Asked by idania
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
Answers
Answered by
PsyDAG
Mean = ∑x/n
Z = (score-mean)/SD
Z = 1.645 for 5% at either extreme.
Z = (score-mean)/SD
Z = 1.645 for 5% at either extreme.
Answered by
PsyDAG
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.
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.
Answered by
Anonymous
Rusta today
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