hal and jose are avid crossword - puzzle solvers. hal solves 48% of his crossword puzzles in under 30…

Explanation:

Step1: Identify the given proportions and sample - sizes

Let $p_H = 0.48$ (Hal's proportion), $n_H=22$ (Hal's sample - size), $p_J = 0.39$ (Jose's proportion), $n_J = 26$ (Jose's sample - size). The mean of the sampling distribution of $\hat{p}_J-\hat{p}H$ is $\mu{\hat{p}_J-\hat{p}_H}=p_J - p_H=0.39 - 0.48=- 0.09$. The standard deviation of the sampling distribution of $\hat{p}_J-\hat{p}H$ is $\sigma{\hat{p}_J - \hat{p}H}=\sqrt{\frac{p_J(1 - p_J)}{n_J}+\frac{p_H(1 - p_H)}{n_H}}$. Substitute the values: [ \begin{align*} \sigma{\hat{p}_J - \hat{p}_H}&=\sqrt{\frac{0.39\times(1 - 0.39)}{26}+\frac{0.48\times(1 - 0.48)}{22}}\ &=\sqrt{\frac{0.39\times0.61}{26}+\frac{0.48\times0.52}{22}}\ &=\sqrt{\frac{0.2379}{26}+\frac{0.2496}{22}}\ &=\sqrt{0.00915 + 0.011345}\ &=\sqrt{0.020495}\approx0.1432 \end{align*} ]

Step2: Calculate the z - score

We want to find $P(\hat{p}_J>\hat{p}_H)$, which is equivalent to $P(\hat{p}_J-\hat{p}_H>0)$. The z - score is $z=\frac{(\hat{p}_J-\hat{p}H)-\mu{\hat{p}_J-\hat{p}H}}{\sigma{\hat{p}_J - \hat{p}_H}}$. Substitute $\hat{p}_J-\hat{p}H = 0$, $\mu{\hat{p}_J-\hat{p}H}=-0.09$ and $\sigma{\hat{p}_J - \hat{p}_H}\approx0.1432$ into the z - score formula: [z=\frac{0-(-0.09)}{0.1432}=\frac{0.09}{0.1432}\approx0.63]

Step3: Find the probability using the z - table

We want $P(Z > 0.63)$. Since the total area under the standard normal curve is 1, $P(Z>0.63)=1 - P(Z\leqslant0.63)$. From the z - table, $P(Z\leqslant0.63)=0.7357$. So $P(Z > 0.63)=1 - 0.7357 = 0.2643\approx0.265$.

Answer:

0.265