Full solution please. I\'m preparing for a final, so please do not post wrong an
ID: 3293773 • Letter: F
Question
Full solution please. I'm preparing for a final, so please do not post wrong answers! Thank you.
At Anytown College, the administration would like the students' grade distribution to be 25% A's, 30% B's, 25% C's, and 20% D's. The school's president thinks that the instructors may not be following this guideline, so he takes a random sample of grades to check his suspicion. A random sample of 30 grades yields 9 A's, 15 B's, 3 C's, and 3 D's. The president wishes to test H_0: p_A = 0.25, p_B = 0.30, p_C = 0.25, p_D = 0.20. Perform the appropriate test at a 5% significance level. Report the value of the test statistic, the critical value(s), and state your decision.Explanation / Answer
The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a chi-square goodness of fit test of the null hypothesis.
Analyze sample data. Applying the chi-square goodness of fit test to sample data, we compute the degrees of freedom, the expected frequency counts, and the chi-square test statistic. Based on the significance level and the degrees of freedom, we determine the critical value of the chi-square test statistic and compare the results to interpret the results .
DF = k-1 = 4 - 1 = 3
Expected frequency is given as,
(Ei) = n * pi where n = 30 and pi is the proportions of the grades in Null Hypothesis.
(E1) = 30 * 0.25 = 7.5
(E2) = 30 * 0.30 = 9
(E3) = 30 * 0.25 = 7.5
(E4) = 30 * 0.20 = 6
Observed frequencies Oi , are 9, 15, 3, 3
Chi square test statistic is,
2 = [ (Oi - Ei)2 / Ei ]
2 = [ (9 - 7.5)2 / 7.5 ] + [ (15 - 9)2 / 9 ] + [ (3 - 7.5)2 / 7.5 ] + [ (3 - 6)2 / 6 ]
2 = 8.5
So, Chi-square test statistic is 8.5
From the ChiSquare table, Critical value of Chisquare test statistic at DF = 3 and 5% significance level is 7.81
As, the observed Chi-square test statistic (8.5) is greater than the critical value (7.81), we reject the null hypothesis and conclude that at 5% significance level, there is insufficient data to show that the proportion of Grades A, B, C and D is 25%, 30% , 25% and 20%, respectively.
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