Respuesta :
Answer:
[tex]t=\frac{13.5-17}{\frac{6.57}{\sqrt{10}}}=-1.68[/tex] Â
[tex]p_v =P(t_{(9)}<-1.68)=0.064[/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 to reject the null hypothesis.
We can say that at 1% of significance the mean lead concentration for all is significantly less than 17 mu​g/g.
Step-by-step explanation:
Data given and notation Â
The mean and sample deviation can be calculated from the following formulas:
[tex]\bar X =\frac{\sum_{i=1}^n x_i}{n}[/tex]
[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)}{n-1}}[/tex]
[tex]\bar X=13.5[/tex] represent the sample mean Â
[tex]s=6.57[/tex] represent the sample standard deviation Â
[tex]n=10[/tex] sample size Â
[tex]\mu_o =17[/tex] represent the value that we want to test Â
[tex]\alpha=0.01[/tex] represent the significance level for the hypothesis test. Â
t 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 population mean is less than 17, the system of hypothesis are : Â
Null hypothesis:[tex]\mu \geq 17[/tex] Â
Alternative hypothesis:[tex]\mu < 17[/tex] Â
Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by: Â
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1) Â
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". Â
Calculate the statistic Â
We can replace in formula (1) the info given like this: Â
[tex]t=\frac{13.5-17}{\frac{6.57}{\sqrt{10}}}=-1.68[/tex] Â
P-value Â
We need to calculate the degrees of freedom first given by:
[tex]df=n-1=10-1=9[/tex]
Since is a one-side left tailed test the p value would given by: Â
[tex]p_v =P(t_{(9)}<-1.68)=0.064[/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 to reject the null hypothesis.
We can say that at 1% of significance the mean lead concentration for all is significantly less than 17 mu​g/g.
