Answer:
Since the P-value is less than the significance level, we can reject the null hypothesis and conclude that her daily tips average more than $80.
The P-value is 0.0020.
Step-by-step explanation:
In this problem we have to perform a hypothesis test with population's standard deviation known.
To conclude that her daily tips average more than $80, we have to reject the null hypothesis (the average daily tips are less than $80). If it is rejected, the former sentence is true.
The null and alternative hypothesis are then:
[tex]H_0: \mu \leq80\\H_1: \mu>80[/tex]
The significance level is 0.01.
The z-value for this sample is
[tex]z=\frac{M-\mu}{\sigma/\sqrt{n}} =\frac{84.85-80}{9.95/\sqrt{35}} =\frac{4.85}{1.6819}=2.8837[/tex]
The P-value for this z is P=0.00197.
Since the P-value is less than the significance level, we can reject the null hypothesis and conclude that her daily tips average more than $80.
