Question

Gardeners on the west coast of the United States are investigating the difference in survival rates of two flowering plants in drought climates. Plant A has a survival rate of 0.75 and plant B has a survival rate of 0.44. The standard error of the difference in proportions is 0.082. What is the margin of error for a 99% confidence interval?

Answer 1

`ME=z_(\alpha/2) SE= 2.58*0.082=0.21156`

The 99% confidence interval would be given (0.09844;0.5216).

We are confident at 99% that the difference between the two proportions is between
`0.09844 \leq p_B -p_A \leq 0.5216`

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".

`p_A` represent the real population proportion for plant A

`\hat p_A =0.75` represent the estimated proportion for plant A

`n_A` is the sample size required for plant A

`p_B` represent the real population proportion for plant b

`\hat p_B =0.44` represent the estimated proportion for plant B

`n_B` is the sample size required for plant B

`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 99% confidence interval the value of
`\alpha=1-0.99=0.01` and
`\alpha/2=0.005`, with that value we can find the quantile required for the interval in the normal standard distribution.

`z_(\alpha/2)=2.58`

The standard error is given by:

`SE=z_(\alpha/2) \sqrt{(\hat p_A(1-\hat p_A))/(n_A) +(\hat p_B (1-\hat p_B))/(n_B)}=0.082`

The margin of error is:

`ME=z_(\alpha/2) SE= 2.58*0.082=0.21156`

And replacing into the confidence interval formula we got:

`(0.75-0.44) - 2.58(0.082)=0.09844`

`(0.75 -0.44) + 2.58(0.082)=0.5216`

And the 99% confidence interval would be given (0.09844;0.5216).

We are confident at 99% that the difference between the two proportions is between
`0.09844 \leq p_B -p_A \leq 0.5216`