Problem 10BSC

In Exercise, conduct the hypothesis test and provide the test statistic, critical value, and/or P-value, and state the conclusion.

Baseball Player Births In his book Outliers, author Malcolm Gladwell argues that more baseball players have birthdates in the months immediately following July 31, because that was the cutoff date for non school baseball leagues. Here is a sample of frequency counts of months of birthdates of American-born major league baseball players starting with January: 387, 329, 366, 344, 336, 313, 313, 503, 421, 434, 398, 371. Using a 0.05 significance level, is there sufficient evidence to warrant rejection of the claim that American-born major league baseball players are born in different months with the same frequency? Do the sample values appear to support Gladwell’s claim?

Answer:

Step 1 of 2

The expected values for each category, Ei , could be determined. With these observed and expected numbers of cases, the hypotheses can be written as

H0 : O = E

H1 : O E

Use a 0.05 significance level

SL. No. |
O |
p = 1/12 |
E = np |
(O - E )2 |
(O - E)2/E |

1 |
387 |
0.083333 |
376.25 |
115.5625 |
0.307143 |

2 |
329 |
0.083333 |
376.25 |
2232.563 |
5.933721 |

3 |
366 |
0.083333 |
376.25 |
105.0625 |
0.279236 |

4 |
344 |
0.083333 |
376.25 |
1040.063 |
2.764286 |

5 |
336 |
0.083333 |
376.25 |
1620.063 |
4.305814 |

6 |
313 |
0.083333 |
376.25 |
4000.563 |
10.63272 |

7 |
313 |
0.083333 |
376.25 |
4000.563 |
10.63272 |

8 |
503 |
0.083333 |
376.25 |
16065.56 |
42.69917 |

9 |
421 |
0.083333 |
376.25 |
2002.563 |
5.322425 |

10 |
434 |
0.083333 |
376.25 |
3335.063 |
8.863953 |

11 |
398 |
0.083333 |
376.25 |
473.0625 |
1.257309 |

12 |
371 |
0.083333 |
376.25 |
27.5625 |
0.073256 |

Sum |
4515 |
1 |
4515 |
35018.25 |
93.07176 |

The first step in conducting the significance test is to compute the expected frequency for each outcome given that the null hypothesis is true. For example, the expected frequency of a "1" is 12 since the probability of a "1" coming up is 1/12 and there were a total of 4515.

Expected frequency (E) = np

= (4515)(1/12)