statistics
posted by Tom .
It is claimed the average age of employee in a large company is 50. A sample of 36 employees is taken and the average age was 49. The standard deviation of the population was 5. Use a level of significance of .05. What do you think of this claim?

Try a onesample ztest on this data:
z = (sample mean  population mean)/(standard deviation divided by the square root of the sample size)
z = (49  50)/(5/√36)
I'll let you finish the calculation.
Check a ztable for your critical or cutoff value at .05 level of significance. Compare the test statistic you calculated to the critical value from the table. Draw your conclusions.
I hope this will help get you started.