Respuesta :
Answer:
The null hypothesis failed to be rejected (P-value=0.014).
Step-by-step explanation:
This is a hypothesis test for the population mean.
The null and alternative hypothesis are:
[tex]H_0: \mu=5\\\\H_a:\mu> 5[/tex]
The significance level is α=0.01.
The sample has a size n=20.
The sample mean is M=9.
As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=7.5.
The estimated standard error of the mean is computed using the formula:
[tex]s_M=\dfrac{s}{\sqrt{n}}=\dfrac{7.5}{\sqrt{20}}=1.677[/tex]
Then, we can calculate the t-statistic as:
[tex]t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{9-5}{1.677}=\dfrac{4}{1.677}=2.385[/tex]
The degrees of freedom for this sample size are:
[tex]df=n-1=20-1=19[/tex]
This test is a right-tailed test, with 19 degrees of freedom and t=2.385, so the P-value for this test is calculated as (using a t-table):
[tex]P-value=P(t>2.385)=0.014[/tex]
As the P-value (0.014) is bigger than the significance level (0.01), the effect is not significant.
The null hypothesis failed to be rejected.
