Question

"An economist wants to estimate the mean per capita income (in thousands of dollars) for a major city in Texas. Suppose that the mean income is found to be $20.3 for a random sample of 2333 people. Assume the population standard deviation is known to be $11.3. Construct the 98% confidence interval for the mean per capita income in thousands of dollars. Round your answers to one decimal place."

Answer 1

Answer: ($19.8, $20.8)

Explanation:

Confidence interval for mean :
`\overline{x}\pm z^*(\sigma)/(√(n))` , where
`\overline{x}` = Sample mean, n= sample size,
`\sigma` = population standard deviation, z* = two tailed critical z-value.

Given:
`\overline{x}` = $20.3

n= 2333

`\sigma` = $11.3

For 98% confidence, z* = 2.326

Then, the98% confidence interval for population mean will be:

`20.3\pm (2.326)(11.3)/(√(2333))\\\\=20.3\pm (2.326)(11.3)/(48.3011387)\\\\=20.3\pm (0.544165224825)\\\\\approx 20.3\pm 0.5\\\\=(20.3-0.5,\ 20.3+0.5)\\\\=(19.8,20.8)`

Hence, the 98% confidence interval for the mean per capita income in thousands of dollars = (19.8, 20.8)