AP Statistics 4.1-4.2: Sampling Distributions for Sample Means

Sampling distribution of the sample mean

For random samples of size $n$ from a population with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of $\bar{x}$ has mean $\mu$ and standard deviation $\dfrac{\sigma}{\sqrt{n}}$, called the standard error of the sample mean.

When is the sampling distribution of $\bar{x}$ approximately normal?

If the population is approximately normal, the sampling distribution of $\bar{x}$ is approximately normal for any sample size. By the central limit theorem, it is also approximately normal for sufficiently large random samples, even when the population is not normal, provided the data are independent and the population does not have extreme skewness or outliers.

Student's t-distribution

A continuous, symmetric, bell-shaped distribution used to make inferences about a population mean when the population standard deviation $\sigma$ is unknown and is estimated with the sample standard deviation $s$. Compared with the standard normal distribution, it has heavier tails.

Why is the Student's t-distribution used instead of the standard normal distribution for a population mean?

When $\sigma$ is unknown, replacing it with $s$ adds uncertainty to the standardized statistic. The t-distribution accounts for this extra variability, especially when the sample size is small.

For a random sample of size $n$ from an approximately normal population with unknown $\sigma$, what is the sampling distribution of the one-sample t-statistic?

The statistic $t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}$ follows a Student's t-distribution with $df=n-1$ degrees of freedom.

Degrees of freedom for a one-sample t procedure

For inference about one population mean, $df=n-1$. One degree of freedom is lost because the sample deviations from $\bar{x}$ must sum to zero.

How does the t-distribution compare with the standard normal distribution?

Both are symmetric about zero and bell-shaped, but the t-distribution has heavier tails and a lower, wider center. As $df$ increases, the t-distribution approaches the standard normal distribution.

What does a t-score measure in a one-sample mean problem?

It measures how many estimated standard errors the sample mean $\bar{x}$ is from the hypothesized or true population mean: $t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}$.

Conditions for using a one-sample t confidence interval for a population mean

The data should come from a random sample or randomized process, observations should be independent, and the population should be approximately normal. For a small sample, strong skewness or outliers can make the procedure unreliable.

Does random sampling guarantee that the population is normally distributed?

No. Random sampling and normality are separate assumptions: random sampling supports independence and representativeness, while approximate normality supports the t-based sampling model, especially for small $n$.

When is the size of the overall population usually relevant to a one-sample t procedure?

It is generally not relevant unless the population is very small relative to the sample. If sampling without replacement, a common additional guideline is that the sample should be no more than about 10% of the population.

One-sample t confidence interval for a population mean

When $\sigma$ is unknown, the confidence interval is $\bar{x}\pm t_{\alpha/2}\dfrac{s}{\sqrt{n}}$, or equivalently $(\bar{x}-EBM,\bar{x}+EBM)$, where $EBM=t_{\alpha/2}\dfrac{s}{\sqrt{n}}$ and $df=n-1$.

What does $t_{\alpha/2}$ represent in a two-sided confidence interval?

$t_{\alpha/2}$ is the positive critical t-value with area $\alpha/2$ to its right, using $df=n-1$. For confidence level $CL$, $\alpha=1-CL$.

How are confidence level and tail area related for a two-sided t interval?

If the confidence level is $CL$, then $\alpha=1-CL$, and each tail has area $\alpha/2$. The area to the left of the positive critical value is $1-\alpha/2$.

How do sample size and confidence level affect the margin of error in a one-sample t interval?

Increasing $n$ decreases the standard error $s/\sqrt{n}$ and usually narrows the interval. Increasing the confidence level increases the critical value and widens the interval.

Why are t-based confidence intervals generally wider than corresponding z-based intervals when $\sigma$ is estimated?

The t critical value is larger than the corresponding z critical value for finite degrees of freedom because the t-distribution has heavier tails. This wider interval reflects the additional uncertainty from estimating $\sigma$ with $s$.

Notation $T\sim t_{df}$

The notation means that the random variable $T$ follows a Student's t-distribution with the specified degrees of freedom. For a one-sample mean with sample size $n$, write $T\sim t_{n-1}$.

A sample contains $20$ observations. What degrees of freedom should be used for a one-sample t interval?

Use $df=n-1=20-1=19$, so the reference distribution is $T\sim t_{19}$.

How can a calculator or computer find a t critical value for a confidence interval?

For a two-sided interval with confidence level $CL$, find the t-value with left-tail area $1-\alpha/2$ and $df=n-1$. On calculators with inverse t functionality, this can be computed as $\operatorname{invT}(1-\alpha/2,df)$.

What does the calculator function $\operatorname{tcdf}(a,b,df)$ calculate?

It calculates the probability that a t random variable with the given degrees of freedom lies between lower bound $a$ and upper bound $b$: $P(a<T<b)$.

How should a one-sample t confidence interval for a population mean be interpreted?

A confidence level describes the long-run success rate of the method: in repeated random sampling, approximately $CL\times100\%$ of intervals would contain the true mean $\mu$. It is not correct to say that the fixed parameter $\mu$ has a $CL$ probability of lying in this particular interval.

A sample has $n=15$, $\bar{x}=8.2267$, $s=1.6722$, and a 95% confidence level. What one-sample t interval results?

Here $df=14$ and $t_{0.025}\approx2.14$. The margin of error is $2.14\left(\dfrac{1.6722}{\sqrt{15}}\right)\approx0.924$, giving the interval $(7.30,9.15)$.

For a 90% confidence interval with $n=20$ observations, what critical-value tail area and degrees of freedom are used?

$\alpha=0.10$, so each tail has area $\alpha/2=0.05$, and $df=20-1=19$. Thus use $t_{0.05}$ with $19$ degrees of freedom.

A 90% one-sample t interval has $\bar{x}=127.45$, $s=25.965$, and $n=20$. Using $t_{0.05,19}=1.729$, what is the interval?

$EBM=1.729\left(\dfrac{25.965}{\sqrt{20}}\right)\approx10.038$. The interval is $127.45\pm10.038$, or approximately $(117.41,137.49)$.

What happens to the t-distribution as the degrees of freedom approach infinity?

It converges to the standard normal distribution $N(0,1)$ because the sample standard deviation becomes a more stable estimate of $\sigma$.

What is the main practical distinction between a z interval and a one-sample t interval for a population mean?

A z interval uses a known population standard deviation $\sigma$, while a t interval is used when $\sigma$ is unknown and replaced by $s$. The t interval uses $df=n-1$ and is typically wider for finite samples.