Solved Examples: Inferential Statistics: Proportion

70% confidence in a 70% confidence interval means that if 100 different confidence intervals are constructed, each based on a different sample of size, n from the same population, N; then it is expected that 70 of the intervals will include the parameter.

$ x = 5 \\[3ex] n = 30 \\[3ex] \hat{p} = \dfrac{x}{n} \\[5ex] \hat{p} = \dfrac{5}{30} \\[5ex] \hat{p} = \dfrac{1}{6} $

$ x = 294 \\[3ex] n = 411 \\[3ex] \hat{p} = \dfrac{x}{n} = \dfrac{294}{411} = 0.7153284672 \\[5ex] \hat{q} = 1 - \hat{p} \\[3ex] \hat{q} = 1 - 0.715328 = 0.2846715328 \\[3ex] For\;\;95\%\;CL,\;\; z_{\dfrac{\alpha}{2}} = 1.96 \\[5ex] E = z_{\dfrac{\alpha}{2}} * \sqrt{\dfrac{\hat{p} * \hat{q}}{n}} \\[5ex] E = 1.96 * \sqrt{\dfrac{0.7153284672(0.2846715328)}{411}} \\[5ex] E = 0.0436274606 \\[3ex] \hat{p} - E = 0.7153284672 - 0.0436274606 = 0.6717010066 \approx 0.672 \\[3ex] \hat{p} + E = 0.7153284672 + 0.0436274606 = 0.7589559278 \approx 0.759 \\[3ex] 0.672 \lt p \lt 0.759 \\[3ex] $ Naija Polls is 95% confident that the population percentage of Nigerians that are suffering is between 67.2% and 75.9%.

$ (a.) \\[3ex] LCL = 0.229 \\[3ex] UCL = 0.691 \\[3ex] n = 1000 \\[3ex] \hat{p} = \dfrac{UCL + LCL}{2} \\[5ex] \hat{p} = \dfrac{0.691 + 0.229}{2} = \dfrac{0.92}{2} = 0.46 \\[5ex] (b.) \\[3ex] E = \dfrac{UCL - LCL}{2} \\[5ex] E = \dfrac{0.691 - 0.229}{2} = \dfrac{0.462}{2} = 0.231 \\[5ex] (c.) \\[3ex] x = \hat{p} * n \\[3ex] x = 0.46 * 1000 \\[3ex] x = 460 $

$ n = \dfrac{0.25 * \left(z_{\dfrac{\alpha}{2}}\right)^2}{E^2} \\[7ex] n \propto \dfrac{1}{E^2} \\[5ex] n = \dfrac{k}{E^2} ...k\;\;is\;\;the\;\;constant\;\;of\;\;proportion \\[5ex] k = n * E^2 \\[3ex] \implies \\[3ex] k = n_1 * E_1^2 = n_2 * E_2^2 \\[3ex] \implies \\[3ex] n_1 * E_1^2 = n_2 * E_2^2 \\[3ex] n_2 * E_2^2 = n_1 * E_1^2 \\[3ex] But\;\;E_2 = \dfrac{E_1}{5} \\[5ex] E_2^2 = \left(\dfrac{E_1}{5}\right)^2 \\[5ex] E_2^2 = \dfrac{E_1^2}{5^2} \\[5ex] E_2^2 = \dfrac{E_1^2}{25} \\[5ex] What\;\;is\;\;n_2\;\;in\;\;terms;\;of\;\;n_1? \\[3ex] \implies \\[3ex] n_2 * \dfrac{E_1^2}{25} = n_1 * E_1^2 \\[5ex] n_2 = n_1 * E_1^2 * \dfrac{25}{E_1^2} \\[5ex] n_2 = n_1 * 25 \\[3ex] n_2 = 25 * n_1 \\[3ex] $ The sample size has to be increased by a factor of 25 to decrease the margin of error by a factor of $\dfrac{1}{5}$

$ x = 2325 \\[3ex] n = 2500 \\[3ex] \hat{p} = \dfrac{x}{n} = \dfrac{2325}{2500} = 0.93 = 93\% \\[5ex] $ We estimate that 93% of Nigerians strongly believe their governors and congressmen do not care for the poor.

