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?

[tex]\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[/tex]

The degrees of freedom are given by:

[tex]df=categories-1=6-1=5[/tex]

And the p value would be given by:

[tex]p_v = P(\chi^2_{5} >14.38)= 0.0133[/tex]

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 [tex]\alpha=0.025[/tex]

The statistic to check the hypothesis is given by:

[tex]\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}[/tex]

The expected frequency for each case is :

[tex] E_i = \frac{200}{6}= 33.333[/tex]

The observed values are:

26, 30, 49, 42, 26, 27

And now we can calculate the statistic:

[tex]\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[/tex]

The degrees of freedom are given by:

[tex]df=categories-1=6-1=5[/tex]

And the p value would be given by:

[tex]p_v = P(\chi^2_{5} >14.38)= 0.0133[/tex]

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

fichoh fichoh · Answer 2 · 2021-12-02T06:40:56+01:00

Using the Chisquare test, the test statistic exceeds the critical value, hence, there is significant evidence to reject the Null.

Defining the hypothesis :

[tex] H_{0} : There \: is \: no \: difference \: in \: behavior [/tex]

[tex] H_{1} : There \: is \: difference \: in \: behavior [/tex]

Using Chisquare statistic :

Test statistic, χ² = [tex]\frac{Observed - Expected}{Expected} [/tex]

Expected score = Total frequency / number of outcomes = 200 / 6 = 33.333

Hence, the χ² can be defined thus :

χ² = [tex]\frac{26 - 33.33}{33.33} + \frac{30 - 33.33}{33.33} + \frac{49 - 33.33}{33.33} + \frac{42 - 33.33}{33.33} + \frac{26 - 33.33}{33.33} + \frac{27 - 33.33}{33.33} = 14.38 [/tex]

Hence, the test statistic = 14.38

The critical value :

Degree of freedom (df) = n-1 = (6 - 1) = 5

[tex]χ²_{5, 0.025} = 12.833[/tex]

Decison Region :

Reject Null if Test statistic > Critical value

Since 14.38 > 12.83 ; then we there is significant evidence to reject the Null and conclude that the loaded die behaves differently.

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