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.
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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.