You are given the sample mean and the population standard deviation. Use this information to construct the $$ 90 \% $$ and $$ 95 \% $$ confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals. A random sample of 55 home theater systems has a mean price of $$ \$ 116.00 $$. Assume the population standard deviation is $$ \$ 18.50 $$. Construct a $$ 90 \% $$ confidence interval for the population mean. The $$ 90 \% $$ confidence interval is ( $$ 111.90,120.10 $$ ). (Round to two decimal places as needed.) Construct a $$ 95 \% $$ confidence interval for the population mean. The $$ 95 \% $$ confidence interval is ( $$ \square . \square) $$ ). (Round to two decimal places as needed.)

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To construct the $$ 95\% $$ confidence interval for the population mean and compare it with the already provided $$ 90\% $$ confidence interval, we'll follow these steps:

### Given Data:
- **Sample Size ($$ n $$)**: 55
- **Sample Mean ($$ \bar{x} $$)**: \$116.00
- **Population Standard Deviation ($$ \sigma $$)**: \$18.50

### Confidence Interval Formula:
When the population standard deviation is known, the confidence interval for the population mean is calculated using the Z-distribution:

$$
\text{Confidence Interval} = \bar{x} \pm z^* \left( \frac{\sigma}{\sqrt{n}} \right)
$$

Where:
- $$ \bar{x} $$ = sample mean
- $$ z^* $$ = Z-score corresponding to the desired confidence level
- $$ \sigma $$ = population standard deviation
- $$ n $$ = sample size

### Step 1: Calculate the Standard Error ($$ \text{SE} $$)
$$
\text{SE} = \frac{\sigma}{\sqrt{n}} = \frac{18.50}{\sqrt{55}} \approx \frac{18.50}{7.416} \approx 2.494
$$

### Step 2: Determine the Z-scores
- **For $$ 90\% $$ Confidence Level**:
  - $$ z^* \approx 1.645 $$
- **For $$ 95\% $$ Confidence Level**:
  - $$ z^* \approx 1.96 $$

### Step 3: Constructing the Confidence Intervals

#### a) **$$ 90\% $$ Confidence Interval**
$$
\text{CI}_{90\%} = 116.00 \pm 1.645 \times 2.494 \approx 116.00 \pm 4.106
$$
$$
\text{CI}_{90\%} = (111.90, 120.10)
$$
*(As provided)*

#### b) **$$ 95\% $$ Confidence Interval**
$$
\text{CI}_{95\%} = 116.00 \pm 1.96 \times 2.494 \approx 116.00 \pm 4.889
$$
$$
\text{CI}_{95\%} = (111.11, 120.89)
$$
*(Rounded to two decimal places)*

### Interpretation of the Confidence Intervals

- **$$ 90\% $$ Confidence Interval (\$111.90, \$120.10)**:
  - We are $$ 90\% $$ confident that the true population mean price of home theater systems lies between \$111.90 and \$120.10.

- **$$ 95\% $$ Confidence Interval (\$111.11, \$120.89)**:
  - We are $$ 95\% $$ confident that the true population mean price of home theater systems lies between \$111.11 and \$120.89.

### Comparison of Confidence Intervals

- **Width of Intervals**:
  - The $$ 95\% $$ confidence interval (\$111.11 to \$120.89) is **wider** than the $$ 90\% $$ confidence interval (\$111.90 to \$120.10).
  
- **Reason for Difference**:
  - A higher confidence level ($$ 95\% $$) requires a larger margin of error to ensure that the interval captures the true population mean with greater certainty. Consequently, the interval becomes wider compared to a lower confidence level ($$ 90\% $$).

### Conclusion

Increasing the confidence level from $$ 90\% $$ to $$ 95\% $$ results in a wider confidence interval. This trade-off between confidence level and interval width is fundamental in statistical inference: higher confidence provides greater assurance that the interval contains the true mean but at the cost of precision.
The $$ 95\% $$ confidence interval for the population mean is ( \$111.11, \$120.89 ). This interval is wider than the $$ 90\% $$ confidence interval of ( \$111.90, \$120.10 ) because a higher confidence level requires a larger margin of error to ensure the true mean is captured with greater certainty.

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