Question

In 2011 and 2015, the study was held to determine the proportion of people who read books. 948 people of 1200 said they read at least one book in the last 3 months in 2011. 1080 people of 1500 said they read at least one book in the last 3 months in 2015. Find the 95% confidence interval for the differance in proportions.

Answer 1

Answer:

$(0.79-0.72) - 1.96 \sqrt{\frac{0.72(1-0.72)}{1200} +\frac{0.79(1-0.79)}{1500}}=0.0373$  

$(0.79-0.72) + 1.96 \sqrt{\frac{0.72(1-0.72)}{1200} +\frac{0.79(1-0.79)}{1500}}=0.103$  

And the 95% confidence interval would be given (0.0373;0.103).  

We are confident at 95% that the difference between the two proportions is between $0.0373 \leq p_A -p_B \leq 0.103$

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 A  

$\hat p_A =\frac{948}{1200}=0.79$ represent the estimated proportion for  A

$n_A=1200$ is the sample size required for  A

$p_B$ represent the real population proportion for B  

$\hat p_B =\frac{1080}{1500}=0.72$ represent the estimated proportion for B

$n_B=1500$ is the sample size required for 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 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:  

$(0.79-0.72) - 1.96 \sqrt{\frac{0.72(1-0.72)}{1200} +\frac{0.79(1-0.79)}{1500}}=0.0373$  

$(0.79-0.72) + 1.96 \sqrt{\frac{0.72(1-0.72)}{1200} +\frac{0.79(1-0.79)}{1500}}=0.103$  

And the 95% confidence interval would be given (0.0373;0.103).  

We are confident at 95% that the difference between the two proportions is between $0.0373 \leq p_A -p_B \leq 0.103$