Question

calculating probabilities for sampling distribution
quiz complete
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in a large local high school, 19% of freshmen have had their wisdom teeth removed and 24% of seniors have had their wisdom teeth removed. suppose that a random sample of 60 freshmen and 50 seniors is selected. let f = the proportion of freshmen who have had their wisdom teeth removed and s = the proportion of seniors who have had their wisdom teeth removed. what is the probability that the proportion of freshmen who have had their wisdom teeth removed is greater than the proportion of seniors?
find the z-table here.
○ 0.139
○ 0.263
○ 0.287
○ 0.737

Explanation:

Step1: Define given values

$p_F=0.19$, $n_F=60$, $p_S=0.24$, $n_S=50$
We need $P(F > S) = P(F - S > 0)$

Step2: Calculate mean of $F-S$

$\mu_{F-S} = p_F - p_S = 0.19 - 0.24 = -0.05$

Step3: Calculate standard error of $F-S$

$$\begin{align*} \sigma_{F-S}&=\sqrt{\frac{p_F(1-p_F)}{n_F}+\frac{p_S(1-p_S)}{n_S}}\\ &=\sqrt{\frac{0.19(1-0.19)}{60}+\frac{0.24(1-0.24)}{50}}\\ &=\sqrt{\frac{0.19\times0.81}{60}+\frac{0.24\times0.76}{50}}\\ &=\sqrt{\frac{0.1539}{60}+\frac{0.1824}{50}}\\ &=\sqrt{0.002565+0.003648}\\ &=\sqrt{0.006213}\approx0.0788 \end{align*}$$

Step4: Calculate z-score

$z=\frac{0 - \mu_{F-S}}{\sigma_{F-S}}=\frac{0 - (-0.05)}{0.0788}\approx0.6345$

Step5: Find corresponding probability

We need $P(Z > 0.6345) = 1 - P(Z \leq 0.6345)$. From z-table, $P(Z \leq 0.63)\approx0.7357$, $P(Z \leq 0.64)\approx0.7389$. Interpolating, $P(Z \leq 0.6345)\approx0.737$. So $P(Z > 0.6345)=1-0.737=0.263$

Answer:

0.263