Suppose that you would like to buy a home priced at $200,000. You plan to make a payment of 10% of the purchase price and pay 1.
2% of the purchase price into an
escrow account annually.
a. Compute the total monthly payment and the total interest paid over the life of the loan for a 30-year mortgage at 4.8% annual interest.
b. Compute the total monthly payment and the total interest paid over the life of the loan for a 20-year mortgage at 4.8% annual interest.
escrow account annually.
a. Compute the total monthly payment and the total interest paid over the life of the loan for a 30-year mortgage at 4.8% annual interest.
b. Compute the total monthly payment and the total interest paid over the life of the loan for a 20-year mortgage at 4.8% annual interest.
1 answer:
Answer:
see below
Step-by-step explanation:
When you must do the same tedious calculation several times with different numbers, it is convenient to let a spreadsheet program do it for you. Here, the spreadsheet function PMT( ) computes the payment amount for the given interest rate, number of payments, and loan amount.
The loan amount is 90% of the purchase price.
The total interest over the life of the loan is the sum of the payments less the original loan amount.
The total monthly payment is the sum of the loan payment and the monthly escrow amount, which is 1/12 of the annual escrow amount.
_____
Here, we computed the total of payments using the unrounded "exact" value of each payment. We take this to be a better approximation of the total amount repaid, since the last payment always has an adjustment for any over- or under-payment due to rounding.
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Answer:
Reject the null hypothesis if the observed "t" value is less than -2.131 or higher than 2.131
Rejection Zone: or
In our case since our calculated value is not on the rejection zone we don't have enough evidence to reject the null hypothesis at 5% of significance.
Step-by-step explanation:
Previous concepts and data given
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
represent the sample mean
represent the sample standard deviation
n=16 represent the sample selected
significance level
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean is different from 3.6, the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
If we analyze the size for the sample is < 30 and we know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
(1)
t-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".
Calculate the statistic
We can replace in formula (1) the info given like this:
Critical values
On this case since we have a bilateral test we need to critical values. We need to use the t distribution with degrees of freedom. The value for
and
so we need to find on the t distribution with 15 degrees of freedom two values that accumulates 0.025 of the ara on each tail. We can use the following excel codes:
"=T.INV(0.025,15)" "=T.INV(1-0.025,15)"
And we got
So the decision on this case would be:
Reject the null hypothesis if the observed "t" value is less than -2.131 or higher than 2.131
Rejection Zone: or
Conclusion
In our case since our calculated value is not on the rejection zone we don't have enough evidence to reject the null hypothesis at 5% of significance.