Problem 11BSC

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

Loaded Die The author drilled a hole in a die and filled it with a lead weight, then proceeded to roll it 200 times. Here are the observed frequencies for the outcomes of 1, 2, 3, 4, 5, and 6, respectively: 27, 31, 42, 40, 28, 32. Use a 0.05 significance level to test the claim that the outcomes are not equally likely. Does it appear that the loaded die behaves differently than a fair die?

Answer:

Step 1 of 2</p>

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 = 1/6 |
E = np |
(O - E )2 |
(O - E)2/E |

1 |
27 |
0.166667 |
33.33333 |
40.11111 |
1.203333 |

2 |
31 |
0.166667 |
33.33333 |
5.444444 |
0.163333 |

3 |
42 |
0.166667 |
33.33333 |
75.11111 |
2.253333 |

4 |
40 |
0.166667 |
33.33333 |
44.44444 |
1.333333 |

5 |
28 |
0.166667 |
33.33333 |
28.44444 |
0.853333 |

6 |
32 |
0.166667 |
33.33333 |
1.777778 |
0.053333 |

Sum |
200 |
1 |
200 |
195.3333 |
5.86 |

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 6 since the probability of a "1" coming up is 1/6 and there were a total of 200.

Expected frequency (E) = np

= 200(1/6)

= 33.33

The calculation continues as follows. Letting E be the expected frequency of an outcome and O be the observed frequency of that outcome.