Respuesta :
Answer:
a) [tex]t=\frac{4.8-4}{\frac{2.3}{\sqrt{15}}}=1.347[/tex]
The first step is calculate the degrees of freedom, on this case:
[tex]df=n-1=15-1=14[/tex]
Since is a two sided test the p value would be:
[tex]p_v =2*P(t_{(14)}>1.347)=0.200[/tex]
b) If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we DON'T have enough evidence to reject the null hypothesis, so we can't conclude that the true mean is different from 4 at 1% of signficance.
Do not reject H0.
Step-by-step explanation:
Data given and notation
[tex]\bar X=4.8[/tex] represent the sample mean
[tex]s=2.3[/tex] represent the sample standard deviation
[tex]n=15[/tex] sample size
[tex]\mu_o =68[/tex] represent the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
Part a
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean is different from 4, the system of hypothesis would be:
Null hypothesis:[tex]\mu = 4[/tex]
Alternative hypothesis:[tex]\mu \neq 4[/tex]
If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{4.8-4}{\frac{2.3}{\sqrt{15}}}=1.347[/tex]
P-value
The first step is calculate the degrees of freedom, on this case:
[tex]df=n-1=15-1=14[/tex]
Since is a two sided test the p value would be:
[tex]p_v =2*P(t_{(14)}>1.347)=0.200[/tex]
Part b: Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we DON'T have enough evidence to reject the null hypothesis, so we can't conclude that the true mean is different from 4 at 1% of signficance.
Do not reject H0.
The test statistic is 1.347 and the p value is 0.200 and the conclusion is that we do not conclude that the true mean is not 4
(a) The test statistic and the p-value.
We have:
Test H0: μ= 4 vs Ha: μ≠4
x= 4.8, s= 2.3, with n= 15.
Start by calculating the degrees of freedom
df = n - 1
df = 15 - 1
df = 14
The test statistic is calculated using:
[tex]t = \frac{\bar x - \mu}{s/\sqrt n}[/tex] ------ for n < 30
So, we have:
[tex]t = \frac{4.8 - 4}{2.3/\sqrt{15}}[/tex]
Evaluate the expression
[tex]t = 1.347[/tex]
Approximate
[tex]t = 1.35[/tex]
For a two-sided test, the p value at a degrees of freedom of 14 for [tex]t = 1.347[/tex] is:
[tex]p =2 * P(t_{14} > 1.35)[/tex]
Using the table of test statistic, we have:
[tex]p =2 * 0.100[/tex]
[tex]p= 0.200[/tex]
Hence, the test statistic is 1.347 and the p value is 0.200
(b) The conclusion
The significance level is given as:
[tex]\alpha = 0.05[/tex]
The calculated p value (0.200) is greater than the significance level (0.05)
The above means that, there is no enough evidence to reject the null hypothesis,
Hence, the conclusion is that we do not conclude that the true mean is not 4
Read more about test of hypothesis at:
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