Freshmen's GPA First-semester GPAs for a random selection of freshmen at a large university are shown below. Estimate the true mean GPA of the freshman class with 99% confidence. Assume s = 0.62.

```
1.9 3.2 2.0 2.9 2.7 3.3 2.8 3.0 3.8 2.7
2.0 1.9 2.5 2.7 2.8 3.2 3.0 3.8 3.1 2.7
3.5 3.8 3.9 2.7 2.0 2.8 1.9 4.0 2.2 2.8
2.1 2.4 3.0 3.4 2.9 2.1
```

#### Solution

**Sample mean**

The sample mean of $X$ is

` $$ \begin{aligned} \overline{x} &=\frac{1}{n}\sum_{i=1}^n x_i\\ &=\frac{101.5}{36}\\ &=2.8194\text{ } \end{aligned} $$ `

The average of GPA is $2.8194$ .

**Sample variance**

Sample variance of $X$ is

` $$ \begin{aligned} s_x^2 &=\dfrac{1}{n-1}\bigg(\sum_{i=1}^{n}x_i^2-\frac{\big(\sum_{i=1}^n x_i\big)^2}{n}\bigg)\\ &=\dfrac{1}{35}\bigg(299.45-\frac{(101.5)^2}{36}\bigg)\\ &=\dfrac{1}{35}\big(299.45-\frac{10302.25}{36}\big)\\ &=\dfrac{1}{35}\big(299.45-286.17361\big)\\ &= \frac{13.27639}{35}\\ &=0.3793 \end{aligned} $$ `

**Sample standard deviation**

The sample standard deviation is

` $$ \begin{aligned} s_x &=\sqrt{s_x^2}\\ &=\sqrt{0.3793}\\ &=0.6159 \text{ } \end{aligned} $$ `

Sample size $n = 20$, sample mean $\overline{X}= 2.8194$, sample standard deviation $s = 0.6159$.

The confidence level is $1-\alpha = 0.95$.

#### Step 1 Specify the confidence level $(1-\alpha)$

Confidence level is $1-\alpha = 0.95$. Thus, the level of significance is $\alpha = 0.05$.

#### Step 2 Given information

Sample size $n =20$, sample mean $\overline{X}=2.8194$, sample standard deviation $s=0.6159$.

#### Step 3 Specify the formula

$100(1-\alpha)$% confidence interval for the population mean $\mu$ is

` $$ \begin{aligned} \overline{X} - E \leq \mu \leq \overline{X} + E \end{aligned} $$ `

where `$E = t_{(\alpha/2,n-1)} \frac{s}{\sqrt{n}}$`

, andand `$t_{\alpha/2, n-1}$`

is the $t$ value providing an area of $\alpha/2$ in the upper tail of the students' $t$ distribution.

#### Step 4 Determine the critical value

The critical value of $t$ for given level of significance and $n-1$ degrees of freedom is $t_{\alpha/2,n-1}$.

Thus `$t_{\alpha/2,n-1} = t_{0.025,20-1}= 2.093$`

.

#### Step 5 Compute the margin of error

The margin of error for mean is

` $$ \begin{aligned} E & = t_{(\alpha/2,n-1)} \frac{s}{\sqrt{n}}\\ & = 2.093 \frac{0.6159}{\sqrt{20}} \\ & = 0.2882. \end{aligned} $$ `

#### Step 6 Determine the confidence interval

$95$% confidence interval estimate for population mean is

` $$ \begin{aligned} \overline{X} - E & \leq \mu \leq \overline{X} + E\\ 2.8194 - 0.288 & \leq \mu \leq 2.8194 + 0.288\\ 2.5312 &\leq \mu \leq 3.1076. \end{aligned} $$ `

Thus, $95$% confidence interval estimate for population mean is $(2.5312,3.1076)$.

#### Interpretation

We can be $95$% confident that the mean number of ounces of coffee that a machine dispenses in 12 ounce cups is between $2.5312$ and $3.1076$.