A manager states that in his factory, the mean number of days per year missed by his employees is less than the national mean of 10. The following data is days missed by some of his employees last year (mean=5.025, standard deviation=3.634). Is there sufficient avidence to belive the manager's statement?
Which "some of his employees"? How many employees were in his sample?
The standard error of the mean (SEm) for a distribution of means is like the standard deviation (SD) for a distribution of scores.
SEm = SD/√n, where n = number of subjects in his sample.
Does this help you answer the question?
To determine if there is sufficient evidence to believe the manager's statement that the mean number of days missed by his employees is less than the national mean of 10, we can conduct a hypothesis test.
Step 1: State the hypotheses:
- Null hypothesis (H0): The mean number of days missed by the manager's employees is greater than or equal to 10. (µ ≥ 10)
- Alternative hypothesis (H1): The mean number of days missed by the manager's employees is less than 10. (µ < 10)
Step 2: Set the significance level:
- This step involves deciding how much evidence we require to reject the null hypothesis. Let's set the significance level (α) at 0.05, which is commonly used.
Step 3: Collect and analyze data:
- You are given the mean (sample mean = 5.025) and the standard deviation (3.634) of the number of days missed by the employees. You can also assume that the population is normally distributed.
Step 4: Calculate the test statistic:
- Since we know the population standard deviation, we can use a z-test. The test statistic formula is:
z = (sample mean - population mean) / (population standard deviation /sqrt(sample size))
Step 5: Determine the critical value:
- The critical value is the value that marks the boundary for rejecting the null hypothesis. In this case, since we are testing a one-tailed hypothesis (the manager's claim is that the mean is less than 10), we need to find the z-score that corresponds to a cumulative probability of 0.05 in the left tail of the standard normal distribution.
Step 6: Make a decision:
- If the test statistic (calculated in Step 4) falls in the rejection region (left tail beyond the critical value), we reject the null hypothesis. If it falls in the non-rejection region, we fail to reject the null hypothesis.
Based on these steps, you can now carry out the calculations to determine the test statistic and compare it to the critical value to make a decision.