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Solution:The null and alternative hypotheses are:\mathrm{HO} : The cube is fairHa: The cube is not fair.The test statistic is:\chi^{2}=\sum \frac{(O-E)^{2}}{E}Therefore, the test statistic is:\chi^{2}=\sum \frac{(O-E)^{2}}{E}=11.2The p-value is:p-\text { value }=P\left(\chi^{2}>11.2\right)=0.0824The excel function to find the \mathrm{p}-value is:= CHIDIST (11.2,6)Where:11.2 is the test statistic and 6 is the dfSince the p-value is greater than the significance level 0.05 , we, therefore, fail to reject the null hypothesis.A goodness-of-fit chi-square test is to be used to test the null hypothesis that the cube is fair. At a significance level of \alpha=0.05, what is the value of chi-square and the appropriate conclusion?\chi^{2}=13.6; fail to reject the null hypothesisx^{2}=11.2; fail to reject the null hypothesis\chi^{2}=27.2; fail to reject the null hypothesis\chi^{2}=27.2 ; reject the null hypothesisx^{2}=20; reject the null hypothesis ...