ACTIVITY 11 1. Find the $$ 95 \% $$ confidence interval if there are 10 students taken the exam with a mean of 100 and standard deviation of 25.

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Given:
- Sample size: $$ n = 10 $$
- Sample mean: $$ \bar{x} = 100 $$
- Sample standard deviation: $$ s = 25 $$
- Confidence level: $$ 95\% $$

Since the sample size is small and the standard deviation is estimated from the sample, we use the $$ t $$-distribution. The confidence interval is given by

$$
\bar{x} \pm t_{\frac{\alpha}{2},\, n-1} \cdot \frac{s}{\sqrt{n}}
$$

where $$ t_{\frac{\alpha}{2},\, n-1} $$ is the $$ t $$-value at $$ \alpha/2 $$ with $$ n-1 $$ degrees of freedom. For a $$ 95\% $$ confidence interval, $$ \alpha = 0.05 $$ and thus $$ \alpha/2 = 0.025 $$.

1. **Determine the degrees of freedom:**

$$
\text{df} = n - 1 = 10 - 1 = 9.
$$

2. **Find the $$ t $$-value:**

For $$ \text{df} = 9 $$ and $$ \frac{\alpha}{2} = 0.025 $$, the $$ t $$-value is approximately

$$
t_{0.025,9} \approx 2.262.
$$

3. **Compute the standard error:**

$$
\text{Standard error} = \frac{s}{\sqrt{n}} = \frac{25}{\sqrt{10}}.
$$

Calculating $$ \sqrt{10} \approx 3.1623 $$,

$$
\frac{25}{3.1623} \approx 7.9057.
$$

4. **Calculate the margin of error:**

$$
\text{Margin of error} = t_{0.025,9} \cdot \text{Standard error} \approx 2.262 \times 7.9057 \approx 17.88.
$$

5. **Determine the confidence interval:**

$$
\text{Lower limit} = \bar{x} - \text{Margin of error} \approx 100 - 17.88 \approx 82.12,
$$
$$
\text{Upper limit} = \bar{x} + \text{Margin of error} \approx 100 + 17.88 \approx 117.88.
$$

Thus, the $$ 95\% $$ confidence interval is

$$
\left(82.12,~117.88\right).
$$
The 95% confidence interval is (82.12, 117.88).

ACTIVITY 11 1. Find the \( 95 \% \) confidence interval if there are 10 students taken the exam with a mean of 100 and standard deviation of 25.

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