Question

ricardo and tammy practice putting golf balls. ricardo makes 47% of his putts and tammy makes 51% of her putts. suppose that ricardo attempts 25 putts and tammy attempts 30 putts. let $r =$ the proportion of putts ricardo makes and $t =$ the proportion of putts tammy makes. what is the probability that ricardo makes a higher proportion of putts than tammy?
find the z-table here.
○ 0.384
○ 0.434
○ 0.484
○ 0.616

Explanation:

Step1: Define given values

$p_R=0.47$, $n_R=25$, $p_T=0.51$, $n_T=30$

Step2: Calculate mean of $\hat{R}-\hat{T}$

$\mu_{\hat{R}-\hat{T}} = p_R - p_T = 0.47 - 0.51 = -0.04$

Step3: Calculate standard error

$$\begin{align*} SE&=\sqrt{\frac{p_R(1-p_R)}{n_R}+\frac{p_T(1-p_T)}{n_T}}\\ &=\sqrt{\frac{0.47(1-0.47)}{25}+\frac{0.51(1-0.51)}{30}}\\ &=\sqrt{\frac{0.47\times0.53}{25}+\frac{0.51\times0.49}{30}}\\ &=\sqrt{\frac{0.2491}{25}+\frac{0.2499}{30}}\\ &=\sqrt{0.009964+0.00833}\\ &=\sqrt{0.018294}\approx0.1353 \end{align*}$$

Step4: Calculate z-score for $\hat{R}>\hat{T}$

We need $P(\hat{R}-\hat{T}>0)$.

$$ z=\frac{0 - \mu_{\hat{R}-\hat{T}}}{SE}=\frac{0 - (-0.04)}{0.1353}=\frac{0.04}{0.1353}\approx0.296 $$

Step5: Find probability from z-table

$P(Z>0.30)=1-P(Z\leq0.30)=1-0.6179=0.3821\approx0.384$

Answer:

0.384