Question

A magazine reports that women trust recommendations from a particular social networking site more than recommendations from any other social network platform. But does trust in this social networking site differ by gender? The following sample data show the number of women and men who stated in a recent sample that they trust recommendations made on this particular social networking site.
Women Men
Sample 150 170
Trust Recommendations Made on the social networking site 123 102

a. What is the point estimate of the proportion of women who trust recommendations made on this particular social networking site?
b. What is the point estimate of the proportion of men who trust recommendations made on this particular social networking site?
c. Provide a 95% confidence interval estimate of the difference between the proportion of women and men who trust recommendations made on this particular social networking site.

Answer 1

Answer:

a) $p_A$ represent the real population proportion for women

$\hat p_A =\frac{123}{150}=0.82$ represent the estimated proportion for women

$n_A=150$ is the sample size selected for women

b) $p_B$ represent the real population proportion for male

$\hat p_B =\frac{102}{170}=0.6$ represent the estimated proportion for male

$n_B=170$ is the sample size selected for male

c) $(0.82-0.6) - 1.96 \sqrt{\frac{0.82(1-0.82)}{150} +\frac{0.6(1-0.6)}{170}}=0.124$  

$(0.82-0.6) + 1.96 \sqrt{\frac{0.82(1-0.82)}{150} +\frac{0.6(1-0.6)}{170}}=0.316$  

And the 95% confidence interval would be given (0.124;0.316).  

We are confident at 95% that the difference between the two proportions is between $0.124 \leq p_A -p_B \leq 0.316$

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".  

Part a

$p_A$ represent the real population proportion for women

$\hat p_A =\frac{123}{150}=0.82$ represent the estimated proportion for women

$n_A=150$ is the sample size selected for women

Part b

$p_B$ represent the real population proportion for male

$\hat p_B =\frac{102}{170}=0.6$ represent the estimated proportion for male

$n_B=170$ is the sample size elected for male

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

Part c

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 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$, with that value we can find the quantile required for the interval in the normal standard distribution.  

$z_{\alpha/2}=1.96$  

And replacing into the confidence interval formula we got:  

$(0.82-0.6) - 1.96 \sqrt{\frac{0.82(1-0.82)}{150} +\frac{0.6(1-0.6)}{170}}=0.124$  

$(0.82-0.6) + 1.96 \sqrt{\frac{0.82(1-0.82)}{150} +\frac{0.6(1-0.6)}{170}}=0.316$  

And the 95% confidence interval would be given (0.124;0.316).  

We are confident at 95% that the difference between the two proportions is between $0.124 \leq p_A -p_B \leq 0.316$