Using the z-distribution, as we are working with a proportion, it is found that the 95% confidence interval for the difference in the proportions of teenagers with access to smartphones in 2017 and in 2013 is of (0.393, 0.507).
In 2017, 450 out of 600 students had access, hence:
[tex]p_1 = \frac{450}{600} = 0.75, s_1 = \sqrt{\frac{0.75(0.25)}{600}} = 0.0177[/tex]
In 2013, 120 out of 400 students had access, hence:
[tex]p_2 = \frac{120}{400} = 0.3, s_2 = \sqrt{\frac{0.3(0.7)}{400}} = 0.0229[/tex]
For the distribution of differences, we have that the mean and the standard error are, respectively, given by:
[tex]p = p_1 - p_2 = 0.75 - 0.3 = 0.45[/tex]
[tex]s = \sqrt{s_1^2 + s_2^2} = \sqrt{0.0177^2 + 0.0229^2} = 0.0289[/tex]
It is given by:
[tex]p \pm zs[/tex]
95% confidence level, hence[tex]\alpha = 0.95[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.95}{2} = 0.975[/tex], so [tex]z = 1.96[/tex].
Then:
[tex]p - zs = 0.45 - 1.96(0.0289) = 0.393[/tex]
[tex]p + zs = 0.45 + 1.96(0.0289) = 0.507[/tex]
The 95% confidence interval for the difference in the proportions of teenagers with access to smartphones in 2017 and in 2013 is of (0.393, 0.507).
More can be learned about the z-distribution at https://brainly.com/question/25890103
