Question

A researcher wants to determine whether there is a relationship between age and the number of text messages sent in a given day. In a random sample of 18- to 25-year-old cell phone users, 100 of 125 users sent texts more often than they made phone calls. In a random sample of 26- to 35-year-old cell phone users, 75 of 120 sent texts more often than they made phone calls. What is the margin of error for a 99% confidence interval for the difference between the proportions of cell phone users in these age groups who text more often than they call?
a. 0.0032
b. 0.0569
c. 0.1114
d. 0.1465
e. 0.2864

Answer 1

Answer:

$ME= 2.58 \sqrt{\frac{0.8(1-0.8)}{125} +\frac{0.625 (1-0.625)}{120}}= 0.1465$

The best option would be:

d. 0.1465

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 for brand 18-25 users

$\hat p_A =\frac{100}{125}=0.8$ represent the estimated proportion for 18-25 users

$n_A=125$ is the sample size

$p_B$ represent the real population proportion for 26-35 users

$\hat p_B =\frac{75}{120}=0.625$ represent the estimated proportion for 26-35 users

$n_B=120$ is the sample size required for Brand B

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

The population proportion have the following distribution  

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

Solution to the problem

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.  

$z_{\alpha/2}=2.58$  

And the margin of error is given by:

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

And if we replace we got:

$ME= 2.58 \sqrt{\frac{0.8(1-0.8)}{125} +\frac{0.625 (1-0.625)}{120}}= 0.1465$

The best option would be:

d. 0.1465