Question 8 of 14, Step 1 of 1 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 $$ \$ 32.6 $$ for a random sample of 1699 people. Assume the population standard deviation is known to be $$ \$ 8.8 $$. Construct the $$ 95 \% $$ confidence interval for the mean per capita income in thousands of dollars. Round your answers to one decimal

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We are given that the sample mean is $$\bar{x} = 32.6$$ (in thousands of dollars), the population standard deviation is $$\sigma = 8.8$$, and the sample size is $$n = 1699$$. For a $$95\%$$ confidence interval, the formula is

$$
\bar{x} \pm z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}.
$$

Since the confidence level is $$95\%$$, the critical value is $$z_{\alpha/2} = 1.96$$.

Step 1: Calculate the standard error

$$
\text{SE} = \frac{\sigma}{\sqrt{n}} = \frac{8.8}{\sqrt{1699}}.
$$

Since $$\sqrt{1699} \approx 41.2$$, we have

$$
\text{SE} \approx \frac{8.8}{41.2} \approx 0.2136.
$$

Step 2: Calculate the margin of error

$$
E = 1.96 \times \text{SE} \approx 1.96 \times 0.2136 \approx 0.4187.
$$

Step 3: Construct the confidence interval

The lower limit is

$$
32.6 - 0.4187 \approx 32.2,
$$

and the upper limit is

$$
32.6 + 0.4187 \approx 33.0.
$$

Thus, the $$95\%$$ confidence interval for the mean per capita income is

$$
(32.2, 33.0)
$$

(thousands of dollars).
The $$95\%$$ confidence interval for the mean per capita income is $$(32.2, 33.0)$$ thousand dollars.

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