Another Statistics question..?

Question

Employer A states the average number of days missed in a year by each staff member is 4.0. Office personnel randomly sampled a portion of their attendance records to determine if this claim was accurate. For the previous year, the following data was obtained from the sample group, regarding absences for that year: 4, 4, 3, 2, 6, 8, 7, 1, 9, 3, 1, and 6. At the .05 level of significance, is there convincing evidence that the number of absences per employee has increased?

Answer 1

No, there is not convincing evidence that the number of absences per employer has increased. While the sample mean is 4.5, a 95% confidence interval (2.797278 6.202722) includes the hypothesized value 4. Thus we would not reject the null hypothesis of increased absences at the 0.05 level.

R output:
> days <- c(4, 4, 3, 2, 6, 8, 7, 1, 9, 3, 1, 6)
> t.test(days, mu=4)

One Sample t-test

data: days
t = 0.6463, df = 11, p-value = 0.5313
alternative hypothesis: true mean is not equal to 4
95 percent confidence interval:
2.797278 6.202722
sample estimates:
mean of x
4.5

Answer 2

I will not do your homework for you, but I will add a couple of comments to the previous answer. First, you need to decide if you should use a one-tailed or a two-tailed test. If there is a logical reason to believe the number of days missed has increased, you should use a one-tailed test. Logical reasons would be things like industry trends, decreased employee morale, or an aging workforce. A higher average this year than last is an observation, not a reason. You should state why you are using a one-tailed or a two-tailed test.

Second, the old average of four days per employee is a sample mean, not a population mean. It would be better to include the data from previous years so that this can be taken into account.

Third, most statistical tests assume you are sampling from an infinite population. If the company does not have a large workforce, and if the variance of days off is greater from one worker to the next than it is from one year to the next for the same worker, you can use a finite population correction. If N is the population size and n is the sample size, multiply the variance over n by (N-n)/(N-1). (Elements of Applied Stochastic Processes by Bhat, U.N.)