This tutorial covers examples on confidence interval for the population mean when the population standard deviation is unknown.
Example 1
A new brand of laptop battery is produced by a company. The company claims that the battery will last for an extended period of time before a recharge is necessary. A sample of 40 batteries is tested for the length of usage time to recharge. The sample results are as follow:
Sample size: 40, Sample Mean: 6.5 hrs, Sample Standard Deviation: 1.3 hrs.
Construct a 99 % confidence interval for the average length of usage time to recharge
Solution
Given that sample size $n = 40$, sample mean $\overline{X}= 6.5$, sample standard deviation $s = 1.3$.
The confidence level is $1-\alpha = 0.99$.
Step 1 Specify the confidence level $(1-\alpha)$
Confidence level is $1-\alpha = 0.99$. Thus, the level of significance is $\alpha = 0.01$.
Step 2 Given information
Given that sample size $n =40$, sample mean $\overline{X}=6.5$, sample standard deviation $s=1.3$.
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 $E = t_{(\alpha/2,n-1)} \frac{s}{\sqrt{n}}$, and $t_{\alpha/2, n-1}$ 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 $t_{\alpha/2,n-1} = t_{0.005,40-1}= 2.708$.
Step 5 Compute the margin of error
The margin of error for mean is
$$ \begin{aligned} E & = t_{(\alpha/2,n-1)} \frac{s}{\sqrt{n}}\\ & = 2.708 \frac{1.3}{\sqrt{40}} \\ & = 0.557. \end{aligned} $$
Step 6 Determine the confidence interval
$99$% confidence interval estimate for population mean is
$$ \begin{aligned} \overline{X} - E & \leq \mu \leq \overline{X} + E\\ 6.5 - 0.557 & \leq \mu \leq 6.5 + 0.557\\ 5.943 &\leq \mu \leq 7.057. \end{aligned} $$
Thus, $99$% confidence interval estimate for population mean is $(5.943,7.057)$.
Interpretation
We can be $99$% confident that the average length of usage time to recharge is between $5.943$ and $7.057$.
Example 2
Find the 95% confidence interval for the mean number of ounces of coffee that a machine dispenses in 12 ounce cups. Assume the variable is normally distributed. The data is shown below:
12.03, 12.10, 12.02, 11.98, 12.00, 12.05, 11.97.
Solution
| $x_i$ | $x_i^2$ | |
|---|---|---|
| 12.03 | 144.72 | |
| 12.10 | 146.41 | |
| 12.02 | 144.48 | |
| 11.98 | 143.52 | |
| 12.00 | 144.00 | |
| 12.05 | 145.20 | |
| 11.97 | 143.28 | |
| Total | 84.15 | 1011.62 |
Sample mean
The sample mean of $X$ is
$$ \begin{aligned} \overline{x} &=\frac{1}{n}\sum_{i=1}^n x_i\\ &=\frac{84.15}{7}\\ &=12.0214\text{ ounce} \end{aligned} $$
The average of ounces of coffee is $12.0214$ ounce.
Sample variance
Sample variance of $X$ is
$$ \begin{aligned} s_x^2 &=\dfrac{1}{n-1}\bigg(\sum_{i=1}^{n}x_i^2-\frac{\big(\sum_{i=1}^n x_i\big)^2}{n}\bigg)\\ &=\dfrac{1}{6}\bigg(1011.6151-\frac{(84.15)^2}{7}\bigg)\\ &=\dfrac{1}{6}\big(1011.6151-\frac{7081.2225}{7}\big)\\ &=\dfrac{1}{6}\big(1011.6151-1011.60321\big)\\ &= \frac{0.01189}{6}\\ &=0.002 \end{aligned} $$
Sample standard deviation
The sample standard deviation is
$$ \begin{aligned} s_x &=\sqrt{s_x^2}\\ &=\sqrt{0.002}\\ &=0.0445 \text{ ounce} \end{aligned} $$
Sample size $n = 20$, sample mean $\overline{X}= 12.0214$, sample standard deviation $s = 0.0445$.
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 $\overline{X}=12.0214$, sample standard deviation $s=0.0445$.
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
$$ \begin{aligned} E & = t_{(\alpha/2,n-1)} \frac{s}{\sqrt{n}}\\ & = 2.093 \frac{0.0445}{\sqrt{20}} \\ & = 0.0208. \end{aligned} $$
Step 6 Determine the confidence interval
$95$% confidence interval estimate for population mean is
$$ \begin{aligned} \overline{X} - E & \leq \mu \leq \overline{X} + E\\ 12.0214 - 0.021 & \leq \mu \leq 12.0214 + 0.021\\ 12.0006 &\leq \mu \leq 12.0422. \end{aligned} $$
Thus, $95$% confidence interval estimate for population mean is $(12.0006,12.0422)$.
Interpretation
We can be $95$% confident that the mean number of ounces of coffee that a machine dispenses in 12 ounce cups is between $12.0006$ and $12.0422$.
Reference
You can read step by step tutorial on Confidence Interval for mean sigma unknown, tutorial will help you to understand how to construct confidence interval for population mean when the population standard deviation is unknown.
You can also use calculator to compute the confidence interval for population mean when the population standard deviation is unknown.
Confidence Interval for mean sigma unknown
Confidence Interval for Mean (t) Calculator
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