Question

Researchers are interested in determining whether more women than men prefer the beach to the mountains. In a random sample of 200 women, 45% prefer the beach, whereas in a random sample of 300 men, 52% prefer the beach. What is the 99% confidence interval estimate for the difference between the percentages of women and men who prefer the beach over the mountains?a. − 0.07 ± 0.1172b. − 0.07 ± 0.0557c. − 0.07 ± 0.0053d. 0.07 ± 0.01172e. 0.07 ± 0.0053

Answer 1

Answer:

$SE=\sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$

 $SE=\sqrt{\frac{0.52(1-0.52)}{300} +\frac{0.45 (1-0.45)}{200}}=0.0455$

The margin of error would be:

$M = z* SE = 2.576*0.0455=0.1172$

$(0.45-0.52) - 0.1172=-0.047$  

$(0.45-0.52) + 0.1172=0.187$  

So then the correct option for this case would be:

a. − 0.07 ± 0.1172

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".  

$p_A$ represent the real population proportion of men

$\hat p_A =0.52$ represent the estimated proportion of men

$n_A= 300$ the random sample for male

$p_B$ represent the real population proportion of female

$\hat p_B =0.45$ represent the estimated proportion of female

$n_B =200$ the random sample for female

$z$ 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  

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

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.  

$z_{\alpha/2}=2.576$  

The standard error is given by:

$SE=\sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$

 $SE=\sqrt{\frac{0.52(1-0.52)}{300} +\frac{0.45 (1-0.45)}{200}}=0.0455$

The margin of error would be:

$M = z* SE = 2.576*0.0455=0.1172$

And replacing into the confidence interval formula we got:  

$(0.45-0.52) - 0.1172=-0.047$  

$(0.45-0.52) + 0.1172=0.187$  

So then the correct option for this case would be:

a. − 0.07 ± 0.1172