7g = -84 ; g = - 12 due in 10 min WHAT IS UP WITH THESE TIMED MATH TESTS T^T
1 answer:
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Answer:
a) If we compare the p value and the significance level given we see that
so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true average is not significantly less than 7.5.
b)
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 that means that the population variance is not significantly higher than 1.96.
Step-by-step explanation:
Assuming this info: "Suppose birth weights follow a normal distribution with mean 7.5 pounds and standard deviation 1.4 pounds"
1) Data given and notation
represent the sample mean
represent the sample standard deviation
represent the population standard deviation
sample size
represent the value that we want to test
represent the significance level for the hypothesis test.
z would represent the statistic (variable of interest)
represent the p value for the test (variable of interest)
2) State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the true mean is less than 7.5, the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:
(1)
z-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".
3) Calculate the statistic
We can replace in formula (1) the info given like this:
4)P-value
Since is a left tailed test the p value would be:
5) Conclusion
Part a
If we compare the p value and the significance level given we see that
so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true average is not significantly less than 7.5.
Part b
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 variance obtained
represent the value that we want to test
Null and alternative hypothesis
On this case we want to check if the population variance increase, 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 9. And since is a right tailed test the p value would be given by:
In order to find the p value we can use the following code in excel:
"=1-CHISQ.DIST(18.367,9,TRUE)"
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 that means that the population variance is not significantly higher than 1.96.