Respuesta :
Answer:
[tex]t=\frac{13.24-13}{\frac{0.4}{\sqrt{9}}}=1.8[/tex]
[tex]p_v =P(t_{8}>1.8)=0.0547[/tex]
If we compare the p value and the significance level given assumed [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we FAIL to reject the null hypothesis, and the the actual mean is not significantly higher than 13 for this case.
Step-by-step explanation:
Data given and notation
[tex]\bar X=13.24[/tex] represent the sample mean
[tex]s=\sqrt{0.16}= 0.4[/tex] represent the standard deviation for the sample
[tex]n=9[/tex] sample size
[tex]\mu_o =13[/tex] represent the value that we want to test
[tex]\alpha[/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)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to determine if the true mean is lower than 13 or not, the system of hypothesis would be:
Null hypothesis:[tex]\mu \leq 13[/tex]
Alternative hypothesis:[tex]\mu > 13[/tex]
We don't know the population deviation, so for this case 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{13.24-13}{\frac{0.4}{\sqrt{9}}}=1.8[/tex]
Calculate the P-value
First we need to calculate the degrees of freedom given by:
[tex]df=n-1=9-1=8[/tex]
Since is a one-side upper test the p value would be:
[tex]p_v =P(t_{8}>1.8)=0.0547[/tex]
Conclusion
If we compare the p value and the significance level given assumed [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we FAIL to reject the null hypothesis, and the the actual mean is not significantly higher than 13 for this case.
The conclusion is that the actual mean is not significantly higher than 13
How to determine the conclusion
Sample size, n = 9
Sample mean, [tex]\bar x = 13.4[/tex]
Population mean, [tex]\mu = 13[/tex]
Sample standard deviation, σ = 0.4
In this case we want to determine if the true mean is less than 13 or not
So, the hypotheses are:
Null [tex]H_o : \mu \le 13[/tex]
Alternate, [tex]H_a : \mu > 13[/tex]
Calculate the degrees of freedom
[tex]df = n - 1[/tex]
[tex]df = 9 - 1 = 8[/tex]
Calculate the standard error
[tex]\sigma_x = \frac{\sigma}{\sqrt n}[/tex]
This gives
[tex]\sigma_x = \frac{0.4}{\sqrt 9}[/tex]
[tex]\sigma_x = 0.4/3[/tex]
The test statistic is then calculated as:
[tex]t = \frac{\bar x - \mu}{\sigma_x}[/tex]
So, we have:
[tex]t = \frac{13.24 - 13}{0.4/3}[/tex]
[tex]t = 0.18[/tex]
Calculate the p value
[tex]p = P(t > 0.18)[/tex]
At a degree of freedom of 9, the p value is:
[tex]p = 0.0547[/tex]
The p value is greater than the level of significance.
This means that we fail to reject the null hypothesis
Hence, the actual mean is not significantly higher than 13
Read more about test statistic at:
brainly.com/question/16628265
