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: 26​, 30​, 49​, 42​, 26​, 27. 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?

Answer 1

Answer:

$\chi^2 = \frac{(26-33.33)^2}{33.33}+\frac{(30-33.33)^2}{33.33}+\frac{(49-33.33)^2}{33.33}+\frac{(42-33.33)^2}{33.33}+\frac{(42-33.33)^2}{33.33}+\frac{(26-33.33)^2}{33.33}+\frac{(27-33.33)^2}{33.33}=14.38$

The degrees of freedom are given by:

$df=categories-1=6-1=5$

And the p value would be given by:

$p_v = P(\chi^2_{5} >14.38)= 0.0133$

Since the p value is lower than the significance level provided we have enough evidence to reject the null hypothesis and we have enough evidence to conclude that  the loaded die behaves differently than a fair​ die

Step-by-step explanation:

System of hypothesis

H0: There is no difference in the outcomes frequency

H1: There is a difference in the outcomes frequency

The level of significance assumed for this case is $\alpha=0.025$

The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

The expected frequency for each case is :

$E_i = \frac{200}{6}= 33.333$

The observed values are:

26​, 30​, 49​, 42​, 26​, 27

And now we can calculate the statistic:

$\chi^2 = \frac{(26-33.33)^2}{33.33}+\frac{(30-33.33)^2}{33.33}+\frac{(49-33.33)^2}{33.33}+\frac{(42-33.33)^2}{33.33}+\frac{(42-33.33)^2}{33.33}+\frac{(26-33.33)^2}{33.33}+\frac{(27-33.33)^2}{33.33}=14.38$

The degrees of freedom are given by:

$df=categories-1=6-1=5$

And the p value would be given by:

$p_v = P(\chi^2_{5} >14.38)= 0.0133$

Since the p value is lower than the significance level provided we have enough evidence to reject the null hypothesis and we have enough evidence to conclude that  the loaded die behaves differently than a fair​ die