To test the hypothesis that the day of the week has an effect on absence, we can use a Chi-Square test for independence. Here are the steps:

1. State the Hypotheses: The null hypothesis (H0) is that the day of the week has no effect on absence. The alternative hypothesis (H1) is that the day of the week does have an effect on absence.

2. Calculate the Expected Frequencies: If the day of the week has no effect on absence, we would expect the absences to be evenly distributed across the days. So, the expected frequency for each day is the total number of absences divided by the number of days. The total number of absences is 15+17+12+10+6 = 60. So, the expected frequency for each day is 60/5 = 12.

3. Calculate the Chi-Square Statistic: The Chi-Square statistic is the sum of the squared differences between the observed (O) and expected (E) frequencies, divided by the expected frequencies, summed over all categories. So, Chi-Square = Σ [ (O-E)^2 / E ] = (15-12)^2/12 + (17-12)^2/12 + (12-12)^2/12 + (10-12)^2/12 + (6-12)^2/12.

4. Find the P-Value: The P-value is the probability of observing a Chi-Square statistic as extreme as, or more extreme than, the observed statistic, under the null hypothesis. You can find this by using a Chi-Square distribution table or a statistical software. The degrees of freedom for this test is the number of categories minus 1, which is 5-1=4.

5. Make a Decision: If the P-value is less than the significance level (0.05), reject the null hypothesis. This would suggest that the day of the week does have an effect on absence. If the P-value is greater than the significance level, do not reject the null hypothesis. This would suggest that the day of the week does not have an effect on absence.

Remember, failing to reject the null hypothesis does not prove it true. It simply suggests that we do not have strong enough evidence against it.

Question

To test the hypothesis that the day of the week has an effect on absence, we can use a Chi-Square test for independence. Here are the steps:

1. State the Hypotheses: The null hypothesis (H0) is that the day of the week has no effect on absence. The alternative hypothesis (H1) is that the day of the week does have an effect on absence.

2. Calculate the Expected Frequencies: If the day of the week has no effect on absence, we would expect the absences to be evenly distributed across the days. So, the expected frequency for each day is the total number of absences divided by the number of days. The total number of absences is 15+17+12+10+6 = 60. So, the expected frequency for each day is 60/5 = 12.

3. Calculate the Chi-Square Statistic: The Chi-Square statistic is the sum of the squared differences between the observed (O) and expected (E) frequencies, divided by the expected frequencies, summed over all categories. So, Chi-Square = Σ [ (O-E)^2 / E ] = (15-12)^2/12 + (17-12)^2/12 + (12-12)^2/12 + (10-12)^2/12 + (6-12)^2/12.

4. Find the P-Value: The P-value is the probability of observing a Chi-Square statistic as extreme as, or more extreme than, the observed statistic, under the null hypothesis. You can find this by using a Chi-Square distribution table or a statistical software. The degrees of freedom for this test is the number of categories minus 1, which is 5-1=4.

5. Make a Decision: If the P-value is less than the significance level (0.05), reject the null hypothesis. This would suggest that the day of the week does have an effect on absence. If the P-value is greater than the significance level, do not reject the null hypothesis. This would suggest that the day of the week does not have an effect on absence.

Remember, failing to reject the null hypothesis does not prove it true. It simply suggests that we do not have strong enough evidence against it.

Knowee AI · Accepted Answer

To test the hypothesis that the day of the week has an effect on absence, we can use a Chi-Square test for independence. Here are the steps:

1. State the Hypotheses: The null hypothesis (H0) is that the day of the week has no effect on absence. The alternative hypothesis (H1) is that the day of the week does have an effect on absence.

2. Calculate the Expected Frequencies: If the day of the week has no effect on absence, we would expect the absences to be evenly distributed across the days. So, the expected frequency for each day is the total number of absences divided by the number of days. The total number of absences is 15+17+12+10+6 = 60. So, the expected frequency for each day is 60/5 = 12.

3. Calculate the Chi-Square Statistic: The Chi-Square statistic is the sum of the squared differences between the observed (O) and expected (E) frequencies, divided by the expected frequencies, summed over all categories. So, Chi-Square = Σ [ (O-E)^2 / E ] = (15-12)^2/12 + (17-12)^2/12 + (12-12)^2/12 + (10-12)^2/12 + (6-12)^2/12.

4. Find the P-Value: The P-value is the probability of observing a Chi-Square statistic as extreme as, or more extreme than, the observed statistic, under the null hypothesis. You can find this by using a Chi-Square distribution table or a statistical software. The degrees of freedom for this test is the number of categories minus 1, which is 5-1=4.

5. Make a Decision: If the P-value is less than the significance level (0.05), reject the null hypothesis. This would suggest that the day of the week does have an effect on absence. If the P-value is greater than the significance level, do not reject the null hypothesis. This would suggest that the day of the week does not have an effect on absence.

Remember, failing to reject the null hypothesis does not prove it true. It simply suggests that we do not have strong enough evidence against it.

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