Question

1. The Bureau of Labor Statistics reported that the average yearly income of dentists in the year 2014 was $116,000. Dr. Smith thinks there has been a significant increase in the average yearly income of dentists. He has a dataset of a sample of 81 dentists, which was taken in 2015, showed an average yearly income of $123,800. Assume the standard deviation of the population of dentists in 2015 is $36,000. Please answer the following questions:

(a)	Develop appropriate null and alternative hypotheses such that rejection of H0 will support Dr. Smith’s argument.
(b)	Compute the test statistic.
(c)	Determine the p-value; and at 95% confidence, test the hypotheses.

2. A sample of 64 account balances from a credit company showed an average daily balance of $1,048. The standard deviation of the population is known to be $240. We are interested in determining if the mean of all account balances (i.e., population mean) is significantly different from $1,000.

Question:

Using the p-value approach at 95% confidence, test the above hypotheses.

Please show the steps of computation to your answer

Answer 1

1)

a)

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 116000
Alternative Hypothesis, Ha: μ > 116000

b)

Test statistic,
z = (xbar - mu)/(sigma/sqrt(n))
z = (123800 - 116000)/(36000/sqrt(81))
z = 1.95

c)

P-value Approach
P-value = 0.0256
As P-value < 0.05, reject the null hypothesis.

2)

Null Hypothesis, H0: μ = 1000
Alternative Hypothesis, Ha: μ ≠ 1000

Test statistic,
z = (xbar - mu)/(sigma/sqrt(n))
z = (1048 - 1000)/(240/sqrt(64))
z = 1.6

P-value Approach
P-value = 0.1096
As P-value >= 0.05, fail to reject null hypothesis.