Question

public accountant financial planner 50.2 48.0 59.8 49.2 57.3 53.1 59.2 55.9 53.2 51.9 56.0 53.6 49.9 49.7 58.5 53.9 56.0 52.8 51.9 48.9

Answer 1

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) = $\frac{x(P)-x(f)}{\sqrt{\left(\frac{(n(p)-1)s(p)^2+(n(f)-1)s(f)^2}{n(p)+n(f)-2}\left(\frac{1}{n(p)}+\frac{1}{n(f)}\right)\right)}}$

t = $\frac{55.2-51.61}{\sqrt{\left(\frac{(10-1)(3.34))^2+(10-1)(2.57)^2}{10+10-2}\left(\frac{1}{10}+\frac{1}{10}\right)\right)}}$

After solving

t = 2.6961

Hence, the value of the test statistic is 2.6961.

To learn more about test statistic link is here

brainly.com/question/14128303

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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.