Question

Conduct the hypothesis test and provide the test statistic and the critical​ value, and state the conclusion. A person 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: 25​, 27​, 45​, 40​, 26​, 37. Use a 0.025 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?

a) The test statistic is

b) The critical value is

c) State the conclusion.

(1)____ Upper H0. There (2)_____ sufficient evidence to support the claim that the outcomes are not equally likely. The outcomes (3)_____ to be equally​ likely, so the loaded die (4)____ to behave differently from a fair die.

Answer 1

a. The test statistics is 10.72

b. The critical value is 12.8333

c. We do not reject the null hypothesis and do not have sufficient evidence to conclude that loaded die behaves differently than a fair​ die.

d. There is not sufficient evidence to support the claim that the outcomes are not equally likely. The outcomes are to be equally​ likely, so the loaded die do not behave differently from a fair die.

Categories	Observed	Expected	(fo-fe)2/fe
1	25	200*0.1666667=33.333	(25-33.333)2/33.333 = 2.083
2	27	200*0.1666667=33.333	(27-33.333)2/33.333 = 1.203
3	45	200*0.1666667=33.333	(45-33.333)2/33.333 = 4.083
4	40	200*0.1666667=33.333	(40-33.333)2/33.333 = 1.333
5	26	200*0.1666667=33.333	(26-33.333)2/33.333 = 1.613
6	37	200*0.1666667=33.333	(37-33.333)2/33.333 = 0.403
Sum =	200	200	10.72

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Null hypothesis

The outcomes are equally likely

Alternative Hypothesis

Ha​: Some of the population proportions differ from the values stated in the null hypothesis

This corresponds to a Chi-Square test for Goodness of Fit.

(2) Rejection Region

The significance level is α=0.025, the number of degrees of freedom is df = 6 - 1 = 5 so then the rejection region for this test is .

(3) Test Statistics

The Chi-Squared statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that , it is then concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is NOT enough evidence to claim that some of the population proportions differ from those stated in the null hypothesis, at the α=0.025 significance level.

Therefore, there is NOT enough evidence to claim that the loaded die behaves differently than a fair​ die.