Question

Use the normal distribution to find a confidence interval for a difference in proportions p1-p2 given the relevant sample results. Assume the results come from random samples.
A 99% confidence interval for p1-p2 given that p^1=0.20 with n1=60 and p^2=0.35 with n2=100

Give the best estimate for p1-p2, the margin of error, and the confidence interval.

Round your answer for the best estimate to two decimal places and round your answers for the margin of error and the confidence interval to three decimal places.

a) Find Best Estimate:
b) Find Margin of Error
c) Find Confidence Interval

Answer 1

Answer:

a) $\hat p_1 -\hat p_2= 0.2-0.35= -0.15$

b) $ME= 2.58 \sqrt{\frac{0.2(1-0.2)}{60} +\frac{0.8(1-0.8)}{100}} =0.169$

c) $(0.2-0.35) - 2.58 \sqrt{\frac{0.2(1-0.2)}{60} +\frac{0.8(1-0.8)}{100}} =-0.319$  

$(0.2-0.35) + 2.58 \sqrt{\frac{0.2(1-0.2)}{60} +\frac{0.8(1-0.8)}{100}} =0.0185$  

And the 99% confidence interval would be given (-0.319;0.0185).  

Step-by-step 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".  

$p_1$ represent the real population proportion 1

$\hat p_1=0.2$ represent the estimated proportion 1

$n_1=60$ is the sample size required 1

$p_2$ represent the real population proportion for 2

$\hat p_2 =0.35$ represent the estimated proportion 2

$n_2=100$ 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  

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

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

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

Part a

The best estimate is given by:

$\hat p_1 -\hat p_2= 0.2-0.35= -0.15$

Part 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 margin of error is given by:

$ME= 2.58 \sqrt{\frac{0.2(1-0.2)}{60} +\frac{0.8(1-0.8)}{100}} =0.169$

Part c

And replacing into the confidence interval formula we got:  

$(0.2-0.35) - 2.58 \sqrt{\frac{0.2(1-0.2)}{60} +\frac{0.8(1-0.8)}{100}} =-0.319$  

$(0.2-0.35) + 2.58 \sqrt{\frac{0.2(1-0.2)}{60} +\frac{0.8(1-0.8)}{100}} =0.0185$  

And the 99% confidence interval would be given (-0.319;0.0185).