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: 2828, 46, 37, 29, 32. 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?
The test statistic is ____ .
(Round to three decimal places as needed.)
The critical value is ___ .
(Round to three decimal places as needed.)
State the conclusion.
______ (Do not reject/ Reject) H0. There ___ (is/ is not) sufficient evidence to support the claim that the outcomes are not equally likely. The outcomes _____ (appear/ do not appear) to be equally likely, so the loaded die ____ (does not appear / appears) to behave differently from a fair die.
Observed data for the given die which was rolled 200 times
X | 1 | 2 | 3 | 4 | 5 | 6 |
Observed value | 28 | 28 | 46 | 37 | 29 | 32 |
When a fair die is rolled 200 times ,the outcome will be equally likely. Hence the expected values for the given die to be fair will be 200*(1/6)=33.33
X | 1 | 2 | 3 | 4 | 5 | 6 |
Expected value | 33.33 | 33.33 | 33.33 | 33.33 | 33.33 | 33.33 |
The level of significance() given is 0.025 or 2.5% which is the probability of rejecting the Null Hypothesis (Ho).
Here,
H0: The outcome of given die is equally likely
H1: The outcome of given die is not equally likely
To test the above hypothesis we would do the square test as the square test is based on comparing the frequencies actually observed with the frequencies expected using the test statistic:
Therefore our test statistic will be:
Solving the above we get the test statistic= 7.540756 i.e approx 7.541.
In chi square table, checking for the value with degree of freedom (v) =5 i.e. (6-1) and probability 0.025 we get the critical value to be 12.83.
This can be presented as:
Since our test statistic is less than the critical value we do not have sufficient evidence to reject our Null Hypothesis Ho.
Therefore the conclusion is:
Do not reject Ho.There is not sufficient evidence to claim that the outcome are not equally likely. The outcomes appear to be equally likely.So the loaded die does not appear to behave differently from a fair die.
Hope I was able to solve your question.
Conduct the hypothesis test and provide the test statistic and the critical value, and state the...
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