STAT 1010 - Fall 2022

- The shape of the distribution of the sample mean and why it is important
- When we can assume a normal distribution for the sample mean
- Control charts and what they are used for
- Find control limits

- SRS
- Stratified sampling
- Cluster sampling
- it would be nice to sample repeatedly to see how the mean values compare

- reduces variation
- more normal than the original distribution

- Find kurtosis(\(K_4\))
- If \(n > 10 |K_4|\), where \(n\) is the sample size, then a normal model adequately approximates the distribution of the sample mean \(\bar{X}\).
- If we know the data come from a normal distribution this is also true.

\[ SD(\bar{X}) = SE(\bar{X}) = \frac{\sigma}{\sqrt{n}} \]

If \(X \sim N(\mu_X, \sigma_X^2)\)

\[ \bar{X} \sim N(\mu = \mu_X, \sigma^2 = \frac{\sigma_X^2}{n}) \]

- Let \(Y \sim N(\mu = 5, \sigma^2 = 16)\), find the distribution of the mean of repeated samples of size 4.
- \(\bar{Y} \sim N(\mu = 5, \sigma^2 = 4)\)

- if mean production is outside certain values, we may need to stop and recallibrate machinery
- these values are called
*control limits* - \(\mu - L \leq \bar{X} \leq \mu + L\)
- \(\mu - L\) and \(\mu + L\) are
*control limits*

- False positives - type 1 error
- act when you should not
- probabiliy of occurence denoted \(\alpha\)

- False negatives - type 2 error
- don’t act when you should
- probabiliy of occurence denoted \(\beta\)

If \(\bar{X} \sim N(\mu = 12, \sigma^2 = 2.3)\) how can we find the control limits?

- Set the control limits and find the \(\alpha\) value
- We want the control limits to be between 10 and 14
- \(Pr(\bar{X} < 10 \textrm{ or } \bar{X} > 14)\)
- \(\begin{aligned}P(\bar{X} < 10) &= P(\frac{\bar{X} - \mu_\bar{X}}{\sigma_\bar{X}} < \frac{10-\mu_\bar{X}}{\sigma_\bar{X}}) \\ & = P(Z < \frac{10-12}{\sqrt{2.3}} = -1.318761)\end{aligned}\)
`pnorm(-1.318761)`

\(\approx 0.09362451\)`pnorm(10, mean = 12, sd = sqrt(2.3))`

If \(\bar{X} \sim N(\mu = 12, \sigma^2 = 2.3)\) how can we find the control limits?

- Set the control limits and find the \(\alpha\) value
- We want the control limits to be between 10 and 14
- \(\begin{aligned}P(\bar{X} > 14) &= P(\frac{\bar{X} - \mu_\bar{X} }{\sigma_\bar{X}} > \frac{14-\mu_\bar{X}}{\sigma_\bar{X}}) \\ & = P(Z > \frac{14-12}{\sqrt{2.3}} = 1.318761)\end{aligned}\)
`1 - pnorm(1.318761)`

\(\approx 0.09362451\)`1 - pnorm(14, mean = 12, sd = sqrt(2.3))`

- the \(\alpha\) value is \(19\%\) which is very high!

- Set the \(\alpha\) value and find the control limits

- We want the \(\alpha\) to be \(0.025\)
- \(Pr(\bar{X} < z_{0.0125} \textrm{ or } \bar{X} > z_{0.0125})\)
`qnorm(0.0125)`

\(\approx -2.241403\)- \(\begin{aligned}-2.241403 =& \frac{X - \mu_\bar{X}}{\sigma_\bar{X}}\\ & = \frac{X - 12}{\sqrt{2.3}} \\ &= -2.241403\sqrt{2.3} +12 = 8.600744 \end{aligned}\)
`qnorm(0.0125, mean = 12, sd = sqrt(2.3))`

- Set the \(\alpha\) value and find the control limits

- We want the \(\alpha\) to be \(0.025\)
- \(Pr(\bar{X} < z_{0.0125} \textrm{ or } \bar{X} > z_{0.0125})\)
`qnorm(1- 0.0125)`

\(\approx 2.241403\)- \(\begin{aligned}2.241403 =& \frac{X - \mu_\bar{X}}{\sigma_\bar{X}}\\ & = \frac{X - 12}{\sqrt{2.3}} \\ &= 2.241403\sqrt{2.3} +12 = 15.39926 \end{aligned}\)
`qnorm(1-0.0125, mean = 12, sd = sqrt(2.3))`

We are not testing once, but multiple times. Assuming independence:

\(\begin{aligned}P(\textrm{within limits for 10 days}) =& P(\textrm{within limits for day 1}) \cdot P(\textrm{within limits for day 2}) \cdot \dots \cdot P(\textrm{within limits for day 10})\\ &= 0.975^{10} \approx 0.7763296 \end{aligned}\)

There is a \(1-0.7763296 = 0.2236704\) percent false positive rate

Management must decide if there is a false positive by checking for mechanical errors and inspecting equipment

Adjust \(\alpha\) value to address this

X-bar charts are slow to detect under or over filling

*S-Chart*tracks the standard deviation from sample to sample*R-Chart*tracks the range from sample to sample