e. The p-value is > 0.1, little or no support for the notion that Woods plays better on par 3 holes.
Step-by-step explanation:
Hypothesis by the sports writer
- On the par 3 holes, Tiger Woods made a birdie in 20 out of 80 attempts.
- On the par 4 holes, he made a birdie in 40 out of 200 attempts.
Null hypothesis: P
3 <= P
4
Alternative hypothesis: P
3 > P
4
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the proportion of birdies on par 3 holes (p
3
) is sufficiently greater than the proportion of birdies on par 4 holes (p
4
)
- Using sample data, we calculate the pooled sample proportion (p) and the standard error (SE). Using those measures, we compute the z-score test statistic (z).
p = (p3 * n
3 + p4 * n4) / (n3 + n) = [(0.25 * 80) + (0.20 * 200)] / (80 + 200) = 50/280 = 0.214
SE = sqrt{ p * ( 1 - p ) * [ (1/n3) + (1/n4) ] }
SE = sqrt [ 0.214 * 0.786 * ( 1/80 + 1/200 ) ]
= sqrt[ 0.214 * 0.786 * 0.0175 }
= sqrt [0.0029548] = 0.0544
z = (p3 - p4) / SE = (0.25 - 0.20)/0.0544 = 0.92
where p3 is the sample proportion of birdies on par 3,
p4 is the sample proportion of birdies on par 4,
n3 is the number of par 3 holes,
and n4 is the number of par 4 holes.
- Since we have a one-tailed test, the P-value is the probability that the z-score is greater than 0.92. We use the Normal Distribution Calculator to find P(z > 0.92) = 0.18. Thus, the P-value = 0.18.
- Since the P-value (0.18) is greater than 0.10, we have little support for the notion that Woods plays better on par 3 holes. In short, we cannot reject the null hypothesis
