Confidence intervals for the mean

The purpose of a confidence interval

The purpose of a confidence interval for the mean is to estimate the population mean $\mu$ using a single sample mean $\bar{x}$. In most real situations, the population standard deviation $\sigma$ is unknown, so we use the sample standard deviation $s$ as an estimate of $\sigma$.

Definitions

Confidence interval for a mean

An interval that gives a range of plausible values for the true population mean $\mu$, based on the sample mean $\bar{x}$. When $\sigma$ is unknown, we use the t-distribution.

Standard error of the mean

An estimate of the standard deviation of the sampling distribution of $\bar{x}$.

$$\begin{array}{rl}\text{SE}& =\frac{s}{\sqrt{n}}\end{array}$$

Margin of error ($\text{ME}$)

The amount we add and subtract from the sample mean to form the confidence interval.

$$\begin{array}{rl}\text{ME}& =t^{\ast }\cdot \text{SE}\end{array}$$

Before using a t-interval, check these conditions:

• Randomness: The sample should be random (ideally a simple random sample).
• Independence: The sample should be no more than $10\%$ of the population.
• Normality (for t-procedures): The population is approximately normal, or the sample size is large enough for the central limit theorem to apply.

A t-confidence interval for a mean has the form:

$$\begin{array}{r}\bar{x}\pm t^{\ast }\cdot \frac{s}{\sqrt{n}}\end{array}$$

Practice problem

Example:

A manufacturer is testing a new laptop model’s battery life. A simple random sample of $12$ laptops were timed and showed a sample mean battery life of $\bar{x}=52.4$ hours with a sample standard deviation $s=3.1$ hours. Assume that the population of battery lives is approximately normally distributed.

Part a.

Give a $95\%$ confidence interval for the true mean battery life of this model.

Method 1: Without a TI-84 calculator

Start by calculating the standard error:

$$\begin{array}{rl}\text{SE}& =\frac{s}{\sqrt{n}}\\ & =\frac{3.1}{\sqrt{12}}\\ & \approx 0.895\end{array}$$

Next, find the critical value from a t-table. Use $t_{0.975,11}\approx 2.2$ because:

• The degrees of freedom are $df=n-1=12-1=11$.
• A $95\%$ confidence interval keeps the middle $95\%$ of the t-distribution, leaving $2.5\%$ in each tail.
• That’s why we use $1-0.025=0.975$ (not $0.95$) when locating the critical value.

Use the standard error and critical value to find the margin of error:

$$\begin{array}{rl}{\text{ME}}_{95}& =t^{\ast }\cdot \text{SE}\\ & \approx 2.2\times 0.895\\ & \approx 1.97\end{array}$$

Construct the interval:

$$\begin{array}{rl}\bar{x}\pm \text{ME}& =52.4\pm 1.97\end{array}$$

Method 2: With a TI-84 calculator

• Choose STAT, then TESTS, then 8: TInterval.
• Choose Stats and enter $\bar{x}=52.4$, $S_x=3.1$, and $n=12$.
• Set C-Level $=0.95$ and select Calculate, then ENTER.

Solution:

Part b.

Based on this interval, do you think the true mean battery life is different from $50$ hours? Explain your reasoning.

To answer this, check whether $50$ falls inside the $95\%$ confidence interval $(50.43,54.37)$.

A TI-84 can also confirm the answer with a one-sample t-test:

• Choose STAT, then TESTS, then 2: T-Test.
• Choose Stats. Input ${\mu }_0=50$, $\bar{x}=52.4$, $S_x=3.1$, $n=12$.
• Choose test type $\ne {\mu }_0$.
• Go to Calculate, then ENTER.

Solution:

Part c.

Determine a $99\%$ confidence interval. What effect does moving from $95\%$ to $99\%$ have on this interval and why?

Method 1: Without a TI-84 calculator

The standard error stays the same:

$$\begin{array}{rl}\text{SE}& \approx 0.895\end{array}$$

Find the new critical value from a t-table. Use $t_{0.995,11}\approx 3.1$ because:

• The degrees of freedom are still $df=n-1=12-1=11$.
• A $99\%$ confidence interval keeps the middle $99\%$ of the distribution, leaving $0.5\%$ in each tail.
• That’s why we use $1-0.005=0.995$ (not $0.99$) when locating the critical value.

Compute the new margin of error:

$$\begin{array}{rl}{\text{ME}}_{99}& =t^{\ast }\cdot \text{SE}\\ & \approx 3.1\times 0.895\\ & \approx 2.779\end{array}$$

Construct the interval:

$$\begin{array}{rl}\bar{x}\pm \text{ME}& =52.4\pm 2.779\end{array}$$

Method 2: With a TI-84 calculator

Follow the same steps as in Part a, but set C-Level $=0.99$.

Moving from $95\%$ to $99\%$ makes the interval wider. A higher confidence level requires capturing more of the t-distribution, which increases the critical value $t^{\ast }$ and therefore increases the margin of error.

Solution:

Part d.

If the sample mean remained $52.4$ and the sample standard deviation remained $3.1$ but the sample size was $24$ instead of $12$, what would the $95\%$ confidence interval be? How does changing the sample size impact the confidence interval and why does this occur?

Method 1: Without a TI-84 calculator

The only change is that $n=24$, which affects the degrees of freedom, the standard error, and the margin of error.

Find the new standard error:

$$\begin{array}{rl}{\text{SE}}_{24}& =\frac{3.1}{\sqrt{24}}\\ & \approx 0.633\end{array}$$

Find the new degrees of freedom:

$$\begin{array}{rl}df& =n-1\\ & =24-1\\ & =23\end{array}$$

From the t-table:

$$\begin{array}{rl}{t}_{0.975,23}& \approx 2.068657\end{array}$$

Compute the margin of error:

$$\begin{array}{rl}\text{ME}& =t^{\ast }\cdot {\text{SE}}_{24}\\ & =2.068657\cdot 0.633\\ & \approx 1.3\end{array}$$

Construct the interval:

$$\begin{array}{rl}\bar{x}\pm \text{ME}& =52.4\pm 1.3\end{array}$$

Method 2: With a TI-84 calculator

Follow the same steps as in Part a, but use $n=24$ instead of $n=12$.

Solution: