Mathematics: 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 $23.4$ 23.4 for a random sample of 976976 people. Assume the population standard deviation is known to be $10$ 10. Construct the 95%95% confidence interval for the mean per capita income in thousands of dollars. Round your answers to one decimal place.

An economist wants to estimate the mean per capita

balakrishnamanoharan

23.01.2022

The 95% confidence interval for the mean per capita income in thousands of dollars is between $22.8 and $24.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.05$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of 1-0.025 = 0.975 , so z = 1.96

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.96*\frac{10}{\sqrt{976}} = 0.6$

The lower end of the interval is the sample mean subtracted by M. So it is 23.4 - 0.6 = $22.8.

The upper end of the interval is the sample mean added to M. So it is 23.4 + 0.6 = $24.

The 95% confidence interval for the mean per capita income in thousands of dollars is between $22.8 and $24.

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lizbethh62

28.07.2021

theires some holes in the house theires some holes in the house...

why dont they fix the holes :(

Step-by-step explanation:

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