Step 1 :Define the null hypothesis as the outcomes are equally likely, which means each outcome (1,2,3,4,5,6) has an expected frequency of 200/6 = 33.33. The alternative hypothesis is that the outcomes are not equally likely.
Step 2 :Calculate the test statistic and compare it with the critical value from the chi-square distribution table with 5 degrees of freedom (6 outcomes - 1) at a significance level of 0.025.
Step 3 :Given observed frequencies are [30, 29, 46, 39, 29, 27] and expected frequencies are [33.333333333333336, 33.333333333333336, 33.333333333333336, 33.333333333333336, 33.333333333333336, 33.333333333333336].
Step 4 :The critical value is 12.832501994030027.
Step 5 :Compare the test statistic with the critical value. If the test statistic is greater than the critical value, reject the null hypothesis and conclude that the outcomes are not equally likely.
Step 6 :The test statistic is \( \boxed{5.400} \) and the critical value is \( \boxed{12.833} \).
Step 7 :There is not sufficient evidence to support the claim that the outcomes are not equally likely. The loaded die does not behave differently from a fair die.