1. **Calculate the sample mean ($\bar{x}$):**

$$
\bar{x} = \frac{\sum x_i}{n} = \frac{71.6 + 81.0 + 88.9 + 80.4 + 78.1 + 72.0}{6} = \frac{472.0}{6} = 78.67
$$

2. **Calculate the sample standard deviation (s):**

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

First, find each deviation from the mean, square it, and sum them up:

$$
\begin{align*}
(71.6 - 78.67)^2 & = 49.77 \
(81.0 - 78.67)^2 & = 5.44 \
(88.9 - 78.67)^2 & = 104.83 \
(80.4 - 78.67)^2 & = 2.99 \
(78.1 - 78.67)^2 & = 0.32 \
(72.0 - 78.67)^2 & = 44.59 \
\end{align*}
$$

Sum of squared deviations:

$$
49.77 + 5.44 + 104.83 + 2.99 + 0.32 + 44.59 = 207.94
$$

Now, calculate the sample standard deviation:

$$
s = \sqrt{\frac{207.94}{5}} = \sqrt{41.59} \approx 6.45
$$

3. **Determine the t-value for a 95% confidence interval with $n-1$ degrees of freedom:**

For $n = 6$, degrees of freedom $df = 6 - 1 = 5$. Using a t-table or calculator, the t-value for a 95% confidence interval and 5 degrees of freedom is approximately 2.571.

4. **Calculate the margin of error (E):**

$$
E = t \times \frac{s}{\sqrt{n}} = 2.571 \times \frac{6.45}{\sqrt{6}} = 2.571 \times 2.63 \approx 6.76
$$

5. **Construct the confidence interval:**

$$
\text{Lower limit} = \bar{x} - E = 78.67 - 6.76 = 71.91
$$
$$
\text{Upper limit} = \bar{x} + E = 78.67 + 6.76 = 85.43
$$

Therefore, the 95% confidence interval for the mean score for all students is:

$$
(71.91, 85.43)
$$

Question

1. **Calculate the sample mean ($\bar{x}$):**

$$
   \bar{x} = \frac{\sum x_i}{n} = \frac{71.6 + 81.0 + 88.9 + 80.4 + 78.1 + 72.0}{6} = \frac{472.0}{6} = 78.67
   $$

2. **Calculate the sample standard deviation (s):**

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

First, find each deviation from the mean, square it, and sum them up:

$$
   \begin{align*}
   (71.6 - 78.67)^2 & = 49.77 \
   (81.0 - 78.67)^2 & = 5.44 \
   (88.9 - 78.67)^2 & = 104.83 \
   (80.4 - 78.67)^2 & = 2.99 \
   (78.1 - 78.67)^2 & = 0.32 \
   (72.0 - 78.67)^2 & = 44.59 \
   \end{align*}
   $$

Sum of squared deviations:

$$
   49.77 + 5.44 + 104.83 + 2.99 + 0.32 + 44.59 = 207.94
   $$

Now, calculate the sample standard deviation:

$$
   s = \sqrt{\frac{207.94}{5}} = \sqrt{41.59} \approx 6.45
   $$

3. **Determine the t-value for a 95% confidence interval with $n-1$ degrees of freedom:**

For $n = 6$, degrees of freedom $df = 6 - 1 = 5$. Using a t-table or calculator, the t-value for a 95% confidence interval and 5 degrees of freedom is approximately 2.571.

4. **Calculate the margin of error (E):**

$$
   E = t \times \frac{s}{\sqrt{n}} = 2.571 \times \frac{6.45}{\sqrt{6}} = 2.571 \times 2.63 \approx 6.76
   $$

5. **Construct the confidence interval:**

$$
   \text{Lower limit} = \bar{x} - E = 78.67 - 6.76 = 71.91
   $$
   $$
   \text{Upper limit} = \bar{x} + E = 78.67 + 6.76 = 85.43
   $$

Therefore, the 95% confidence interval for the mean score for all students is:

$$
   (71.91, 85.43)
   $$

Knowee AI · Accepted Answer