Respuesta :
Answer:
The p-value for this hypothesis test is P=0.015.
Step-by-step explanation:
In this case we have hypothesis test for the mean, with standard deviation of the population unknown.
The null hypothesis we want to test is
[tex]H_0: \mu\leq80[/tex]
To work with this test we have a sample of size n=20, sample mean=91 and sample standard deviation=21.
First, we estimate the standard deviation of the population
[tex]s_M=s/\sqrt{n}=21/\sqrt{20}=4.696[/tex]
Then, because we have an estimated standard deviation, we have to calculate the statistics t.
[tex]t=\frac{M-\mu}{s_M}=\frac{91-80}{4.696} =2.342[/tex]
We can look up this value of t in a t-table to know the probability of this value, taking into account 19 degrees of freedom:
[tex]df=n-1=20-1=19[/tex]
The p-value or the probability of P(t>2.342) is 0.01511.
This value P=0.0151 is compared to the significance level (0.05). Since the probability value (0.0151) is less than the significance level (0.05) the effect is statistically significant. Since the effect is significant, the null hypothesis is rejected.
