the monthly income for 12 randomly selected people, each with abachelors degree in economics,are shown below. Assuming the population is normally distributed,construct a 90% confidence interval for the average income of a person with a bachelors degree in economics. 4450.34, 4596.25, 4366.14, 4455.21, 4151.45, 3727.34, 4283.94, 4527.76, 4407.52, 3946.75, 4023.77, 4221.16

3 answers

Mean = ∑x/n

Z = (score-mean)/SD

Z = 1.645 for 5% at either extreme.
Sorry, I forgot.

Find the mean first = sum of scores/number of scores

Subtract each of the scores from the mean and square each difference. Find the sum of these squares. Divide that by the number of scores to get variance.

Standard deviation = square root of variance

I'll let you do the calculations.
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