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An economist wants to estimate the mean per capita income (in thousands of dollars) for a major city in Texas. He believes that the mean income is $31.3, and the standard deviation is known to be $9.4. How large of a sample would be required in order to estimate the mean per capita income at the 85% level of confidence with an error of at most $0.46

Respuesta :

The sample size which is required in order to estimate the mean per capita income at the 85% level of confidence with an error of at most $0.46 is 866.

Given mean income of $31.3 and standard  deviation of $9.4.

We have to calculate the  sample size of confidence interval of 85% and at an error of $0.46.

Margin of error is basically the difference between actual values and calculated values.

Mean =$31.3

Standard deviation=$9.4

Confidence level=85%

Margin of error=$0.46

Margin of error=z*σ/[tex]\sqrt{n}[/tex]

0.46=1.44*9.4/[tex]\sqrt{n}[/tex]

[tex]\sqrt{n}[/tex]=1.44*9.4/0.46

[tex]\sqrt{n}[/tex]=29.42

squaring both sides

n=865.53

Round off and we get the following:

n=866.

Hence the sample size needed to estimate the mean per capita income at the 85% level of confidence is 866.

Learn more about margin of error at https://brainly.com/question/24289590

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