Step-by-step explanation:
An arithmetic sequence is given by relation as follows :
For the first term, put n = 1. So,
For fourth term, put n = 4. So,
For tenth term, put n = 10. So,
Hence, the correct option is (C).
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1 answer:
Answer:
for new car loans at 2.79% the present value is $15879.04
for old car loans at 3.29 the present value is $15721.34
Step-by-step explanation:
firstly we use the present value annuity formula for this problem as there is future consistent payments that will be done and so we want to evaluate the present value for these two interest rates that are given respectively which is 2.79% and 3.29% which is compounded annually. therefore we will start by calculating the present value for payments on a new car loan.
we are given that she does not want to pay more than $350 per month therefore this gives us the future payments per month of $350 and now we know that the interest rates must be adjusted to monthly compounding so we will use this present value annuity formula:
we will substitute values as follows ;
Pv is the present value we are looking for for the new car loan
C is the periodic payments which is the $350 per month she can afford.
i is the interest rate adjusted to monthly rate which is 2.79%/12
n is the number of payments made so 4years x 12 months = 48 months that she will be repaying the loan.
Pv = 350[(1-(1+(2.79%/12))/(2.79%/12)] then compute on calculator.
Pv = $15879.04 this is the maximum amount she can take for a new car loan.
then for the old car loan we only change i from the above interest to 3.29%/12 then substitute all the other values the same to the above mentioned present value formula.
Pv = 350[(1-(1+(3.29%/12))/(3.29%/12)]
Pv= $15721.34 so this is the maximum amount she can take for an old car loan.
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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.