Respuesta :
Answer:
a. Null hypothesis:[tex]\mu \leq 48400[/tex]
Alternative hypothesis:[tex]\mu > 48400[/tex]
b. [tex]z=\frac{50000-48400}{\frac{8000}{\sqrt{100}}}=2[/tex]
c. [tex]P(Z>a)=0.05[/tex]
And the value of a that satisfy this is a=1.64. So our critical region is: [tex] (1.64,\infty)[/tex]
d. For this case since our calculated value is higher than the critical value we have enough evidence to reject the null hypothesis at 5% of significance.
e. [tex]p_v =P(Z>2)=0.02275[/tex]
Step-by-step explanation:
Data given and notation
[tex]\bar X=5000[/tex] represent the sample mean
[tex]\sigma=800[/tex] represent the standard deviation for the population
[tex]n=100[/tex] sample size
[tex]\mu_o =48400[/tex] represent the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
a. State the null and alternative hypotheses to be tested
We need to conduct a hypothesis in order to determine if the mean salary for graduates its higher than 48400, the system of hypothesis would be:
Null hypothesis:[tex]\mu \leq 48400[/tex]
Alternative hypothesis:[tex]\mu > 48400[/tex]
b. Compute the test statistic
We know the population deviation, so for this case is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:
[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (1)
z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
We can replace in formula (1) the info given like this:
[tex]z=\frac{50000-48400}{\frac{8000}{\sqrt{100}}}=2[/tex]
c. The null hypothesis is to be tested at 95% confidence. Determine the critical value for this test.
For this case we need one critical value since we are conducting a one right tailed test. We have this equality:
[tex]P(Z>a)=0.05[/tex]
And the value of a that satisfy this is a=1.64. So our critical region is: [tex] (1.64,\infty)[/tex]
d. What do you conclude?
For this case since our calculated value is higher than the critical value we have enough evidence to reject the null hypothesis at 5% of significance.
e. Compute the p-value
Since is a one right tailed test the p value would be:
[tex]p_v =P(Z>2)=0.02275[/tex]
If we compare the p value and a significance level assumed [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we can reject the null hypothesis, and the actual true mean for the salary is significantly higher from 48400.
