Question

6.127 Team Batting Average in Baseball The dataset BaseballHits gives 2014 season statistics for all Major League Baseball teams. We treat this as a sample of all MLB teams in all years. Computer output of descriptive statistics for the variable giving the batting average is shown: Descriptive Statistics: BattingAvg \(\begin{array}{rrrr}\mathrm{N} & \text { Mean } & \text { SE Mean } & \text { StDev } \\ 30 & 0.25110 & 0.00200 & 0.01096\end{array}\) Variable BattingAvg \(\begin{array}{rrrrr}\text { Minimum } & \text { Q1 } & \text { Median } & \text { Q3 } & \text { Maximum } \\ 0.22600 & 0.24350 & 0.25300 & 0.25675 & 0.27700\end{array}\) (a) How many teams are included in the dataset? What is the mean batting average? What is the standard deviation? (b) Use the descriptive statistics above to conduct a hypothesis test to determine whether there is evidence that average team batting average is different from 0.260 . Show all details of the test. (c) Compare the test statistic and p-value you found in part (b) to the computer output below for the same data: One-Sample T: BattingAvg Test of \(m u=0.26\) vs not \(=0.26\) \(\begin{array}{lrrr}\text { Variable } & \text { N } & \text { Mean } & \text { StDev } \\ \text { BattingAvg } & 30 & 0.25110 & 0.01096 \\ \text { SE Mean } & 95 \% \text { Cl } & \text { T } & \text { P } \\ 0.00200 & \\{0.24701,0.25519) & -4.45 & 0.000\end{array}\)

Short answer

The dataset includes statistics from 30 teams. The mean batting average is \(0.25110\) while the standard deviation is \(0.01096\). By conducting a hypothesis test, it was found that the mean batting average was significantly different from \(0.260\) with a t-value of \(-4.475372208025383\) and a p-value lower than \(0.05\), as seen from both manually conducted analysis and the computer output presented.

Step-by-step solution

Interpret Descriptive Statistics

The descriptive statistics table provides us with various pieces of information, including number of teams (\(N=30\)), mean batting average (\(0.25110\)), standard deviation (\(0.01096\)), minimum batting average (\(0.22600\)), first quartile (\(0.24350\)), median (\(0.25300\)), third quartile (\(0.25675\)), and maximum batting average (\(0.27700\)).

Conduct Hypothesis Test

We conduct a hypothesis test to determine whether there is evidence that average team batting average differs from \(0.260\). The null hypothesis is that the mean team batting average is equal to \(0.260\) (\(H0: \mu = 0.260\)), and the alternative hypothesis is that it differs (\(H1: \mu ≠ 0.260\)). The test statistic is calculated using the formula \(t = \frac{ (\bar{x}- \mu) }{ \frac{s}{\sqrt{n}} }\), where \(\bar{x}\) = sample mean, \(\mu\) = population mean, \(s\) = standard deviation, and \(n\) = number of observations. Inserting the values, the test statistic \(t = \frac{(0.25110 - 0.260)}{0.01096/\sqrt{30}} = -4.475372208025383\). The exact p-value cannot be calculated without additional statistical tables or software, but we know it will be very small since the magnitude of the t-score is quite large.

Compare to Computer Output

Comparing our calculated t-value with the computer output, we can see that our manually calculated t-value matches the one calculated by the computer (\(-4.475372208025383\) approx to \(-4.45\)). The p-value reported in the computer output is \(0.000\), providing strong evidence against the null hypothesis. This indicates that the mean batting average is significantly different from \(0.260\) with a confidence level of \(95\%.