Question

An economist wants to estimate the mean per capita income (in thousands of dollars) for a major city in California. Suppose that the mean income is found to be $22.7$⁢22.7 for a random sample of 183183 people. Assume the population standard deviation is known to be $6.3$⁢6.3. Construct the 85%85% confidence interval for the mean per capita income in thousands of dollars. Round your answers to one decimal place.

Answer 1

Answer:

(22.0297, 23.3703)

Step-by-step explanation:

Given that an economist wants to estimate the mean per capita income (in thousands of dollars) for a major city in California.

Let X be per capita income (in thousands of dollars) for a major city in California.

Mean = 22.7

n = 183

Population std dev = 6.3

Since population std dev is known we can use Z critical value.

Std error = $\frac{6.3}{\sqrt{183} } \\=0.4657$

Z critical =1.44

Marginof error = ±1.44*0.4657=0.6706

Confidence interval 85%

=$(22.7-0.6703, 22.7+0.6703)\\= (22.0297, 23.3703)$