Sampling Distributions¶

Sampling Distributions for Sample Means¶

Take all possible samples of size $n$ from a population and calculate the sample mean $\overline{x}$ for each, all possible values of the sample mean are calculated

Parameters of Sampling Distributions for Sample Means¶

• Mean = $\mu$
• Standard deviation = $\dfrac{\sigma}{\sqrt{n}}$
• Standardized $z$-score = $\dfrac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

Normality of Sampling Distributions for Sample Means¶

• If the population is normally distributed, the sampling distribution for sample means is normally distributed

Central Limit Theorem¶

• If a population is not normally distributed
• A large enough random sample of size $n \geq 30$ is taken while sample values are independent
• Then the sampling distribution for sample means is approximately normally distributed

Sampling Distributions for Differences in Sample Means¶

One-Sample Problem¶

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

Two-Sample Problem¶

If one random sample of size $n_1$ is taken from one population, then a different random sample of size $n_2$ is taken from a different population that is independent to the first population

Parameters of Sampling Distribution for Differences in Sample Means¶

• Mean = $\mu_1 - \mu_2$
• Standard deviation = $\sqrt{\dfrac{\sigma_1^2}{n_1} + \dfrac{\sigma_2^2}{n_2}}$
  • Sampling with replacement or each sample size is less than 10% of the population size
  • Otherwise, $\sigma$ will be smaller
• Standardized $z$-score = $\dfrac{(\overline{x_1} - \overline{x_2}) - (\mu_1 - \mu_2)}{\sqrt{\dfrac{\sigma_1^2}{n_1} + \dfrac{\sigma_2^2}{n_2}}}$

Normality of Sampling Distributions for Differences in Sample Means¶

• If two independent populations are normally distributed, the sampling distribution for differences in sample means is also normally distributed

Sampling Distributions for Sample Proportions¶

• Population proportion, $p$, is the percentage of success
• $n$ is the sample size
• $X$ is the number of successes in a sample, following binomial distribution
• Sample proportion $\hat{p} = \dfrac{X}{n}$

If

• $np \geq 10$
• $n(1-p) \geq 10$

Then $\hat{p}$

• Normally distributed
• Mean = $p$
• Standard deviation = $\sqrt{\dfrac{p(1-p)}n}$
• $z$-score = $\dfrac{\hat{p} - p}{{\sqrt{\dfrac{(1-p)}{n}}}}$

Sampling Distributions for Differences in Sample Proportions¶

If

• $n_1p_1 \geq 10$
• $n_1(1-p_1) \geq 10$
• $n_2p_2 \geq 10$
• $n_2(1-p_2) \geq 10$

Then $\hat{p_1} - \hat{p_2}$

• Normally distributed
• Mean = $p_1 - p-2$
• Standard deviation = $\sqrt{\dfrac{p_1(1-p_1)}{n_1} + \dfrac{p_2(1-p_2)}{n_2}}$
• Standardized $z$-score = $\dfrac{(\hat{p_1} - \hat{p_2}) - (p_1 - p_2)}{\sqrt{\dfrac{p_1(1-p_1)}{n_1} + \dfrac{p_2(1-p_2)}{n_2}}}$

Biased & Unbiased Estimators¶

Estimator is used to estimate the population parameter

To know if an estimator is a good predictor

• All possible estimates from all possible samples of size $n$ must be generated
• Check to see if, on average, estimates are centered around the value of the population parameter
• An estimator is said to be unbiased if the mean of its sampling distribution equals the population parameter being estimated

Unbiased Estimators¶

• Sample mean $\overline{x}$
• Sample proportion $\hat{p}$
• Sample standard deviation $s$

Affects on Unbiased Estimators¶

• $n$ does not affect $\overline{x}$
• Greater $n$ gives more accurate $S = \dfrac{\sigma}{\sqrt{n}}$

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