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
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.
Rusta today
To construct the 90% confidence interval for the average income of a person with a bachelor's degree in economics, we can follow these steps:
Step 1: Calculate the sample mean.
To find the sample mean, add up all the incomes and divide by the number of observations. In this case, the sum of the incomes is:
4450.34 + 4596.25 + 4366.14 + 4455.21 + 4151.45 + 3727.34 + 4283.94 + 4527.76 + 4407.52 + 3946.75 + 4023.77 + 4221.16 = 51157.43
Since we have 12 observations, the sample mean (X̄) is:
X̄ = 51157.43 / 12 ≈ 4263.12
Step 2: Calculate the sample standard deviation.
To find the sample standard deviation, you need to compute the differences between each observation and the sample mean, square them, sum them up, divide by the number of observations minus 1 (in this case, 12 - 1 = 11), and take the square root of the result. It looks like this:
√((∑(Xi - X̄)^2) / (n - 1))
First, compute the squared differences for each observation:
(4450.34 - 4263.12)^2 ≈ 33183.66
(4596.25 - 4263.12)^2 ≈ 11100.05
(4366.14 - 4263.12)^2 ≈ 10601.93
(4455.21 - 4263.12)^2 ≈ 3668.61
(4151.45 - 4263.12)^2 ≈ 12505.93
(3727.34 - 4263.12)^2 ≈ 289546.47
(4283.94 - 4263.12)^2 ≈ 455.59
(4527.76 - 4263.12)^2 ≈ 6977.60
(4407.52 - 4263.12)^2 ≈ 2067.35
(3946.75 - 4263.12)^2 ≈ 1009.14
(4023.77 - 4263.12)^2 ≈ 5697.27
(4221.16 - 4263.12)^2 ≈ 1759.22
Then, sum up all the squared differences:
33183.66 + 11100.05 + 10601.93 + 3668.61 + 12505.93 + 289546.47 + 455.59 + 6977.60 + 2067.35 + 1009.14 + 5697.27 + 1759.22 = 388270.42
Finally, calculate the sample standard deviation:
√(388270.42 / 11) ≈ 294.46
Step 3: Determine the critical value.
At a 90% confidence level, we need to find the critical value from the t-distribution with (n-1) degrees of freedom. Since we have 12 observations, we have 11 degrees of freedom. Using a t-distribution table or calculator, the critical value for a 90% confidence level with 11 degrees of freedom is approximately 1.796.
Step 4: Calculate the margin of error.
The margin of error (E) is calculated by multiplying the critical value with the standard deviation divided by the square root of the number of observations:
E = (critical value) × (sample standard deviation / √n)
E = 1.796 × (294.46 / √12) ≈ 241.34
Step 5: Calculate the confidence interval.
The confidence interval is calculated by subtracting the margin of error from the sample mean and adding it to the sample mean:
Lower Limit = X̄ - E
Lower Limit = 4263.12 - 241.34 ≈ 4021.78
Upper Limit = X̄ + E
Upper Limit = 4263.12 + 241.34 ≈ 4504.46
Therefore, the 90% confidence interval for the average income of a person with a bachelor's degree in economics is approximately $4021.78 to $4504.46.