Respuesta :
Answer:
[tex]p_v = P(\chi^2_{29}>42.963)=1-0.954=0.0459[/tex]
b. between .025 and .05
Step-by-step explanation:
Previous concepts and notation
The chi-square test is used to check if the standard deviation of a population is equal to a specified value. We can conduct the test "two-sided test or a one-sided test".
[tex]\bar X [/tex] represent the sample mean
n = 30 sample size
s= 0.2 represent the sample deviation
[tex]\sigma_o =\sqrt{0.027}=0.164[/tex] the value that we want to test
[tex]p_v [/tex] represent the p value for the test
t represent the statistic
[tex]\alpha=[/tex] significance level
State the null and alternative hypothesis
On this case we want to check if the population standard deviation is less than 0.027, so the system of hypothesis are:
H0: [tex]\sigma \leq 0.027[/tex]
H1: [tex]\sigma >0.027[/tex]
In order to check the hypothesis we need to calculate the statistic given by the following formula:
[tex] t=(n-1) [\frac{s}{\sigma_o}]^2 [/tex]
This statistic have a Chi Square distribution distribution with n-1 degrees of freedom.
What is the value of your test statistic?
Now we have everything to replace into the formula for the statistic and we got:
[tex] t=(30-1) [\frac{0.2}{0.164}]^2 =42.963[/tex]
What is the approximate p-value of the test?
The degrees of freedom are given by:
[tex] df=n-1= 30-1=29[/tex]
For this case since we have a right tailed test the p value is given by:
[tex]p_v = P(\chi^2_{29}>42.963)=1-0.954=0.0459[/tex]
And the best option would be:
b. between .025 and .05
