Respuesta :
Answer:
A) Null and alternative hypothesis
[tex]H_0: \mu=2\\\\H_a:\mu\neq 2[/tex]
B) M = 2.2 hours
C) s = 0.52 hours
D) P-value = 0.255
E) At a significance level of 0.05, there is not enough evidence to support the claim that the mean tree-planting time significantly differs from two hours.
Step-by-step explanation:
This is a hypothesis test for the population mean.
The claim is that the mean tree-planting time significantly differs from two hours.
Then, the null and alternative hypothesis are:
[tex]H_0: \mu=2\\\\H_a:\mu\neq 2[/tex]
The significance level is 0.05.
The sample has a size n=10.
The sample mean is M=2.2.
As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=0.52.
The estimated standard error of the mean is computed using the formula:
[tex]s_M=\dfrac{s}{\sqrt{n}}=\dfrac{0.52}{\sqrt{10}}=0.1644[/tex]
Then, we can calculate the t-statistic as:
[tex]t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{2.2-2}{0.1644}=\dfrac{0.2}{0.1644}=1.216[/tex]
The degrees of freedom for this sample size are:
[tex]df=n-1=10-1=9[/tex]
This test is a two-tailed test, with 9 degrees of freedom and t=1.216, so the P-value for this test is calculated as (using a t-table):
[tex]\text{P-value}=2\cdot P(t>1.216)=0.255[/tex]
As the P-value (0.255) is bigger than the significance level (0.05), the effect is not significant.
The null hypothesis failed to be rejected.
At a significance level of 0.05, there is not enough evidence to support the claim that the mean tree-planting time significantly differs from two hours.
Sample mean and standard deviation:
[tex]M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{10}(1.7+1.5+2.6+. . .+2.3)\\\\\\M=\dfrac{22}{10}\\\\\\M=2.2\\\\\\s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{9}((1.7-2.2)^2+(1.5-2.2)^2+(2.6-2.2)^2+. . . +(2.3-2.2)^2)}\\\\\\s=\sqrt{\dfrac{2.4}{9}}\\\\\\s=\sqrt{0.27}=0.52\\\\\\[/tex]
