Answer:
Part a) The length of the rectangle is
Part b) The area of rectangle is
Step-by-step explanation:
step 1
Find the length
Let
L-----> the length of the rectangle
W---> the width of the rectangle
we know that
The perimeter of rectangle is
so
----> equation A
-----> equation B
substitute equation B in equation A
solve for W
Find the value of L
step 2
Find the area
The area of rectangle is equal to
substitute the values
Using the formula of test statistic, the value of the test statistic is 2.6961.
The mean of the public accountant is
Mean x(p) =50.2+59.8+57.3+59.2+53.2+56.0+49.9+58.5+56.0+51.9/10
x(p) = 55.2
Now the standard deviation of public accountant is
SD(p) = √{∑(x-x(p))^2/n-1}
SD(p) = √(50.2-55.2)^2+(59.8-55.2)^2+..................+(51.9-55.2)^2/n-1
After solving;
SD(p) = 3.34
The mean of the financial planner is
Mean x(F) =48.0+49.2+53.1+55.9+51.9+53.6+49.7+53.9+52.8+48.9/10
x(F) = 51.6
Now the standard deviation of financial planner is
SD(F) = √{∑(x-x(p))^2/n-1}
SD(F) = √(48.0-51.6)^2+(49.2-51.6)^2+..................+(48.9-51.6)^2/n-1
After solving;
SD(F) = 2.57
Test Statistic (t) =
t =
After solving
t = 2.6961
Hence, the value of the test statistic is 2.6961.
To learn more about test statistic link is here
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The right question is
public accountant 50.2 59.8 57.3 59.2 53.2 56.0 49.9 58.5 56.0 51.9
financial planner 48.0 49.2 53.1 55.9 51.9 53.6 49.7 53.9 52.8 48.9
Use a 0.05 level of significance and test the hypothesis that there is no difference between the starting annual salaries of public accountants and financial planner
Find the value of the test statistic.