A company manager wishes to test a union leader\'s claim that absences occur on

Question

A company manager wishes to test a union leader's claim that absences occur on the different work days with the same frequencies, Test this claim at the 0.05 level of significance if the following sample data have been compiled. Use a chi^2 test to the claim that in the given contingency table, the row variable and the column variable are independent. 6) Responses to a survey question are broken down according to gender and the sample results are given below. At the 0.05 significance level test the claim that response and gender are independent.

Explanation / Answer

Solution:-

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 chi-square statistic and the degrees of freedom, we determine the P-value.

DF = k - 1 = 5 - 1 = 4

(Ei) = n * pi

(E1) = (E2) = (E3) = (E4) = (E5) = 130 * 0.20 = 26

2 = [ (Oi - Ei)2 / Ei ]
2 = [ (37 - 26)2 / 26 ] + [ (15 - 26)2 / 26 ] + [ (12 - 26)2 / 26 ] + [ (23 - 26)2 / 26 ] + [ (43 - 26)2 / 26 ]
2 = (121 / 26) + (121 / 26) + (196 / 26) + (9 / 26) + (289 / 26)

= 28.31

where DF is the degrees of freedom, k is the number of levels of the categorical variable, n is the number of observations in the sample, Ei is the expected frequency count for level i, Oi is the observed frequency count for level i, and 2 is the chi-square test statistic.

The P-value is the probability that a chi-square statistic having 4 degrees of freedom is more extreme than 28.31.

We use the Chi-Square Distribution Calculator to find P(2 > 28.31) = 0.0001.

Interpret results. Since the P-value (0.0001) is less than the significance level (0.05), we cannot accept the null hypothesis.

Conclusion. The union leader's claim that absences occur on the different week days with the same frequency is not supported in this test.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.