Respuesta :
Answer:
[tex]z=\frac{9.25-10}{\frac{2}{\sqrt{35}}}=-2.219[/tex] Â
[tex]p_v =P(z<-2.219)=0.0132[/tex] Â
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 have enough evidence to reject the null hypothesis, so we can conclude that the true mean is significantly lower than 10 minutes. Â Â
Step-by-step explanation:
Data given and notation Â
[tex]\bar X=9.25[/tex] represent the sample mean Â
[tex]\sigma=2[/tex] represent the population standard deviation
[tex]n=35[/tex] sample size Â
[tex]\mu_o =10[/tex] represent the value that we want to test Â
[tex]\alpha=0.05[/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 less than 10 minutes, the system of hypothesis would be: Â
Null hypothesis:[tex]\mu \geq 10[/tex] Â
Alternative hypothesis:[tex]\mu < 10[/tex] Â
Since we know the population deviation, 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] (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 Â
We can replace in formula (1) the info given like this: Â
[tex]z=\frac{9.25-10}{\frac{2}{\sqrt{35}}}=-2.219[/tex] Â
P-value Â
Since is a left tailed test the p value would be: Â
[tex]p_v =P(z<-2.219)=0.0132[/tex] Â
Conclusion Â
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 have enough evidence to reject the null hypothesis, so we can conclude that the true mean is significantly lower than 10 minutes. Â Â