
Respuesta :
Answer:
[tex]z=\frac{26.5-25.1}{\frac{5.25}{\sqrt{40}}}=1.687[/tex] Â Â
[tex]p_v =P(z>1.687)=0.0458[/tex] Â
If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is higher than 25.1 (significant increase) at 1% of signficance. Â
Step-by-step explanation:
Data given and notation Â
[tex]\bar X=26.5[/tex] represent the sample mean
[tex]\sigma=5.25[/tex] represent the population standard deviation for the sample Â
[tex]n=40[/tex] sample size Â
[tex]\mu_o =25.1[/tex] represent the value that we want to test
[tex]\alpha=0.01[/tex] represent the significance level for the hypothesis test. Â
z would represent the statistic (variable of interest) Â
[tex]p_v[/tex] represent the p value for the test (variable of interest) Â
State the null and alternative hypotheses. Â
We need to conduct a hypothesis in order to check if the mean is higher than 25.1, the system of hypothesis would be: Â
Null hypothesis:[tex]\mu \leq 25.1[/tex] Â
Alternative hypothesis:[tex]\mu > 25.1[/tex] Â
If we analyze the size for the sample is > 30 and we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by: Â
[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] Â (1) Â
z-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". Â
Calculate the statistic
We can replace in formula (1) the info given like this: Â
[tex]z=\frac{26.5-25.1}{\frac{5.25}{\sqrt{40}}}=1.687[/tex] Â Â
P-value
Since is a one right side test the p value would be: Â
[tex]p_v =P(z>1.687)=0.0458[/tex] Â
Conclusion Â
If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is higher than 25.1 (significant increase) at 1% of signficance. Â