Probability and Inference II Lab

We are again using Sean Lahman’s Baseball Database, focusing on players’ batting statistics for players who had 502 or more plate appearances in the 1962-2019 seasons (the MLB threshold to qualify for the batting title).

Question 1

What is the mean batting average? Is the mean batting average different from .300 at the 95% confidence level?

Mean = 0.2767106 Yes, the mean batting average is different from .300 at the 95% confidence level.

mean(batting_names$BA)

## [1] 0.2767106

t.test(batting_names$BA, mu = 0.3 , alternative = "two.sided")

## 
##  One Sample t-test
## 
## data:  batting_names$BA
## t = -71.951, df = 7161, p-value < 2.2e-16
## alternative hypothesis: true mean is not equal to 0.3
## 95 percent confidence interval:
##  0.2760760 0.2773451
## sample estimates:
## mean of x 
## 0.2767106

Question 2

What is the mean number of home runs? Is the mean number of home runs greater than 15 at the 95% confidence level?

Mean = 17.72843 Yes, the mean number of home runs greater than 15 at the 95% confidence level.

mean(batting_names$HR)

## [1] 17.72843

t.test(batting_names$HR, mu = 15 , alternative = "two.sided")

## 
##  One Sample t-test
## 
## data:  batting_names$HR
## t = 21.099, df = 7161, p-value < 2.2e-16
## alternative hypothesis: true mean is not equal to 15
## 95 percent confidence interval:
##  17.47493 17.98193
## sample estimates:
## mean of x 
##  17.72843

Question 3

Many people argue that left-handed hitters perform better than right-handed hitters. One reason is that batters can see and track the ball better out of the pitcher’s hand when the pitcher and batter’s handedness does not match (i.e., left-handed hitter vs. right-handed pitcher). Since more pitchers are right-handed, left-handed hitters are more likely to enjoy this handedness advantage on a per-at-bat basis.

Say we want to evaluate this claim in the case of batting averages.

• What are the null and alternative hypotheses? Null hypothesis - Left-handed hitters do not preform better than right-handed hitters on average. Alternative hypothesis - Left-handed hitters preform better on average than right-handed hitters due to the batters view of the pitcher which is advantageous.
• Perform the appropriate statistical test to evaluate the hypotheses. Report the test statistic and p-value. (NOTE: Only use right-handed and left-handed hitters, ignore switch-hitters for the time being) Test Statistic = 6.6512, P-Value = 1
• What do you conclude from the data? The p-value in this context would lay the groundwork for the conclusion that the null hypothesis should not be rejected as the p-value demonstrates no statistical significance and the alternative hypothesis is invalid as the p-value represents no statistical significance.

t.test(batting_names$BA[which(batting_names$bats=='L')],batting_names$BA[which(batting_names$throws=='R')], alternative = "less")

## 
##  Welch Two Sample t-test
## 
## data:  batting_names$BA[which(batting_names$bats == "L")] and batting_names$BA[which(batting_names$throws == "R")]
## t = 6.6512, df = 3898.9, p-value = 1
## alternative hypothesis: true difference in means is less than 0
## 95 percent confidence interval:
##         -Inf 0.005715039
## sample estimates:
## mean of x mean of y 
## 0.2801965 0.2756148

Question 4

Many people argue that batting performance was artificially inflated by the use of steroids during the “Steroid Era” (approximately 1994-2005). Mainly, people suggest that steroids helped batters build strength to hit baseballs further, such that home run totals soared.

Say we want to test this claim about the Steroid Era and home run totals.

• What are the null and alternative hypotheses? Null hypothesis - During 1994-2005, the inflation of batting performance is not attributed to steroid use. Alternative hypothesis - The inflation of batting performance between 1994-2005 is due to steroid use.
• Perform the appropriate statistical test to evaluate the hypotheses. Report the test statistic and p-value. Test Statistic = 2.1061, P-Value = 0.01764
• What do you conclude from the data? The p-value in this context would lay groundwork for the conclusion that the null hypothesis should be rejected and the alternative hypothesis is valid as the p-value represents statistical significance.

steroidHR <- batting_names$HR[which(batting_names$yearID >= 1994 & batting_names$yearID <= 2005)]
modernHR <- batting_names$HR[which(batting_names$yearID >= 2005 & batting_names$yearID <= 2024)]
t.test(steroidHR, modernHR, alternative = 'greater')

## 
##  Welch Two Sample t-test
## 
## data:  steroidHR and modernHR
## t = 2.1061, df = 3130, p-value = 0.01764
## alternative hypothesis: true difference in means is greater than 0
## 95 percent confidence interval:
##  0.172979      Inf
## sample estimates:
## mean of x mean of y 
##  20.73477  19.94404