Answer:
C. ± 2.326 years.
Step-by-step explanation:
We have the standard deviation for the sample. So we use the t-distribution to solve this question.
We have that to find our 
level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Ztable as such z has a pvalue of 
.
So it is z with a pvalue of 
, so ![z = 2.326/tex]Now, find the width of the interval[tex]W = z*\frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=z%20%3D%202.326%2Ftex%5D%3C%2Fp%3E%3Cp%3E%3Cstrong%3ENow%3C%2Fstrong%3E%2C%20find%20the%20width%20of%20the%20interval%3C%2Fp%3E%3Cp%3E%5Btex%5DW%20%3D%20z%2A%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
In which 
is the standard deviation of the population and n is the size of the sample.
In this question:
So
The correct answer is:
C. ± 2.326 years.
Answer:
Step-by-step explanation:
It continues to go up in the positive X and Y Axis. Like an exponential function
Answer: point estimate = 3.88
Margin of error = 0.63
Step-by-step explanation:
Confidence interval is written in the form,
(Point estimate - margin of error, Point estimate + margin of error)
The sample mean, x is the point estimate for the population mean. Let m represent the margin of error. Since the confidence interval is given as (3.25 , 4.51), it means that
x - m = 3.25
x + m = 4.51
Adding both equations, it becomes
2x = 7.76
x = 7.76/2
x = 3.88
Substituting x = 3.88 into x + m = 4.51, it becomes
3.88 + m = 4.51
m = 4.51 - 3.88
m = 0.63
y+2 = 3x
subtract 2 from each side
y = 3x-2
Answer:
B
Step-by-step explanation:
1.6/0.2=8
10⁵-10²=10³
