11.6 **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, 31, 47, 42, 26, 29. Use a 0.10 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.](#) ### Statistical Analysis Example**The test statistic is** [ ]. *(Round to three decimal places as needed.)***The critical value is** [ ]. *(Round to three decimal places as needed.)***The P-value is** [ ]. *(Round to four decimal places as needed.)*### State the Conclusion\[ \square \] \(H_0\). There \[ \triangle \] sufficient evidence to support the claim that the outcomes are not equally likely. The outcomes \[ \diamond \] to be equally likely, so the loaded die \[ \pentagon \] to behave differently from a fair die.### Instructions:1. **Test Statistic**: Input the calculated test statistic value and round it to three decimal places.2. **Critical Value**: Enter the critical value for the test, also rounded to three decimal places.3. **P-Value**: Provide the P-value calculated from the test and round it to four decimal places.### Conclusion Options:- Choose whether to reject or fail to reject \(H_0\).- State whether there is sufficient or insufficient evidence.- Indicate whether the outcomes seem or do not seem to be equally likely.- Decide whether the loaded die appears or does not appear to behave differently from a fair die.This framework aids in conducting hypothesis tests to draw conclusions about statistical data distributions and their fairness.

Answer Given The level of significance = 0.10degree of freedom = n-1 = 6-1 = 5Total sample size [N] = 200Expected value:E = N*pi200* 1/6 = 33.3333200* 1/6 = 33.3333200* 1/6 = 33.3333200* 1/6 = 33.3333200* 1/6 = 33.3333200* 1/6 = 33.3333Reject H0. There is sufficient evidence to support the claim that the outcomes are not equally likely.The outcomes are not to be equally likely, so the loaded die appears to behave differently from a fair dieFinal answer Test statistics = 12.280P-value = 0.0311Reject the H0,Sufficient evidence.