A sports reporter conducts a study of the fans' overall satisfaction with the sporting event after the event is completed. He surveys 158 randomly selected fans on their way out of the event and asks them to rate their satisfaction with the event on a scale from 0 to 10. The average satisfaction rating is 6.2. It is known from previous studies of this type that the standard deviation in satisfaction level is 2.8. Calculate the margin of error and construct the 80% confidence interval for the true mean satisfaction level for the sporting event.

Standard Normal Distribution Table

E=

Round to three decimal places if necessary

  <  μ  <  

Round to three decimal places if necessa

To calculate the margin of error (E) and construct the 80% confidence interval for the true mean satisfaction level, we'll follow these steps:

### Step 1: Determine the Z-score for an 80% confidence level
For an 80% confidence level, the corresponding Z-score (z) can be found from a standard normal distribution table. The middle 80% leaves 20% in the tails, so 10% is in each tail. We need the Z-score for which the cumulative probability is \( 1 - 0.10 = 0.90 \).

From the standard normal distribution table, the Z-score for a 90% cumulative probability is approximately:

\[
z = 1.282
\]

### Step 2: Calculate the standard error of the mean
The standard error of the mean (SE) is given by the formula:

\[
SE = \frac{\sigma}{\sqrt{n}}
\]

Where:
- \( \sigma = 2.8 \) is the standard deviation,
- \( n = 158 \) is the sample size.

\[
SE = \frac{2.8}{\sqrt{158}} \approx \frac{2.8}{12.57} \approx 0.223
\]

### Step 3: Calculate the margin of error (E)
The margin of error is calculated using the formula:

\[
E = z \times SE
\]

Substitute the values for \( z \) and \( SE \):

\[
E = 1.282 \times 0.223 \approx 0.286
\]

### Step 4: Construct the confidence interval
The confidence interval for the true mean \( \mu \) is given by:

\[
\text{CI} = \bar{x} \pm E
\]

Where \( \bar{x} = 6.2 \) is the sample mean. So, the confidence interval is:

\[
6.2 - 0.286 < \mu < 6.2 + 0.286
\]

\[
5.914 < \mu < 6.486
\]

### Final Answer:
- **Margin of Error (E):** \( 0.286 \)
- **80% Confidence Interval:** \( (5.914, 6.486) \)

Let me know if you'd like more details on any of the steps.

Here are 5 related questions to explore further:
1. How does increasing the sample size affect the margin of error?
2. What would be the confidence interval at a 95% confidence level?
3. How does the margin of error change with different confidence levels?
4. What if the standard deviation was different? How would that impact the interval?
5. How would the confidence interval change if the average satisfaction was higher?

**Tip:** As confidence levels increase, the Z-score increases, resulting in a wider confidence interval, indicating more uncertainty.

A sports reporter conducts a study of the fans' overall satisfaction with the sporting event after the event is completed. He surveys 158 randomly selected fans on their way out of the event and asks them to rate their satisfaction with the event on a scale from 0 to 10. The average satisfaction rating is 6.2. It is known from previous studies of this type that the standard deviation in satisfaction level is 2.8. Calculate the margin of error and construct the 80% confidence interval for the true mean satisfaction level for the sporting event.

Learn how to calculate the margin of error and construct an 80% confidence interval for a sports fan satisfaction survey. The study involved 158 fans, with an average satisfaction rating of 6.2 and a standard deviation of 2.8. This solution walks through the statistical concepts and formulas required to solve the problem.

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