Question

In the baseball data set discussed in the last exercise, it was found that out of the 59 baseball players, 15 were left-handed. Is this odd, since the proportion of left-handed males in America is about \(11 \% ?\) Answer by using \((4.2 .7)\) to construct a \(95 \%\) approximate confidence interval for \(p\), the proportion of left-handed professional baseball players.

Short answer

Based on our analysis, the 95% confidence interval for the proportion of left-handed professional baseball players is [0.197032982, 0.311441594]. This includes the proportion of left-handed males in America (11% or 0.11), so it doesn't appear to be unusually high that 15 out of 59 baseball players are left-handed.

Step-by-step solution

Understanding the Problem

In order to solve this exercise, it's necessary to understand the problem. We have a data set with the information that 15 out of 59 baseball players are left-handed. The task is to check if this proportion is unusual, considering that the proportion of left-handed men in America is about 11%.

Calculate the Proportion (p)

The proportion \(p\) is the number of left-handed players out of the total number of players. In this case, \(p = 15 / 59 = 0.254237288\).

Calculating the standard deviation of the sample proportion

Standard deviation of the sample proportion can be calculated using the following formula: \[SE = \sqrt{ p (1 - p) / n } = \sqrt{ 0.254237288 * (1 - 0.254237288) / 59 } = 0.0608932206 \]

Construct the approximate 95% confidence interval for \(p\)

A 95% confidence interval indicates that 95 out of 100 times, the true proportion will fall within the calculated interval. Using the formula given \((4.2 .7)\), we can compute the interval as follows: \[ CI = p \pm Z*SE \] Using 0.95 for the Z value (since it's a 95% confidence interval), calculated value for \(p\) and \(SE\) from step 2 and 3, we get: \[ CI = 0.254237288 +- 0.95*0.0608932206 = [0.197032982, 0.311441594]. \]