`

`(0.31-0.25) +1.64 \sqrt{(0.31(1-0.31))/(180) +(0.25(1-0.25))/(250)}=0.132`

And the 90% confidence interval would be given (-0.0122;0.132).

c)
`(0.46-0.61) - 1.64 \sqrt{(0.46(1-0.46))/(100) +(0.61(1-0.61))/(120)}=-0.260`

`(0.46-0.61) +1.64 \sqrt{(0.46(1-0.46))/(100) +(0.61(1-0.61))/(120)}=-0.0404`

And the 90% confidence interval would be given (-0.260;-0.0404).

Explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Part a

`p_1` represent the real population proportion for sample 1

`\hat p_1 =0.67` represent the estimated proportion for sample 1

`n_A=400` is the sample size required for sample 1

`p_2` represent the real population proportion for sample 2

`\hat p_2 =0.56` represent the estimated proportion for sample 2

`n_2=400` is the sample size required for sample 2

`z` represent the critical value for the margin of error

The population proportion have the following distribution

`p \sim N(p,\sqrt{(p(1-p))/(n)})`

The confidence interval for the difference of two proportions would be given by this formula

`(\hat p_1 -\hat p_2) \pm z_(\alpha/2) \sqrt{(\hat p_1(1-\hat p_1))/(n_1) +(\hat p_2 (1-\hat p_2))/(n_2)}`

For the 90% confidence interval the value of
`\alpha=1-0.9=0.1` and
`\alpha/2=0.05`, with that value we can find the quantile required for the interval in the normal standard distribution.

`z_(\alpha/2)=1.64`

And replacing into the confidence interval formula we got:

`(0.67-0.56) - 1.64 \sqrt{(0.67(1-0.67))/(400) +(0.56(1-0.56))/(400)}=0.054`

`(0.67-0.56) + 1.64 \sqrt{(0.67(1-0.67))/(400) +(0.56(1-0.56))/(400)}=0.166`

And the 90% confidence interval would be given (0.054;0.166).

Part b

`(0.31-0.25) - 1.64 \sqrt{(0.31(1-0.31))/(180) +(0.25(1-0.25))/(250)}=-0.0122`

`(0.31-0.25) +1.64 \sqrt{(0.31(1-0.31))/(180) +(0.25(1-0.25))/(250)}=0.132`

And the 90% confidence interval would be given (-0.0122;0.132).

Part c

`(0.46-0.61) - 1.64 \sqrt{(0.46(1-0.46))/(100) +(0.61(1-0.61))/(120)}=-0.260`

`(0.46-0.61) +1.64 \sqrt{(0.46(1-0.46))/(100) +(0.61(1-0.61))/(120)}=-0.0404`

And the 90% confidence interval would be given (-0.260;-0.0404).