## Confidence Interval for the population mean when standard deviation is unknown

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

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

.

### 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 $12.0422$.