Question In a recent survey of 200 elementary students (80 girls and 120 boys), 60 of the 80 girls surveyed and 80 of the 120 boys surveyed declared to prefer Math to English. A. What is the difference between the proportion of girls and the proportion of boys who said they prefer Math to English? ["0.0042", "0.0833", "0.4097", "0.5"] B. What is the standard error of the difference between the proportion of girls and the proportion of boys who said they prefer Math to English? ["0.0833", "0.0648", "0.0042", "0.4097"] C. What is an 95% confidence interval for the difference between the two proportions? (0.0288 , 0.1378)", "(-0.0437 , 0.2103)", "(0.0305 , 0.1362)", "(0.0798 , 0.0869)"] D Suppose that the 90% confidence interval for the difference between the two proportions is (-0.0233, 0.1900). Is the following interpretation of the confidence interval true? "For all elementary students, I have 90% confidence that the true difference between the proportion of girls and boys who prefer Math to English is in the interval (-0.0233, 0.1900)." ["True", "False"] E. Based on this data, can we make the conclusion that, among elementary students, girls like math more than boys do? ["False", "True"]

Pi: Proportion of Boys Prefered Math to EnglishPx: Proportion of qirls Prefered Math to EnalishWe haxe x_{1}=80, n_{1}=120, x_{2}=60, n_{2}=80\hat{p}_{2}=\frac{x_{1}}{n_{1}}=\frac{80}{120}=0.6667, \hat{\rho}_{7}=\frac{x_{2}}{n_{2}}=\frac{60}{80}=0.75A). \quad \hat{p}_{1}-\hat{p}_{2}=0.75-0.6667=0.0833B). stundard frrrs ( S \cdot E )\begin{array}{c}S \cdot E=\sqrt{\frac{\hat{P}_{1}\left(1-\hat{P}_{1}\right)}{n_{1}}+\frac{\hat{P}_{2}\left(1-\hat{P}_{2}\right)}{n_{L}}}=\sqrt{\frac{0.75(1-0.75)}{80}+\frac{0.6667(1-0.6667)}{120}} \\S \cdot E=0.0648\end{array}c) 95 \% confidence interval, \alpha=0.05\begin{array}{l}z_{c}=z_{1-\frac{\alpha}{2}}=z_{1}-\frac{0.05}{2}=z_{0.975}=1.96 \\\left(\hat{p}_{1}-\hat{p}_{2}-Z_{c} S \cdot E, \hat{p}_{1}-\hat{p}_{2}+z_{c} S \cdot E\right) \\(0.75-0.6667-1.96 * 0.0648,0.75-0.6667+1.96 * 0.0648) \\(-0.0437,0.2103)\end{array}D). - TeveIf 90 \% confidence interval is q'xencs (-0.0233,0.1900) Then inteipietasion is For all elementary students I hare 90 \% confidonce that the teve differance beteaen the Moportion of gitls is boy's Who Prefere math to Enqlish is in interval (-0.0233,0.1906)E). Falsefince confidence interul contuin value ' o ' it mears We ae 95 \%-contident that Tire differance x p_{1}-8 p_{2} Will be 'o' i.e p_{1}-p_{2}=0 \Rightarrow p_{1}=p_{2}\therefore We can't conclude that giels Jike math mae than boy's do ...