In: Statistics and Probability
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
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.
Salaries |
57860 |
66863 |
91982 |
66979 |
66940 |
82976 |
67073 |
72006 |
73496 |
72972 |
66169 |
65983 |
55646 |
62758 |
58012 |
63756 |
75536 |
60403 |
70445 |
61507 |
66555 |
∑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.