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

Answer:

$ME=z_{\alpha/2} SE= 2.58*0.082=0.21156$

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 $0.09844 \leq p_B -p_A \leq 0.5216$

Step-by-step 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  

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$  

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

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

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:

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

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 $0.09844 \leq p_B -p_A \leq 0.5216$