Manufacturers are testing a die to make sure that it is fair (has a uniform distribution). They roll the die 66 times and record
the outcomes. They conduct a chi-square Goodness-of-Fit hypothesis test at the 5% significance level. (a) The null and alternative hypotheses are: H0: The die has the uniform distribution. Ha: The die does not have the uniform distribution. (b) χ20=15.091. (c) χ20.05=11.070. (d) What conclusions can be made? Select all that apply. Select all that apply: We should reject H0. We should not reject H0. At the 5% significance level, there is sufficient evidence to conclude that the die is not fair. At the 5% significance level, there is not enough evidence to conclude that the die is not fair.
At the 5% significance level, there is sufficient evidence to conclude that the die is not fair.
Step-by-step explanation:
Null hypothesis is a statement that is to be tested against the alternative hypothesis and then decision is taken whether to accept or reject the null hypothesis. The critical value is 15.091 and test statistic is 11.070. The null hypothesis is rejected or accepted on the basis of level of significance. When the p-value Test statistics is greater than level of significance we fail to reject the null hypothesis and null hypothesis is then accepted. In the given case p-value is less than critical value then we should reject the null hypothesis.
