Employers want to know which days of the week employees are absent in a five-day work week. Most employers would like to believe that employees are absent equally during the week. Suppose a random sample of 61 managers were asked on which day of the week they had the highest number of employee absences. The results were distributed as in Table. For the population of employees, do the days for the highest number of absences occur with equal frequencies during a five-day work week? Test at a 0.05% significance level. Day Observed Frequency Monday 6 Tuesday 14 Wednesday 15 Thursday 10 Friday 16What is the chi-square test-statistic for this data? (Report answer accurate to three decimal places.)χ2=What are the degrees of freedom for this test?d.f.=What is the p-value for this sample? (Report answer accurate to four decimal places.)p-value=

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Employers want to know which days of the week employees are absent in a five-day work week. Most employers would like to believe that employees are absent equally during the week. Suppose a random sample of 61 managers were asked on which day of the week they had the highest number of employee absences. The results were distributed as in Table. For the population of employees, do the days for the highest number of absences occur with equal frequencies during a five-day work week? Test at a 0.05% significance level. Day  Observed Frequency  Monday 6 Tuesday 14 Wednesday 15 Thursday 10 Friday 16What is the chi-square test-statistic for this data? (Report answer accurate to three decimal places.)χ2=What are the degrees of freedom for this test?d.f.=What is the p-value for this sample? (Report answer accurate to four decimal places.)p-value=

Lesly Fernandez · Accepted Answer

Given information:In order to know the days of week the employees are absent in a five-day work week, the number of absences on each day of the week are tabulated as follows: Day  Frequency, Oi Monday 6 Tuesday 14 Wednesday 15 Thursday 10 Friday 16 Total(N) 61It is believed that the employees are absent equally during the week.The test is performed at 5% level of significance.The null and alternate hypothesis for testing are:H0: The employees are absent equally during the week.Hα: The employees are not equally absent during the week.The chi-square test statistic for testing is:χ2=∑i(Oi−Ei)2Eiwhere, Oi are observed frequencies, Ei are expected frequencies.The expected frequencies are computed as N5, since it is believed that the employees are absent equally during the week. Day  Observed Frequency  Expected Frequency (Oi−Ei)2Ei Monday 612.23.1508 Tuesday 1412.20.2656 Wednesday 1512.20.6426 Thursday 1012.20.3967 Friday 1612.21.1836 Total(N) 61615.6393Thus, the chi-squares test statistic is:χ2=∑i(Oi−Ei)2Ei=5.6393∼5.639Therefore, the chi-square test-statistic for this data is 5.639.The degrees of freedom (dof) for the test is:dof=(k-1)where,k is the number of categories under study.Thus, dof=(k-1)=(5-1)=4So, the degrees of freedom for this test is 4.The p-value for this sample is:p-value=P(χ2&amp;gt;x)=P(χ2&amp;gt;5.639)=0.2278Thus, the p-value for the chi-square test statistic is 0.2278.

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