Respuesta :
Answer:
[tex] (0.79-0.72) -1.96 \sqrt{\frac{0.79(1-0.79)}{1200} +\frac{0.72(1-0.72)}{1500}} =0.0376[/tex]
[tex] (0.79-0.72) +1.96 \sqrt{\frac{0.79(1-0.79)}{1200} +\frac{0.72(1-0.72)}{1500}} =0.1024[/tex]
And the 95% confidence interval for the difference of the two proportions is given by:
[tex] 0.0376 \leq p_1 -p_2 \leq 0.1024[/tex]
Step-by-step explanation:
For this case we have the following info given:
[tex] X_1 = 948[/tex] number of people that they read at least one book in the last 3 months in 2011
[tex]n_1 = 1200[/tex] the sample size selected for 2011
[tex] X_2 = 1080[/tex] number of people that they read at least one book in the last 3 months in 2015
[tex]n_2 = 1500[/tex] the sample size selected for 2015
The estimated proportions people that they read at least one book in the last 3 months for each year are given by:
[tex]\hat p_1 = \frac{948}{1200}= 0.79[/tex]
[tex]\hat p_2 = \frac{1080}{1500}= 0.72[/tex]
And the confidence interval for the true difference of proportions is given by:
[tex](\hat p_1 -\hat p_2) \pm z_{\alpha/2} \sqrt{\frac{\hat p_1 (1-\hat p_1)}{n_1} +\frac{\hat p_2 (1-\hat p_2)}{n_2}}[/tex]
The confidence level is 95% so then the significance is 0.05 or 5% and [tex]\alpha/2 =0.025[/tex] and the critical value for this case using the normal standard distribution is:
[tex]z_{\alpha/2}=\pm 1.96[/tex]
And replacing into the confidence interval formula we got:
[tex] (0.79-0.72) -1.96 \sqrt{\frac{0.79(1-0.79)}{1200} +\frac{0.72(1-0.72)}{1500}} =0.0376[/tex]
[tex] (0.79-0.72) +1.96 \sqrt{\frac{0.79(1-0.79)}{1200} +\frac{0.72(1-0.72)}{1500}} =0.1024[/tex]
And the 95% confidence interval for the difference of the two proportions is given by:
[tex] 0.0376 \leq p_1 -p_2 \leq 0.1024[/tex]
