Respuesta :
Answer:
[tex]p_v =P(z>z_{calc})=0.067[/tex]
Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.1[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis at 10% of significance.
So then the answer would be
a. true
Step-by-step explanation:
Data given and notation
[tex]\bar X[/tex] represent the sample mean
[tex]\sigma[/tex] represent the population standard deviation
[tex]n[/tex] sample size
[tex]\mu_o =800[/tex] represent the value that we want to test
[tex]\alpha=0.1[/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)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean is less or equal than 800 for the null hypothesis:
Null hypothesis:[tex]\mu \leq 800[/tex]
Alternative hypothesis:[tex]\mu > 800[/tex]
The z 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".
Calculate the statistic
We can replace in formula (1) and we assume that she got a calculated value [tex] z_{calc}[/tex]
P-value
Since is a right tailed test the p value would be:
[tex]p_v =P(z>z_{calc})=0.067[/tex]
Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.1[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis at 10% of significance.
So then the answer would be
a. true
