Respuesta :
Answer:
a. P-value = 0.039.
The null hypothesis is rejected.
At a significance level of 0.05, there is enough evidence to support the claim that the population mean significantly differs from 100.
b. P-value = 0.013.
The null hypothesis is rejected.
At a significance level of 0.05, there is enough evidence to support the claim that the population mean significantly differs from 100.
c. P-value = 0.130.
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 population mean significantly differs from 100.
Step-by-step explanation:
This is a hypothesis test for the population mean.
The claim is that the population mean significantly differs from 100.
Then, the null and alternative hypothesis are:
[tex]H_0: \mu=100\\\\H_a:\mu\neq 100[/tex]
The significance level is 0.05.
The sample has a size n=65.
The degrees of freedom for this sample size are:
[tex]df=n-1=65-1=64[/tex]
a. The sample mean is M=103.
As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=11.5.
The estimated standard error of the mean is computed using the formula:
[tex]s_M=\dfrac{s}{\sqrt{n}}=\dfrac{11.5}{\sqrt{65}}=1.4264[/tex]
Then, we can calculate the t-statistic as:
[tex]t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{103-100}{1.4264}=\dfrac{3}{1.4264}=2.103[/tex]
This test is a two-tailed test, with 64 degrees of freedom and t=2.103, so the P-value for this test is calculated as (using a t-table):
[tex]\text{P-value}=2\cdot P(t>2.103)=0.039[/tex]
As the P-value (0.039) is smaller than the significance level (0.05), the effect is significant.
The null hypothesis is rejected.
At a significance level of 0.05, there is enough evidence to support the claim that the population mean significantly differs from 100.
b. The sample mean is M=96.5.
As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=11.
The estimated standard error of the mean is computed using the formula:
[tex]s_M=\dfrac{s}{\sqrt{n}}=\dfrac{11}{\sqrt{65}}=1.3644[/tex]
Then, we can calculate the t-statistic as:
[tex]t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{96.5-100}{1.3644}=\dfrac{-3.5}{1.3644}=-2.565[/tex]
This test is a two-tailed test, with 64 degrees of freedom and t=-2.565, so the P-value for this test is calculated as (using a t-table):
[tex]\text{P-value}=2\cdot P(t<-2.565)=0.013[/tex]
As the P-value (0.013) is smaller than the significance level (0.05), the effect is significant.
The null hypothesis is rejected.
At a significance level of 0.05, there is enough evidence to support the claim that the population mean significantly differs from 100.
c. The sample mean is M=102.
As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=10.5.
The estimated standard error of the mean is computed using the formula:
[tex]s_M=\dfrac{s}{\sqrt{n}}=\dfrac{10.5}{\sqrt{65}}=1.3024[/tex]
Then, we can calculate the t-statistic as:
[tex]t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{102-100}{1.3024}=\dfrac{2}{1.3024}=1.536[/tex]
This test is a two-tailed test, with 64 degrees of freedom and t=1.536, so the P-value for this test is calculated as (using a t-table):
[tex]\text{P-value}=2\cdot P(t>1.536)=0.130[/tex]
As the P-value (0.13) 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 population mean significantly differs from 100.
