Question

Assume that a sample is used to estimate a population mean μ. Find the 98% confidence interval for a sample of size 401 with a mean of 78.1 and a standard deviation of 14.4. Enter your answer as a tri-linear inequality accurate to one decimal place (because the sample statistics are reported accurate to one decimal place).

___ < μ < ___

Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places.

Answer 1

To find the 98% confidence interval for the population mean μ, we can use the following formula for the confidence interval:

\[
\text{CI} = \bar{x} \pm z^* \left( \frac{s}{\sqrt{n}} \right)
\]

Where:
- \(\bar{x}\) is the sample mean,
- \(z^*\) is the critical value for a 98% confidence level,
- \(s\) is the sample standard deviation,
- \(n\) is the sample size.

### Step 1: Identify the Sample Statistics

- Sample size \(n = 401\)
- Sample mean \(\bar{x} = 78.1\)
- Sample standard deviation \(s = 14.4\)

### Step 2: Find the Critical Value \(z^*\)

For a 98% confidence interval, we need the critical value that captures the central 98% of the standard normal distribution. This leaves 1% in each tail.

Using a standard normal distribution table or calculator:

\[
z^* \approx 2.33
\]

### Step 3: Calculate the Standard Error (SE)

The standard error (SE) of the sample mean is calculated as follows:

\[
SE = \frac{s}{\sqrt{n}} = \frac{14.4}{\sqrt{401}}
\]

Calculating \(\sqrt{401}\):

\[
\sqrt{401} \approx 20.0249843945
\]

Now, calculate SE:

\[
SE = \frac{14.4}{20.0249843945} \approx 0.719
\]

### Step 4: Compute the Margin of Error (ME)

The margin of error (ME) is given by:

\[
ME = z^* \cdot SE = 2.33 \cdot 0.719
\]

Calculating that:

\[
ME \approx 1.674
\]

### Step 5: Calculate the Confidence Interval

Now we can calculate the lower and upper bounds of the confidence interval:

\[
\text{Lower bound} = \bar{x} - ME = 78.1 - 1.674 \approx 76.426
\]

\[
\text{Upper bound} = \bar{x} + ME = 78.1 + 1.674 \approx 79.774
\]

### Step 6: Round the Results

To one decimal place for the final values:

- Lower bound: \( 76.4 \)
- Upper bound: \( 79.8 \)

### Final Result

Thus, the 98% confidence interval for μ can be expressed as:

\[
76.4 < μ < 79.8
\]

This gives us the final answer in a tri-linear inequality format.