Question 1204073: what is the sollution of A school principal would like to know which days of the week students are most likely to
be absent. The principal expects that students will be equally during the 5-day school
week. The principal selects a random sample of 100 teachers asking them which day of
the week they had the highest number of student absences. The observed and expected
results are shown in the table below. Based on these results, do the days for the highest
number of absences occur with equal frequencies? (Use a 5% significance level).
Monday Tuesday Wednesday Thursday Friday
Observed absences 23 16 14 19 28
Expected Absences 20 20 20 20 20
Answer by asad1975(2)   (Show Source):
You can put this solution on YOUR website! To answer this question, we can use a chi-square test of independence, which is a statistical test that compares the observed frequencies of two categorical variables to the expected frequencies under the assumption of independence. The null hypothesis is that the days for the highest number of absences occur with equal frequencies, and the alternative hypothesis is that they do not.
To perform the chi-square test of independence, we need to calculate the chi-square test statistic using the formula:
$$X^2 = \sum \frac{(O-E)^2}{E}$$
where O is the observed frequency and E is the expected frequency for each cell in the contingency table. The degrees of freedom for the test are given by:
$$df = (r-1)(c-1)$$
where r is the number of rows and c is the number of columns in the contingency table. The p-value for the test is the probability of obtaining a chi-square value equal to or greater than the calculated test statistic under the null hypothesis. We can use a chi-square distribution table or a calculator to find the p-value.
Using the data given in the question, we can calculate the chi-square test statistic as follows:
$$X^2 = \frac{(23-20)^2}{20} + \frac{(16-20)^2}{20} + \frac{(14-20)^2}{20} + \frac{(19-20)^2}{20} + \frac{(28-20)^2}{20}$$
$$X^2 = 0.45 + 0.8 + 1.8 + 0.05 + 3.2$$
$$X^2 = 6.3$$
The degrees of freedom for the test are:
$$df = (1-1)(5-1)$$
$$df = 0 \times 4$$
$$df = 0$$
Since the degrees of freedom are zero, we cannot use a chi-square distribution table or a calculator to find the p-value. However, we can use a logical argument to conclude that the p-value must be very large, since any value of X^2 will be equal to or greater than zero under the null hypothesis. Therefore, we fail to reject the null hypothesis at the 5% significance level.
We can conclude that there is not enough evidence to suggest that the days for the highest number of absences occur with unequal frequencies. The principal's expectation that students will be equally absent during the 5-day school week is not contradicted by the data.
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