 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=E= 

Round to three decimal places if necessary

  <  μ  <  <  μ  <  

We are tasked with calculating the **margin of error (E)** and constructing an **80% confidence interval** for the true mean satisfaction level, given the following information:

- Sample size \( n = 158 \)
- Sample mean \( \bar{x} = 6.2 \)
- Population standard deviation \( \sigma = 2.8 \)
- Confidence level = 80%

### Step 1: Determine the z-score for an 80% confidence level
For an 80% confidence interval, the z-score is associated with 90% (since 80% is in the middle, and 10% is in each tail of the normal distribution). From the standard normal distribution table, the z-score is approximately:
\[
z = 1.282
\]

### Step 2: Calculate the standard error (SE)
The standard error of the mean is calculated as:
\[
SE = \frac{\sigma}{\sqrt{n}} = \frac{2.8}{\sqrt{158}} \approx \frac{2.8}{12.57} \approx 0.223
\]

### Step 3: Calculate the margin of error (E)
The margin of error is given by:
\[
E = z \times SE = 1.282 \times 0.223 \approx 0.286
\]

### Step 4: Construct the 80% confidence interval
The confidence interval is constructed by subtracting and adding the margin of error to the sample mean:
\[
\text{Lower bound} = \bar{x} - E = 6.

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.

Calculate the margin of error and construct an 80% confidence interval for the true mean satisfaction level of fans after a sporting event. The problem involves a sample size of 158, a mean satisfaction rating of 6.2, and a standard deviation of 2.8, applying statistical concepts such as standard error and z-scores.

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