QuestionA study was conducted to determine the proportion of people who dream in black and white instead of color. Among 322 people over the age of 55,78 dream in black and white, and among 285 people under the age of 25,16 dream in black and white. Use a 0.05 significance level to test the claim that the proportion of people over 55 who dream in black and white is greater than the proportion for those under 25. Complete parts (a) through (c) below. $$H_{1}: p_{1}=p_{2}$$ D. $$\begin{array}{l} H_{0}: p_{1} \leq p_{2} \\ H_{1}: p_{1} \neq p_{2} \end{array}$$ $$H_{1}: p_{1} \neq p_{2}$$ E. $H_{0}: p_{1}=p_{2}$ $H_{1}: p_{1}>p_{2}$ $$H_{1}: p_{1} \neq p_{2}$$ F. $H_{0}: p_{1}=p_{2}$ $H_{1}: p_{1}<p_{2}$
Identify the test statistic. $$z=6.32$$ (Round to two decimal places as needed.) Identify the P -value. P -value $=0$ (Round to three decimal places as needed.) What is the conclusion based on the hypothesis test? The P -value is $\square$ less than the signticance level of $\alpha=0.05$, so $\square$ reject the null hypothesis. There is $\square$ sufficient evidence to support the claim that the proportion of people over 55 who dream in black and white is greater than the proportion for those under 25. b. Test the claim by constructing an appropriate confidence interval.
The 90\% confidence interval is $\square$ $<\left(\mathrm{p}_{1}-\mathrm{p}_{2}\right)<$ $\square$ $\square$. (Round to three decimal places as needed.) n example Get more help - Clear all Check answe