8. A sample of 16 gas stations shows that the mean price for regular gasoline is x= $3.25. It is known that the standard deviation is o = $.08. a. Find a 95% confidence interval. b. If we had used a 99% confidence interval, what effect if any would this have on the length of the confidence interval.

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Chapter10: Statistics
Section10.3: Measures Of Spread
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**Statistical Analysis: Confidence Intervals**

**Problem Statement:**

8. A sample of 16 gas stations shows that the mean price for regular gasoline is \( \bar{x} = \$3.25 \). It is known that the standard deviation is \( \sigma = \$0.08 \).

   a. Find a 95% confidence interval.
   
   b. If we had used a 99% confidence interval, what effect, if any, would this have on the length of the confidence interval?

**Solution:**

**Part a: Finding a 95% Confidence Interval**

To find the 95% confidence interval for the mean price of regular gasoline, we use the formula for the confidence interval of the mean for a sample:

\[ CI = \bar{x} \pm Z \left( \frac{\sigma}{\sqrt{n}} \right) \]

Where:
- \( \bar{x} \) is the sample mean
- \( Z \) is the Z-value that corresponds to the desired confidence level
- \( \sigma \) is the population standard deviation
- \( n \) is the sample size

Given:
- \( \bar{x} = \$3.25 \)
- \( \sigma = \$0.08 \)
- \( n = 16 \)

For a 95% confidence level, the Z-value is 1.96.

So, we calculate:

\[ CI = 3.25 \pm 1.96 \left( \frac{0.08}{\sqrt{16}} \right) \]
\[ CI = 3.25 \pm 1.96 \left( \frac{0.08}{4} \right) \]
\[ CI = 3.25 \pm 1.96 \left( 0.02 \right) \]
\[ CI = 3.25 \pm 0.0392 \]

Thus, the 95% confidence interval is:

\[ (\$3.2108, \$3.2892) \]

**Part b: Effect of Using a 99% Confidence Interval**

If we use a 99% confidence interval, the Z-value would change. For a 99% confidence interval, the Z-value is approximately 2.576.

The formula to recalculate the confidence interval with a 99% confidence level would be:

\[ CI = 3.25 \pm 2.576 \
Transcribed Image Text:**Statistical Analysis: Confidence Intervals** **Problem Statement:** 8. A sample of 16 gas stations shows that the mean price for regular gasoline is \( \bar{x} = \$3.25 \). It is known that the standard deviation is \( \sigma = \$0.08 \). a. Find a 95% confidence interval. b. If we had used a 99% confidence interval, what effect, if any, would this have on the length of the confidence interval? **Solution:** **Part a: Finding a 95% Confidence Interval** To find the 95% confidence interval for the mean price of regular gasoline, we use the formula for the confidence interval of the mean for a sample: \[ CI = \bar{x} \pm Z \left( \frac{\sigma}{\sqrt{n}} \right) \] Where: - \( \bar{x} \) is the sample mean - \( Z \) is the Z-value that corresponds to the desired confidence level - \( \sigma \) is the population standard deviation - \( n \) is the sample size Given: - \( \bar{x} = \$3.25 \) - \( \sigma = \$0.08 \) - \( n = 16 \) For a 95% confidence level, the Z-value is 1.96. So, we calculate: \[ CI = 3.25 \pm 1.96 \left( \frac{0.08}{\sqrt{16}} \right) \] \[ CI = 3.25 \pm 1.96 \left( \frac{0.08}{4} \right) \] \[ CI = 3.25 \pm 1.96 \left( 0.02 \right) \] \[ CI = 3.25 \pm 0.0392 \] Thus, the 95% confidence interval is: \[ (\$3.2108, \$3.2892) \] **Part b: Effect of Using a 99% Confidence Interval** If we use a 99% confidence interval, the Z-value would change. For a 99% confidence interval, the Z-value is approximately 2.576. The formula to recalculate the confidence interval with a 99% confidence level would be: \[ CI = 3.25 \pm 2.576 \
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