Question

In​ 2003, an organization surveyed 1 comma 508 adult Americans and asked about a certain​ war, "Do you believe the United States made the right or wrong decision to use military​ force?" Of the 1 comma 508 adult Americans​ surveyed, 1 comma 085 stated the United States made the right decision. In​ 2008, the organization asked the same question of 1 comma 508 adult Americans and found that 575 believed the United States made the right decision. Construct and interpret a​ 90% confidence interval for the difference between the two population​ proportions, p 2003 minus p 2008.

Answer 1

`(0.719-0.381) - 1.64 \sqrt{(0.719(1-0.719))/(1508) +(0.381(1-0.381))/(1508)}=0.310`

`(0.719-0.381) + 1.64 \sqrt{(0.719(1-0.719))/(1508) +(0.381(1-0.381))/(1508)}=0.366`

And the 90% confidence interval would be given (0.310;0.366).

We are confident at 90% that the difference between the two proportions is between
`0.310 \leq p_A -p_B \leq 0.366`

Explanation:

Previous concepts

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".

`p_A` represent the real population proportion for 2003

`\hat p_A =(1085)/(1508)=0.719` represent the estimated proportion for 2003

`n_A=1508` is the sample size required for 2003

`p_B` represent the real population proportion for 2008

`\hat p_B =(575)/(1508)=0.381` represent the estimated proportion for 2008

`n_B=1508` is the sample size required for 2008

`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_A -\hat p_B) \pm z_(\alpha/2) \sqrt{(\hat p_A(1-\hat p_A))/(n_A) +(\hat p_B (1-\hat p_B))/(n_B)}`

For the 90% confidence interval the value of
`\alpha=1-0.90=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:

`(0.719-0.381) - 1.64 \sqrt{(0.719(1-0.719))/(1508) +(0.381(1-0.381))/(1508)}=0.310`

`(0.719-0.381) + 1.64 \sqrt{(0.719(1-0.719))/(1508) +(0.381(1-0.381))/(1508)}=0.366`

And the 90% confidence interval would be given (0.310;0.366).

We are confident at 90% that the difference between the two proportions is between
`0.310 \leq p_A -p_B \leq 0.366`