Question

The Bureau of Labor Statistics reported that the average yearly income of dentists in the year 2005 was $110,000. A sample of 81 dentists, which was taken in 2006, showed an average yearly income of $120,000. Assume the standard deviation of the population of dentists in 2006 is $36,000.
a. We want to test to determine if there has been a significant increase in the average yearly income of dentists. Provide a null and the alternative hypotheses.b. Compute the test statistic.c. Determine the p-value; and at 95% confidence, test the hypotheses.

Answer 1

Answer:

a) We want to test to determine if there has been a significant increase in the average yearly income of dentists, the system of hypothesis would be:  

Null hypothesis:$\mu \leq 110000$  

Alternative hypothesis:$\mu > 110000$  

b) $z=\frac{120000-110000}{\frac{36000}{\sqrt{81}}}=2.5$    

c) $p_v =P(Z>2.5)=0.0062$  

We see that the p value is lower than the significance level of $1-0.95=0.05$ so then we can conclude that the true mean is significantly higher than 110000 because we reject the null hypothesis.

Step-by-step explanation:

Information given

$\bar X=12000$ represent the sample mean for the yearly income in 2006 for the dentists

$\sigma= 36000$ represent the sample population deviation

$n=81$ sample size  

$\mu_o =110000$ represent the value that we want to test

$\alpha=0.05$ represent the significance level

z would represent the statistic

$p_v$ represent the p value for the test

a) Hypothesis to verify

We want to test to determine if there has been a significant increase in the average yearly income of dentists, the system of hypothesis would be:  

Null hypothesis:$\mu \leq 110000$  

Alternative hypothesis:$\mu > 110000$  

Since we know the population deviation the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$  (1)  

b) Statistic

When we replace the data given we got:

$z=\frac{120000-110000}{\frac{36000}{\sqrt{81}}}=2.5$    

c) P value

Now we can find the p value using the fact that we are conducting a right tailed test:

$p_v =P(Z>2.5)=0.0062$  

We see that the p value is lower than the significance level of $1-0.95=0.05$ so then we can conclude that the true mean is significantly higher than 110000 because we reject the null hypothesis.