Question

In: Statistics and Probability

The annual salaries (in dollars) for a random sample of 21 accounting executives in Toronto, Ontario...

The annual salaries (in dollars) for a random sample of 21 accounting executives in Toronto, Ontario are listed below:

       57,860    66,863   91,982   66,979   66,940   82,976   67,073

        72,006    73,496   72,972   66,169   65,983   55,646   62,758

       58,012    63,756   75,536   60,403   70,445   61,507   66,555

  1. What is the best estimate of the average annual salaries?

  1. Determine the 95% confidence interval for the average Salaries. Interpret its meaning.

  1. An Analyst claims that the mean annual Salary for advertising account executives in Toronto, Ontario is more than the national mean, $67,800.

Assume the population is normally distributed. At α = 0.05, is there enough evidence to support the Analyst’s claim? Answer in the following steps.

  1. Which type of error (I or II) might have occurred in your decision above? Explain the error in that situation.
Salaries
57860
66863
91982
66979
66940
82976
67073
72006
73496
72972
66169
65983
55646
62758
58012
63756
75536
60403
70445
61507
66555

Solutions

Expert Solution

∑x = 1425917

∑x² = 98268264689

n = 21

Mean , x̅ = Ʃx/n = 1425917/21 = 67900.8095

Standard deviation, s = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(98268264689-(1425917)²/21)/(21-1)] = 8506.8974

a) Best estimate of the average annual salaries = $ 67900.8095

b) 95% Confidence interval :

At α = 0.05 and df = n-1 = 20, two tailed critical value, t-crit = T.INV.2T(0.05, 20) = 2.086

Lower Bound = x̅ - t-crit*s/√n = 67900.8095 - 2.086 * 8506.8974/√21 = 64028.516

Upper Bound = x̅ + t-crit*s/√n = 67900.8095 + 2.086 * 8506.8974/√21 = 71773.103

64028.516 < µ < 71773.103

There is 95% chance that the population mean is between 64028.516 and 71773.103.

c) As the confidence interval contain 67800, we fail to reject the null hypothesis.

we reject the Analyst’s claim that mean annual Salary for advertising account executives in Toronto, Ontario is more than the national mean, $67,800.

d) We might have made a Type II error.

Type II error is failing to reject the null hypothesis when it is not true.


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