Your statistics instructor would like to use a 6-side die (with each of its six faces showing a differentnumber from 1 to 6) to choose the next quiz question. You suspect that the die is not fair, and oneor more numbers show up more frequently than others, so you are conducting a Goodness-of-Fithypothesis test for the claim that the 6-side die is not fair at the significant level of a = 0.01.Step 1) State claims identify the parameter:Your friend Alicia helped you with the null and alternative hypotheses...1Ho: p1 = p2 = p3 = P4 = P5 = P6HA: At least one of the proportion is different from the others.Step 2) Identify the procedureFor this study, we should use Select an answer ♥Step 3) Gather a sample and summarize with statisticsYou drew a sample of size n=87, and the observed frequencies are recorded below. Please complete thetable using the exact answer (e.g., whole numbers, fractions or decimals).NumberObservedExpectedFrequencyRolledFrequency1162173164552013 The degree of freedom :The test-statistic for this data =(Please show your answer to three decimal places.)The p-value for this sample =(Please show your answer to four decimal places.)Step 4) State the conclusionThe p-value is Select an answerBased on this, we should Select an answerhypothesisAs such, the final conclusion is that...Based on the sample data, there is not sufficient evidence to conclude the claim that not all 6categories are equally likely to be selected at the significant level of a = 0.01.Based on the sample data, there is sufficient evidence to conclude the claim that not all 6categories are equally likely to be selected at the significant level of a = 0.01.

The chi - square (χ2) goodness of fit - test statistics formula is defined by Oi = Observed frequency Ei = Expected frequency Step 2: For this study, we should use chi - square χ2 test. Because chi - square (χ2) goodness of fit – test is appropriate to determine proportion in each category differ from the expected proportion Step 3: Degrees of freedom = k – 1 [ k = No of categories] = 6 -1 = 5 Thus, the Degrees of freedom (df ) is 5 Test statistic : Expected frequency = sample size (n) / 6 = 87/6 = 14.5 Number Rolled Observed Frequency (Oi) Expected frequency (Ei) 1 16 14.5 2 17 14.5 3 16 14.5 4 5 14.5 5 20 14.5 6 13 14.5 The value of test statistic is Therefore, the value of test statistic is 9.205 Calculation of p value: From χ2 distribution area table values, If chi square test χ2 = 9.205 with df = 5 then p value is 0.1012 i.e., P(χ2 > 9.205) at 5 df = 0.1012 [ since, chi square is right tailed distribution] [ For p value, use technology (or) CHISQ.DIST. RT (x, df) excel function] Therefore, the p value is 0.1012 Step 4 - Conclusion: Given level of significance α = 0.01 The p value (0.1012) is ≥ α (0.01) Decision rule: If p value is lower than (<) 0.05 then reject null hypothesis H0. Otherwise do no reject the null hypothesis Ho Based on this, we should fail to reject the null hypothesis. Conclusion is, based on the sample data, there is not sufficient evidence to conclude the claim that not all 6 categories are equally likely to be selected at the significance level of α = 0.01 Option (A) is correct