Find the product of 3.14 and 0.056
1 answer:
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Answer:
a)
b) For this case since we have a two tailed test we have two critical values, and we can find the two values since we need that on each tail the area would be , the distribution is the F with 9 df for the numerator and denominator, and we can use the following excel codes to find the critical values:
"=F.INV(0.025,9,9)"
"=F.INV(1-0.025,9,9)"
The two critical values are 0.248 and 4.026, so the rejection zone would be: since our calculated value is not on the rejection zone we fail to rejec the null hypothesis
Step-by-step explanation:
Data given and notation
represent the sampe size for the Miller's Stores
represent the sample size for the Albert's stores
represent the sample mean for Miller's store
represent the sample mean for Albert's store
represent the sample deviation for the Miller's store
represent the sample deviation for the Albert's stores
represent the sample variance for the utility stocks
represent the significance level provided
Confidence =0.95 or 95%
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:
Solution to the problem
System of hypothesis
We want to test if the variation for th two groups is the same, so the system of hypothesis are:
H0:
H1:
a) Calculate the statistic
Now we can calculate the statistic like this:
Now we can calculate the p value but first we need to calculate the degrees of freedom for the statistic. For the numerator we have and for the denominator we have
and the F statistic have 9 degrees of freedom for the numerator and 9 for the denominator. And the P value is given by:
P value
And we can use the following excel code to find the p value:"=2*(1-F.DIST(1.727,9,9,TRUE))"
b) Critical value
For this case since we have a two tailed test we have two critical values, and we can find the two values since we need that on each tail the area would be , the distribution is the F with 9 df for the numerator and denominator, and we can use the following excel codes to find the critical values:
"=F.INV(0.025,9,9)"
"=F.INV(1-0.025,9,9)"
The two critical values are 0.248 and 4.026, so the rejection zone would be: since our calculated value is not on the rejection zone we fail to rejec the null hypothesis
Since our calcu
Conclusion
Since the we have enough evidence to FAIL to reject the null hypothesis. And we can say that we don't have enough evidence to conclude that the two deviations are different at 5% of significance.