Do employees take more sick leave in the year before retirement? They may well h
ID: 3311570 • Letter: D
Question
Do employees take more sick leave in the year before retirement? They may well have an incentive to do so if their accumulated paid sick leave (the number of days they are entitled to be away with full pay) is about to expire. Indeed, this appears to happen with government workers. One evaluation of this issue looked at statistics gathered by the U.S. General Accounting Office (GAO).25 The study concluded,
[What if] the bulge in sick days was just an aberration in the GAO sample rather than a real symptom of goofing off? In zeroing in on this question, we note that the 714 retirees in the GAO sample averaged 30 sick days in their last year instead of the “expected” 14 days. So in a work year of 251 days (average for federal employees), the retirees were finding themselves indisposed 12.0% of the time instead of 5.6%. Could that happen by chance? The science of statistics tells us that the probability of any such swing in so large a sample is low. To be precise, one in 200,000.
a. Identify the population and the sample.
b. Identify the hypotheses being tested, in terms of percent of time indisposed.
c. Identify the p-value here.
d. Which hypothesis (if any) has been rejected? Which has been accepted?
e. How significant (statistically) is the result?
Explanation / Answer
Solution:-
a. Identify the population and the sample.
Population - U.S. General Accounting Office retiring persons
Sample is 714 retirees.
b. Identify the hypotheses being tested, in terms of percent of time indisposed.
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: P < 0.056
Alternative hypothesis: P > 0.056
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected only if the sample proportion is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.
Analyze sample data. Using sample data, we calculate the standard deviation () and compute the z-score test statistic (z).
= sqrt[ P * ( 1 - P ) / n ]
= 0.008605
z = (p - P) /
z = 7.44
where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.
Since we have a one-tailed test, the P-value is the probability that the z-score is greater than 7.44.
c) Thus, the P-value = less than 0.0001
Interpret results. Since the P-value (almost 0) is less than the significance level (0.05), we cannot accept the null hypothesis.
d) Reject the null hypothesis, accept the alternate hypothesis.
e) The result is very much significant because the p-value is very less than 0.05.
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