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

a. Null and alternative hypotheses: no significant increase vs. significant increase in average yearly income of dentists. b. Calculate the test statistic using the z-score formula. c. Determine the p-value and test the hypotheses at 95% confidence level.

Step-by-step explanation:

a. Null Hypothesis (H0): There has been no significant increase in the average yearly income of dentists. Alternative Hypothesis (Ha): There has been a significant increase in the average yearly income of dentists.

b. Test Statistic: In order to determine if there has been a significant increase in the average yearly income of dentists, we can calculate the z-test statistic using the formula: z = (sample mean - population mean) / (population standard deviation / sqrt(sample size)). Plugging in the given values, we have: z = (120000 - 110000) / (36000 / sqrt(81)).

c. P-value and Hypothesis Test: The p-value can be determined by looking up the calculated test statistic in a standard normal distribution table. If the p-value is less than the chosen significance level (in this case, 0.05), we reject the null hypothesis. At 95% confidence, if the p-value is less than 0.05, we can conclude that there has been a significant increase in the average yearly income of dentists.

Answer 2

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=(120000-110000)/((36000)/(√(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=(\bar X-\mu_o)/((\sigma)/(√(n)))` (1)

b) Statistic

When we replace the data given we got:

`z=(120000-110000)/((36000)/(√(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.