Sampling Distributions for Differences in Sample Means (College Board AP® Statistics): Study Guide

Sampling distributions for differences in sample means

What is a one-sample problem?

• When one random sample of size $n$ has been taken from one population

  • with population mean $\mu$ and population standard deviation $\sigma$

    • The sample mean is $\overline{x}$

    • This is a one-sample problem

What is a two-sample problem?

• If one random sample of size $n_1$ is taken from one population with population mean ${\mu }_1$ and population standard deviation ${\sigma }_1$

  • then a different random sample of size $n_2$is taken from a different population (that is independent to the first population) with population mean ${\mu }_2$ and population standard deviation ${\sigma }_2$

    • then this is a two-sample problem

    • The sample means are ${\overline{x}}_1$ and ${\overline{x}}_2$

What is the difference in sample means?

• In a two-sample problem you can compare the sample means from separate samples of two independent populations

  • You can look at the difference in sample means, ${\overline{x}}_1-{\overline{x}}_2$

    • e.g. if ${\overline{x}}_1-{\overline{x}}_2>0$ then the mean of the first sample is greater than the mean of the second sample

What is the sampling distribution for differences in sample means?

• You can find the differences in sample means, if

  • you take all possible samples of size $n_1$ from the first population and calculate their sample means, ${\overline{x}}_1$

  • then take all possible samples of size $n_2$ from the second population and calculate their sample means, ${\overline{x}}_2$

  • then work out all the possible values that the difference ${\overline{x}}_1-{\overline{x}}_2$ can take

    • The collection of all these values is called the sampling distribution for differences in sample means

What are the mean and standard deviation of the sampling distribution for differences in sample means?

• If the first population has a population mean of ${\mu }_1$ and a population standard deviation of ${\sigma }_1$

  • and the second independent population has a population mean of ${\mu }_2$ and population standard deviation of ${\sigma }_2$

• Then the sampling distribution for differences in sample means, ${\overline{x}}_1-{\overline{x}}_2$

  • has a mean of ${\mu }_1-{\mu }_2$

  • and a standard deviation of $\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}}$

  • where $n_1$ is the size of the first sample

  • and $n_2$ is the size of the second sample

• The standard deviation of $\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}}$ assumes sampling was done with replacement

• If the sampling is carried out without replacement then the sample must satisy:

  • The randomization conditon

    • a random sampling method should be used for both samples

  • The 10% condition

    • each sample size is less than 10% of the population size

    • $n_1\le 10\%N_1$ and $n_2\le 10\%N_2$

What conditions are needed for normality?

• If in addition to the above, the two independent populations are also known to be normally distributed

  • then the sampling distribution for differences in sample means is also normally distributed

    • with mean ${\mu }_1-{\mu }_2$ and standard deviation $\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}}$

• You can use these properties to calculate probabilities involving differences in sample means, ${\overline{x}}_1-{\overline{x}}_2$, as they follow a normal distribution

  • Its standardized z-statistic is $\frac{\left({\overline{x}}_1-{\overline{x}}_2\right)-\left({\mu }_1-{\mu }_2\right)}{{\displaystyle \sqrt{\frac{{\displaystyle {\sigma }_1^2}}{{\displaystyle n_1}}+\frac{{\displaystyle {\sigma }_2^2}}{{\displaystyle n_2}}}}}$

    • ${\mu }_1$, ${\mu }_2$, ${\sigma }_1$ and ${\sigma }_2$ will be given in the question

What do I do if the populations are not normally distributed?

• If the populations are not normally distributed, then the sampling distribution for differences in sample means is not guaranteed to be normally distributed

• However, despite not knowing its shape, the sampling distribution for differences in sample means still has a

  • mean of ${\mu }_1-{\mu }_2$ and a standard deviation of $\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}}$

    • i.e. you can always write these down, even though the distribution is unknown

Can I use the Central Limit theorem if the populations are not normally distributed?

• If the populations are not normally distributed, but both sample sizes are greater than or equal to 30 ($n_1\ge 30$ and $n_2\ge 30$)

  • then the Central Limit theorem can be applied

  • meaning the sampling distribution for differences in sample means is approximately normally distributed with the parameters above

    • i.e. mean ${\mu }_1-{\mu }_2$ and standard deviation $\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}}$

• You can use these properties to estimate probabilities involving differences in sample means, ${\overline{x}}_1-{\overline{x}}_2$, as they follow an approximate normal distribution

  • Its standardized z-statistic is $\frac{\left({\overline{x}}_1-{\overline{x}}_2\right)-\left({\mu }_1-{\mu }_2\right)}{{\displaystyle \sqrt{\frac{{\displaystyle {\sigma }_1^2}}{{\displaystyle n_1}}+\frac{{\displaystyle {\sigma }_2^2}}{{\displaystyle n_2}}}}}$

    • ${\mu }_1$, ${\mu }_2$, ${\sigma }_1$ and ${\sigma }_2$ will be given in the question