A sports writer hypothesized that Tiger Woods plays better on par 3 holes than on par 4 holes. He reviewed Woods' performance in a random sample of golf tournaments. On the par 3 holes, 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. How would you interpret this result?

a. The p-value is < 0.001, very strong evidence that Woods plays better on par 3 holes.
b. The p-value is between 0.001 and 0.01, strong evidence that Woods plays better on par 3 holes.
c. The p-value is between 0.01 and 0.05, moderate evidence that Woods plays better on par 3 holes.
d. The p-value is between 0.05 and 0.1, some evidence that Woods plays better on par 3 holes.
e. The p-value is > 0.1, little or no support for the notion that Woods plays better on par 3 holes.

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

9 x 34 = (10 – 1 ) x 34 = 340 – 34 = 11 x 34