Answer:
Since p value <0.1 accept the claim that oven I repair costs are more
Step-by-step explanation:
The data given for two types of ovens are summarised below:
Group Group One Group Two
Mean 85.7900 78.6700
SD 15.1300 17.8400
SEM 1.9533 2.3840
N 60 56
Alpha = 10%
(Right tailed test)
The mean of Group One minus Group Two equals 7.1200
df = 114
standard error of difference = 3.065
t = 2.3234
p value = 0.0219
If p value <0.10 reject null hypothesis
4) Since p value <0.1 accept the claim that oven I repair costs are more
There is a 73.5% chance that Nathan is allergic to penicillin and the test predicts it.
75% * 98% = 73.5%
.75 * .98 = .735
Monthly payments, P = {R/12*A}/{1- (1+R/12)^-12n}
Where R = APR = 4.4% = 0.044, A = Amount borrowed = $60,000, n = Time the loan will be repaid
For 20 years, n = 20 years
P1 = {0.044/12*60000}/{1- (1+0.044/12)^-12*20} = $376.36
Total amount to be paid in 20 years, A1 = 376.36*20*12 = $90,326.30
For 3 years early, n = 17 year
P2 = {0.044/12*60,000}/{1-(1+0.044/12)^-12*17} = $418.22
Total amount to be paid in 17 years, A2 = 418.22*17*12 = $85,316.98
The saving when the loan is paid off 3 year early = A1-A2 = 90,326.30 - 85,316.98 = $5,009.32
Therefore, the approximate amount of savings is A. $4,516.32. This value is lower than the one calculated since the time of repaying the loan does not change. After 17 years, the borrower only clears the remaining amount of the principle amount.
