Math / Data & Statistics

QuestionA sample of size $n = 66$ is drawn from a normal population whose standard deviation is $\sigma = 6.2$ . The sample mean is $\overline{x} = 37.77$ .
Part: 0 / 2
Part 1 of 2
(a) Construct a 99% confidence interval for $\mu$ . Round the answer to at least two decimal places.
A 99% confidence interval for the mean is $\boxed{\phantom{00}} < \mu < \boxed{\phantom{00}}$ .

Studdy Solution

STEP 1

1. The sample is drawn from a normal population.
2. The population standard deviation $\sigma$ is known.
3. The sample size $n = 66$ .
4. The sample mean $\overline{x} = 37.77$ .
5. We are constructing a 99% confidence interval for the population mean $\mu$ .

STEP 2

1. Identify the critical value for a 99% confidence interval.
2. Calculate the standard error of the mean.
3. Construct the confidence interval using the sample mean, critical value, and standard error.

STEP 3

Identify the critical value for a 99% confidence interval. Since the population standard deviation is known and the sample size is large, use the standard normal distribution (Z-distribution). The critical value $Z$ for a 99% confidence interval is approximately:
$Z = 2.576$

STEP 4

Calculate the standard error of the mean (SEM) using the formula:
$\text{SEM} = \frac{\sigma}{\sqrt{n}}$
Substitute the given values:
$\text{SEM} = \frac{6.2}{\sqrt{66}}$
Calculate:
$\text{SEM} \approx \frac{6.2}{8.124} \approx 0.763$

STEP 5

Construct the confidence interval using the formula:
$\overline{x} \pm Z \times \text{SEM}$
Substitute the values:
$37.77 \pm 2.576 \times 0.763$
Calculate the margin of error:
$2.576 \times 0.763 \approx 1.965$
Calculate the confidence interval:
$37.77 - 1.965 \approx 35.81$
$37.77 + 1.965 \approx 39.73$
Thus, the 99% confidence interval for $\mu$ is:
$\boxed{35.81} < \mu < \boxed{39.73}$

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