Answer :
Answer:
a) F test is a statistical test that uses a F Statistic to compare two population variances, with the sample deviations s1 and s2. The F statistic is always positive number since the variance it's always higher than 0. The statistic is given by:
[tex]F=\frac{s^2_2}{s^2_1}[/tex]
b) H0: [tex] \sigma^2_1 = \sigma^2_2[/tex]
H1: [tex] \sigma^2_1 \neq \sigma^2_2[/tex]
c) [tex]\alpha=0.05[/tex] represent the significance level provided
Confidence =0.95 or 95%
d) We need to calculate the degrees of freedom first. For the numerator we have [tex]n_2 -1 =12-1=11[/tex] and for the denominator we have
We can find the critical value on the F table or with the following excel code: "=F.INV(1-0.025,11,11)"
And we got [tex]F_{critc}=3.474[/tex]
e) [tex]F=\frac{s^2_2}{s^2_1}=\frac{0.0056^2}{0.0019^2}=8.68[/tex]
f) Since our calculated value 8.68 is higher than the critical value of 3.474 we have enough evidence to reject the null hypothesis and at 5 % of significance we can conclude that the two variances are different.
Step-by-step explanation:
Data given and notation
We can calculate the mean and deviation with the following formulas:
[tex]\bar X =\frac{\sum_{i=1^n X_i}}{n}[/tex]
[tex] s= \frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}[/tex]
[tex]n_1 = 12 [/tex] represent the sampe size for the new drill
[tex]n_2 =12[/tex] represent the sample size for the old drill
[tex]\bar X_1 =0.500[/tex] represent the sample mean for the new drill
[tex]\bar X_2 =0.500[/tex] represent the sample mean for the old drill
[tex]s_1 = 0.0019[/tex] represent the sample deviation for the new drill
[tex]s^2_1 = 3.76x10^{-6}[/tex] represent the sample variance for the new drill
represent the sample deviation for the old drill
[tex]s^2_2 = 3.18x10^{-5}[/tex] represent the sample variance for the old drill
Part a
F test is a statistical test that uses a F Statistic to compare two population variances, with the sample deviations s1 and s2. The F statistic is always positive number since the variance it's always higher than 0. The statistic is given by:
[tex]F=\frac{s^2_2}{s^2_1}[/tex]
Solution to the problem
Part B: System of hypothesis
We want to test if the variation for oil stocks it's higher than the variation for utility stocks, so the system of hypothesis are:
H0: [tex] \sigma^2_1 = \sigma^2_2[/tex]
H1: [tex] \sigma^2_1 \neq \sigma^2_2[/tex]
Part c: Significance level
[tex]\alpha=0.05[/tex] represent the significance level provided
Confidence =0.95 or 95%
Part D: Critical value
We need to calculate the degrees of freedom first. For the numerator we have [tex]n_2 -1 =12-1=11[/tex] and for the denominator we have
We can find the critical value on the F table or with the following excel code: "=F.INV(1-0.025,11,11)"
And we got [tex]F_{critc}=3.474[/tex]
Part E :Calculate the statistic
Now we can calculate the statistic like this:
[tex]F=\frac{s^2_2}{s^2_1}=\frac{0.0056^2}{0.0019^2}=8.68[/tex]
Part F: Decision
Since our calculated value 8.68 is higher than the critical value of 3.474 we have enough evidence to reject the null hypothesis and at 5 % of significance we can conclude that the two variances are different.