Q3. a. Simplify the following ratios. 360:120:60
1 answer:
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Answer:
Step-by-step explanation:
So we have the expression:
And we want to find the value of the expression when t is -3.
So, substitute -3 for t:
Multiply:
Add:
Evaluate:
So, our answer is 9.
And we're done!
Answer:
In order to calculate the p value we need to have in count the degrees of freedom , on this case 27. And since is a left tailed test the p value would be given by:
In order to find the p value we can use the following code in excel:
"=CHISQ.DIST(22.314,27,TRUE)"
The critical value can be founded with the following code:
=CHISQ.INV(0.05,27)
And we got :
If we compare the p value and the significance level provided we see that so on this case we have enough evidence in order to FAIL reject the null hypothesis at the significance level provided. And we can conclude that the true deviation is not lower than 22 at 5% of significance.
Step-by-step explanation:
Notation and previous concepts
A chi-square test is "used to test if the variance of a population is equal to a specified value. This test can be either a two-sided test or a one-sided test. The two-sided version tests against the alternative that the true variance is either less than or greater than the specified value"
represent the sample size
represent the confidence level
represent the sample deviation obtained
represent the value that we want to test
Null and alternative hypothesis
On this case we want to check if the population deviation specification is lower than 22, so the system of hypothesis would be:
Null Hypothesis:
Alternative hypothesis:
Calculate the statistic
For this test we can use the following statistic:
And this statistic is distributed chi square with n-1 degrees of freedom. We have eveything to replace.
Calculate the p value
In order to calculate the p value we need to have in count the degrees of freedom , on this case 27. And since is a left tailed test the p value would be given by:
In order to find the p value we can use the following code in excel:
"=CHISQ.DIST(22.314,27,TRUE)"
The critical value can be founded with the following code:
=CHISQ.INV(0.05,27)
And we got :
Conclusion
If we compare the p value and the significance level provided we see that so on this case we have enough evidence in order to FAIL reject the null hypothesis at the significance level provided. And we can conclude that the true deviation is not lower than 22 at 5% of significance.