Answer:
12.68%
Step-by-step explanation:
To calculate effective annual interest rate we need to use the following formula:
Where, 'i' is the effective annual interest rate
'r' is the annual rate of interest
'm' is the frequency of compounding.
When there is continuous compounding the effective annual rate uses the following formula:
In our case we would are assume that there is continuous compounding since no information regarding the frequency of compounding is given:
Plugging r=12%=0.12, we get:
Therefore, the effective annual interest rate is 12.74%.
Answer:
<em>x=25/16</em>
Step-by-step explanation:
<u>Approximate Roots of Functions:</u>
We are required to find the root of
Let's construct a function
Reducing:
The equation will have a solution when f(x)=0. We'll start from the approximate crossing point given in the graph (x=1.5)
We'll move up on x=1.52
We are closer to the solution, let's try some more up
This value x=1.54 is close enough to the solution of the original equation
From the options given, is the closest to our solution
Answer:
A. As we are trying to test if the two processes yield different average errors, we are interested in any of the two posibilities: the average error of process A being statistically significant lower or bigger than the average error of Process B. That is why this is a two-tailed test.
B. t=-3.144
C. The critical value for a significance level of 0.05, a two tailed test with 4 degrees of freedom is t=±2.064.
D. P-value = 0.004
E. Reject H0
F. The probability of making a Type I error is equal to the significance level: P(Type I error) = 0.05.
Step-by-step explanation:
This is a hypothesis test for the difference between populations means.
The claim is that the two processes yield different average errors.
As we are trying to test if the two processes yield different average errors, we are interested in any of the two posibilities: the average error of process A being statistically significant lower or bigger than the average error of Process B. That is why this is a two-tailed test.
Then, the null and alternative hypothesis are:
being μ1: average error for Process A and μ2: average error fo Process B.
The significance level is 0.05.
The sample 1, of size n1=12 has a mean of 2 and a standard deviation of 1.
The sample 1, of size n1=14 has a mean of 3 and a standard deviation of 0.5.
The difference between sample means is Md=-1.
The estimated standard error of the difference between means is computed using the formula:
Then, we can calculate the t-statistic as:
The degrees of freedom for this test are:
The critical value for a significance level of 0.05, a two tailed test with 4 degrees of freedom is t=±2.064.
This test is a two-tailed test, with 24 degrees of freedom and t=-3.144, so the P-value for this test is calculated as (using a t-table):
As the P-value (0.004) is smaller than the significance level (0.05), the effect is significant.
The null hypothesis is rejected.
There is enough evidence to support the claim that the two processes yield different average errors.
Answer:
hope. It helps. What. I Understood
