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 2.1Solution
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 $3.1076$.
Further Reading
- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators
