Rewrite your survey question from Discussion 0. List all the class responses to your survey question in ascending order. How many times have you travelled by airplane last year ? 0,0,0,1,1,1,1,3,3,4, 5,

What is the best point estimate for your survey question? Be sure to include both the correct notation (symbol) and the value.

Construct a confidence interval for the responses to your survey. Assume that you do not know the population standard deviation.
Indicate your level of confidence and the corresponding critical value.

What is the margin of error?

If we would increase the sample size to 500 students, what would happen to your margin of error?

### Survey Question:
"How many times have you traveled by airplane in the last year?"

### Class Responses:
- List of responses in ascending order: 0, 0, 0, 1, 1, 1, 1, 3, 3, 4, 5

### Best Point Estimate:
The best point estimate for the average number of times the class traveled by airplane last year is the sample mean (\(\bar{x}\)).

\[
\bar{x} = \frac{\sum x_i}{n} = \frac{0 + 0 + 0 + 1 + 1 + 1 + 1 + 3 + 3 + 4 + 5}{11} = \frac{19}{11} \approx 1.727
\]

Thus, the best point estimate is \(\bar{x} \approx 1.727\).

### Confidence Interval:
To construct a confidence interval for the mean, we need the sample mean, the sample standard deviation, and the critical value for the desired confidence level.

1. **Sample Mean (\(\bar{x}\)):** \(1.727\)
2. **Sample Standard Deviation (s):**

\[
s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}
\]

First, calculate each squared deviation from the mean:

\[
\begin{aligned}
(0 - 1.727)^2 &= 2.983 \\
(0 - 1.727)^2 &= 2.983 \\
(0 - 1.727)^2 &= 2.983 \\
(1 - 1.727)^2 &= 0.528 \\
(1 - 1.727)^2 &= 0.528 \\
(1 - 1.727)^2 &= 0.528 \\
(1 - 1.727)^2 &= 0.528 \\
(3 - 1.727)^2 &= 1.618 \\
(3 - 1.727)^2 &= 1.618 \\
(4 - 1.727)^2 &= 5.188 \\
(5 - 1.727)^2 &= 10.70 \\
\end{aligned}
\]

Sum of squared deviations:

\[
\sum (x_i - \bar{x})^2 = 30.167
\]

Sample variance:

\[
s^2 = \frac{30.167}{10} = 3.017
\]

Sample standard deviation:

\[
s = \sqrt{3.017} \approx 1.737
\]

3. **Critical Value (\(t^*\)):**
   For a 95% confidence level with 10 degrees of freedom (n-1 = 11-1), the critical value \(t^*\) is approximately 2.228.

4. **Margin of Error (ME):**

\[
\text{ME} = t^* \cdot \frac{s}{\sqrt{n}} = 2.228 \cdot \frac{1.737}{\sqrt{11}} \approx 1.167
\]

5. **Confidence Interval:**

\[
\text{CI} = \bar{x} \pm \text{ME} = 1.727 \pm 1.167
\]

Thus, the 95% confidence interval is approximately:

\[
[0.560, 2.894]
\]

### Margin of Error:
The margin of error is approximately \(1.167\).

### Effect of Increasing Sample Size:
If we increase the sample size to 500 students, the margin of error would decrease because the margin of error is inversely proportional to the square root of the sample size. Mathematically:

\[
\text{ME} \propto \frac{1}{\sqrt{n}}
\]

Increasing the sample size to 500:

\[
\text{New ME} = t^* \cdot \frac{s}{\sqrt{500}}
\]

Since \(\sqrt{500} > \sqrt{11}\), the margin of error would decrease, leading to a narrower confidence interval.

### Questions:
1. What is the best point estimate for your survey question?
2. How do you calculate the sample standard deviation?
3. What is the critical value for a 95% confidence level?
4. How does increasing the sample size affect the margin of error?
5. Why do we use the t-distribution instead of the normal distribution for this confidence interval?

### Tip:
When constructing a confidence interval, ensure that the sample size is sufficiently large to justify the use of the t-distribution, especially when the population standard deviation is unknown.

Learn how to construct confidence intervals and point estimates in statistics with a detailed example. Calculate the best point estimate, sample standard deviation, margin of error, and construct a confidence interval. Understand the impact of sample size on margin of error.

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