Question

An association of Christmas tree growers in Indiana sponsored a sample survey of 500 randomly selected Indiana households to help improve the marketing of Christmas trees. One question the researchers asked was, "Did you have a Christmas tree this year?" Respondents who had a tree during the holiday season were asked whether the tree was natural or artificial. Respondents were also asked if they lived in an urban area or in a rural area. The tree growers want to know if there is a difference in preference for natural trees versus artificial trees between urban and rural households. Among the 160 who lived in rural areas, 64 had a natural tree. Among the 261 who lived in an urban area, 89 had a natural tree. Construct and interpret a 95% confidence interval for the difference in the proportion of rural and urban Indiana residents who had a natural Christmas tree this year.

Answer 1

`(0.4-0.341) - 1.96 \sqrt{(0.4(1-0.4))/(160) +(0.341(1-0.341))/(261)}=-0.0362`

`(0.4-0.341) + 1.96 \sqrt{(0.4(1-0.4))/(160) +(0.341(1-0.341))/(261)}=0.154`

And the 95% confidence interval would be given (-0.0362;0.154).

We are confident at 95% that the difference between the two proportions is between
`-0.0362 \leq p_A -p_B \leq 0.154`

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 rural

`\hat p_A =(64)/(160)=0.4` represent the estimated proportion for rural with natural tree

`n_A=160` is the sample size required for rural with natural tree

`p_B` represent the real population proportion for urban with natural tree

`\hat p_B =(89)/(261)=0.341` represent the estimated proportion for urban with natural tree

`n_B=261` is the sample size required for Brand B

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

Solution to the problem

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 95% confidence interval the value of
`\alpha=1-0.95=0.05` and
`\alpha/2=0.025`, with that value we can find the quantile required for the interval in the normal standard distribution.

`z_(\alpha/2)=1.96`

And replacing into the confidence interval formula we got:

`(0.4-0.341) - 1.96 \sqrt{(0.4(1-0.4))/(160) +(0.341(1-0.341))/(261)}=-0.0362`

`(0.4-0.341) + 1.96 \sqrt{(0.4(1-0.4))/(160) +(0.341(1-0.341))/(261)}=0.154`

And the 95% confidence interval would be given (-0.0362;0.154).

We are confident at 95% that the difference between the two proportions is between
`-0.0362 \leq p_A -p_B \leq 0.154`