Question

In: Statistics and Probability

Conduct the hypothesis test and provide the test​ statistic, critical value and​ P-value, and state the...

Conduct the hypothesis test and provide the test​ statistic, critical value and​ P-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

200200

times. Here are the observed frequencies for the outcomes of​ 1, 2,​ 3, 4,​ 5, and​ 6, respectively:

2626​,

3030​,

4848​,

4141​,

2828​,

2727.

Use a

0.010.01

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?Click here to view the chi-square distribution table.

LOADING...

The test statistic is

nothing .

​(Round to three decimal places as​ needed.)

Solutions

Expert Solution

Given Die is rolled for 200 times

the probabilities of getting the number 1 to 6 is equally likely which is equal to 1/6 which mean the expected frequency for each number = 200/6 = 33.3

Observed frequencies Oi Expected frequencies Ei (Oi - Ei)^2 ((Oi - Ei)^2)/Ei
26.0 33.3 53.8 1.6
30.0 33.3 11.1 0.3
48.0 33.3 215.1 6.5
41.0 33.3 58.8 1.8
28.0 33.3 28.4 0.9
27.0 33.3 40.1 1.2

the test statistic

= 1.6 + 0.3 + 6.5 +1.8 + 0.9 + 1.2 = 12.220

degrees of freedom = number of outcomes -1 = 6 - 1 = 5

given level of significance = 0.01

Hypothesis

Null Hypothesis: the outcomes are equally likely.

Alternative Hypothesis : the outcomes are not equally likely.

critical value = 15.09

Since critical value is greater than test statistic we fail to reject null hypothesis and conclude that there is no reason to believe that the outcomes are not equally likely, which means the outcomes are equally likely, the loaded die does not behaves differently than a fair​ die


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