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

What is the best estimate of the average annual salaries?

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

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.

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.

orchestra answered 1 year ago

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