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"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."

Respuesta :

JeanaShupp

Answer: ($19.8, $20.8)

Step-by-step explanation:

Confidence interval for mean : [tex]\overline{x}\pm z^*\dfrac{\sigma}{\sqrt{n}}[/tex] , where [tex]\overline{x}[/tex] = Sample mean, n= sample size, [tex]\sigma[/tex] = population standard deviation, z* = two tailed critical z-value.

Given:  [tex]\overline{x}[/tex] = $20.3

n= 2333

[tex]\sigma[/tex] = $11.3

For  98% confidence,  z* = 2.326

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

[tex]20.3\pm (2.326)\dfrac{11.3}{\sqrt{2333}}\\\\=20.3\pm (2.326)\dfrac{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)[/tex]

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

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