How would I begin to find the answer using nature of estimation used in statistics?

Employees in a large accounting firm claim that the mean salary of the firm’s accountant’s is less than that of its competitor’s, which is $45,000. A random sample of 30 of the firm’s accountants has a mean salary of $43,500 with a standard deviation of $5200. At alpha = 0.05, test the employee’s claim.

How would I begin to find the answer using nature of estimation used in statistics?
This is a hypothesis test for the population mean.





H0: mu = 45,000


HA: mu %26lt; 45,000





Since n %26gt;= 30, then you can use a z test statistic.





z = (xbar-mu)/{sigma/sqrt(n)}


= (43500-45000)/{5200/sqrt(30)}


= -1.58





Keep in mind that I used s (the sample standard deviation) instead of sigma (the population standard deviation) in this expression. That is because the sample size is %26gt;= 30, so the sample sd is going to be close enough to the population sd, and we can substitute.





The one tailed p-value is 0.0571 from the standard normal table. Since this is greater than alpha, you would not reject the null hypothesis. The evidence does not suggest that the mean salary at this firm is less than $45,000.
Reply:Ho: μ=45000


Ha: μ%26lt;45000





x-bar=43500


n=30


σ=5200


Assuming all of the simple conditions are met, use the equation to find your z-score


z=(x-bar - μ)/(σ/√n)





z=(43500-45000)/(5200/√30)


z=-1.57997





Look up the p-value in a standard normal probabilities table


The p-value is 0.0571





You can conclude that the employee's claim is true is it is less that the α-level. Since the p-value is greater than the α-level, you cannot conclude that the employee's claim is true. (You fail to reject the null hypothesis.)