$ (a.) \\[3ex] x = 323 \\[3ex] n = 380 \\[3ex] \hat{p} = \dfrac{x}{n} = \dfrac{323}{380} = 0.85 \\[5ex] \hat{q} = 1 - \hat{p} = 1 - 0.85 = 0.15 \\[3ex] For\;\;99\%\;CL;\;\; z_{\dfrac{\alpha}{2}} = 2.576 \\[5ex] E = z_{\dfrac{\alpha}{2}} * \sqrt{\dfrac{\hat{p} * \hat{q}}{n}} \\[5ex] E = 2.576 * \sqrt{\dfrac{0.85 * 0.15}{380}} \\[5ex] E = 0.0471855643 \\[3ex] \hat{p} - E = 0.85 - 0.0471855643 = 0.8028144357 \approx 0.803 \\[3ex] \hat{p} + E = 0.85 + 0.0471855643 = 0.8971855643 \approx 0.897 \\[3ex] 0.803 \lt p \lt 0.897 \\[3ex] (b.) \\[3ex] \hat{p} - E \gt 0.5 \\[3ex] \hat{p} + E \gt 0.5 \\[3ex] $ Hence, the method appear to be effective because the proportion of girls in both cases is greater than 0.5 (50%).

It is within the margin of error for Nahum Joel to receive between 45% and 51% of the popular vote.
It is also within the margin of error for Micah Haggai to receive between 43% and 49% of the popular vote.
The two confidence intervals overlap.
Hence, the poll cannot predict the winner. The race is too close to call.

(a.) There is 95% confidence that the population proportion favoring the death penalty is between 0.577315 and 0.619930

(b.) ​Yes, it is a plausible claim because the confidence interval contains values above​ 50%.

(a.) With​ 95% confidence, the population proportion of people who think marriage is becoming obsolete is between 0.35214 and 0.39417

(b.) No, it is not a plausible claim because the confidence interval does not contain values above​ 50%.

$ (a.) \\[3ex] x = 724 \\[3ex] n = 1067 \\[3ex] \hat{p} = \dfrac{x}{n} \\[5ex] \hat{p} = \dfrac{724}{1067} \\[5ex] \hat{p} = 0.6785379569 \\[3ex] $ The proportion of the randomly selected adults who favor stricter gun​ laws is 0.6785379569 or 67.85379569%

(b.) Verify the conditions for using the Central Limit Theorem to find a confidence interval
(I.) Condition 1: Simple Random Sample.
724 out of 1067 randomly selected adults... verified

(II.) Condition 2: Independent Trials.
We do not know the population size.
The Independent Trials condition can reasonably be assumed to hold.

(III.) Condition 3: Large Sample Size

$ n = 1067 \\[3ex] \hat{p} = 0.6785379569 \\[3ex] n * \hat{p} = 1067(0.6785379569) = 724 \ge 10 \\[3ex] \hat{q} = 1 - \hat{p} \\[3ex] \hat{q} = 1 - 0.6785379569 \\[3ex] \hat{q} = 0.3214620431 \\[3ex] n * \hat{q} = 1067(0.3214620431) = 343 \ge 10 \\[3ex] $ There are at least ten successes and at least ten failures... verified

$ (c.) \\[3ex] For\;\;95\%\;CL,\;\; z_{\dfrac{\alpha}{2}} = 1.96 \\[5ex] E = z_{\dfrac{\alpha}{2}} * \sqrt{\dfrac{\hat{p} * \hat{q}}{n}} \\[5ex] E = 1.96 * \sqrt{\dfrac{0.6785379569(0.3214620431)}{1067}} \\[5ex] E = 0.02802372 \\[3ex] \hat{p} - E = 0.6785379569 - 0.02802372 = 0.6505142369 \approx 0.651 \gt 0.5 \\[3ex] \hat{p} + E = 0.6785379569 + 0.02802372 = 0.7065616769 \approx 0.707 \gt 0.5 \\[3ex] 0.651 \lt p \lt 0.707 \\[3ex] $ The​ 95% confidence interval for the population proportion who favor stricter gun laws is (0.651, 0.707)

(d.) Because the confidence interval only includes values greater than​ 50%, a majority of adults in the country favor stricter gun​ laws.