Question 3 of 14, Step 1 of 1 A research company desires to know the mean consumption of milk per week among males over age 25. A sample of 650 males over age 25 was drawn and the mean milk consumption was 3.4 liters. Assume that the population standard deviation is known to be 1.3 liters. Construct the $$ 95 \% $$ confidence interval for the mean consumption of milk among males over age 25. Round your answers to one decimal place.

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UpStudy AI · Accepted Answer

The confidence interval is computed using the formula

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

where:
- $$\bar{x} = 3.4$$ liters is the sample mean,
- $$\sigma = 1.3$$ liters is the population standard deviation,
- $$n = 650$$ is the sample size,
- $$z_{\alpha/2} = 1.96$$ for a $$95\%$$ confidence level.

**Step 1: Compute the standard error**

$$
SE = \frac{\sigma}{\sqrt{n}} = \frac{1.3}{\sqrt{650}}.
$$

Calculating $$\sqrt{650} \approx 25.5$$, we have

$$
SE \approx \frac{1.3}{25.5} \approx 0.051.
$$

**Step 2: Compute the margin of error**

$$
\text{Margin of Error} = z_{\alpha/2} \cdot SE \approx 1.96 \times 0.051 \approx 0.1.
$$

**Step 3: Construct the confidence interval**

$$
\text{CI} = \bar{x} \pm \text{Margin of Error}
$$

Calculating the endpoints:

$$
\text{Lower limit} = 3.4 - 0.1 = 3.3,
$$
$$
\text{Upper limit} = 3.4 + 0.1 = 3.5.
$$

Thus, the $$95\%$$ confidence interval for the mean consumption of milk among males over age 25 is

$$
(3.3,\, 3.5) \text{ liters}.
$$
The 95% confidence interval for the mean milk consumption is (3.3, 3.5) liters.

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