Respuesta :
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:[tex]\mu \leq 110000[/tex]
Alternative hypothesis:[tex]\mu > 110000[/tex]
b) [tex]z=\frac{120000-110000}{\frac{36000}{\sqrt{81}}}=2.5[/tex]
c) [tex]p_v =P(Z>2.5)=0.0062[/tex]
We see that the p value is lower than the significance level of [tex]1-0.95=0.05[/tex] 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
[tex]\bar X=12000[/tex] represent the sample mean for the yearly income in 2006 for the dentists
[tex]\sigma= 36000[/tex] represent the sample population deviation
[tex]n=81[/tex] sample size
[tex]\mu_o =110000[/tex] represent the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level
z would represent the statistic
[tex]p_v[/tex] 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:[tex]\mu \leq 110000[/tex]
Alternative hypothesis:[tex]\mu > 110000[/tex]
Since we know the population deviation the statistic is given by:
[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (1)
b) Statistic
When we replace the data given we got:
[tex]z=\frac{120000-110000}{\frac{36000}{\sqrt{81}}}=2.5[/tex]
c) P value
Now we can find the p value using the fact that we are conducting a right tailed test:
[tex]p_v =P(Z>2.5)=0.0062[/tex]
We see that the p value is lower than the significance level of [tex]1-0.95=0.05[/tex] so then we can conclude that the true mean is significantly higher than 110000 because we reject the null hypothesis.
