MT5762 Lecture 8

• Ultimately we are aiming for confidence intervals
• These are one of our main inferential tools
• We can see what these mean by simulation (so you can see where we want to get to)

(refer to confidence interval PPT)

We have a sample - how do we convey information about the true value? Why do it this way?

• centre - the sample mean is our best guess at the population mean.
• unbiased estimate - true value is just as likely to be above our estimate as below
• precision information should be included
• have an idea of shape of disribution for the parameter (CLT)
• need some magical multiplier that makes us correct X% of the time

• The sample mean Calcium level for our samples as 34038.12
• The sample standard deviation was 9069.678
• The standard error therefore:

\[ se(\bar{x})=\frac{s_x}{\sqrt{n}}= \frac{9069.678}{\sqrt{24}}= 1851.340 \]

• The CLT states that for large samples the distribution of sample means is closely Normal
• We know about this distribution:
  • 2 standard errors (we're talking distribution of means) about the mean captures the central 95%
  • Values >2 SE from the mean are getting rare: <5% or <1 in 20
  • Random \( \bar{x} \) more than 3 SE from \( \mu \), are even rarer: about 0.3%

So - transfer this logic to centre on the sample mean we have drawn

• It is likely the true mean lies within a 2 standard error interval:

\[ \bar{x} \pm 2\times se(\bar{x}) \]

A two-standard-error interval

\[ \begin{align*} estimate ~\pm ~~&~~~~ 2 \times standard~ error\\ \bar{x}- 2\times se(\bar{x}), ~~&~~\bar{x}+2\times se(\bar{x})\\ 30208.33-2 \times \frac{9069.678}{\sqrt{24}},~~&~~ 30208.33+2 \times \frac{9069.678}{\sqrt{24}}\\ (26505.65,~~ &~~ 33911.01) \end{align*} \]

So, the true mean Calcium level for our potting mix population, is likely (about 19/20) to be between 26505 and 33911 units.

• This inferential approach basically applies to all means
• This includes proportions (mean of a binary variable) and many model estimates e.g. regression coefficients
• We do need to be more exact than 2 SE though

Applying this to proportions

• In this study, the proportion of cannabis seeds that germinated was very low - only about 46% of seeds germinated
• How accurate is \( \hat{p}=0.46 \) as a measure of \( p \)?
• Note: proportions can be the mean of a binary variable (here germinate=1, not-germinated=0).

• We estimated this \( p \) from 100 seeds - so:
  • fixed number of trials (n = 100),
  • 46 successes
• So lends itself to a binomial distribution

• For large samples, the distribution of \( \hat{p} \) is approximately Normal with:

\[ mean~ = p ~\textrm{and standard deviation} = \sqrt{\frac{{p}(1-{p})}{n}} \]

but since we never know \( p \), we use the standard error of the sample proportion:

\[ \text{Std. Err.}~{\hat{p}} = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \]

• The sample proportion \( \hat{p} \) is an unbiased estimate of \( {p} \)

Apply the 2-SE interval:

\[ \begin{align*} estimate \pm ~~&~~2 \times standard~error\\ \hat{p}- 2\times se(\hat{p}),~~&~~\hat{p}+ 2\times se(\hat{p})\\ \hat{p}- 2\times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}},~~&~~\hat{p}+2\times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\\ 0.46-2 \times \sqrt{\frac{{0.46}(1-{0.46})}{100}},~~&~~0.46+2 \times \sqrt{\frac{{0.46}(1-{0.46})}{100}}\\ (0.3603,~~&~~0.5597) \end{align*} \]

• So quite likely (about 19/20) \( p \) lies somewhere between 36% and 56%

• We can further extend this idea to other problems
• The difference between means also gives rise to a distribution of means
• So we can inferentially estimate differences between populations

Summary statistics for Calcium values in Cannabis leaves

  CaSummary <- CaData %>% group_by(Group) %>% 
    summarise(SD = sd(Ca), Q25 = quantile(Ca, 0.25), 
              median = median(Ca), Q75 = quantile(Ca, 0.75),
              mean = mean(Ca), n = n())

  CaSummary

# A tibble: 3 x 7
  Group     SD    Q25 median    Q75   mean     n
  <fct>  <dbl>  <dbl>  <dbl>  <dbl>  <dbl> <int>
1 bhb   11587. 58000. 60000. 76000. 63615.    13
2 nth    9185. 47000. 56000. 62000. 54111.     9
3 pm     9070. 22750. 30000. 35000. 30208.    24

• Average Calcium levels in Blockhouse bay appear to be about double the potting mix and higher than Northland
• The spread in each group is fairly similar
• Potting mix and Northland appears symmetrical, whereas Blockhouse Bay is right-skewed

• We'll compare average Calcium levels for potting mix and Blockhouse Bay
• It appears Blockhouse Bay is higher than potting mix, but sampling uncertainty exists
• The parameter we want to estimate is the true mean difference between Calcium levels in Blockhouse Bay and potting mix (ie. \( \mu_{bhb}-\mu_{pm} \)).

• We have an estimate for this parameter:

\[ \begin{align*} \bar{x}_{bhb}-\bar{x}_{pm}= &63620-30210\\ = &33410 \end{align*} \]

• To build our 2-SE interval (i.e. plausible values of the true difference), we need an estimate of precision – the standard error of this estimate: we need \( se(\bar{x}_{bhb}-\bar{x}_{pm} \)
• We can find the standard errors for each estimate separately, but we need to combine them.

We can combine these standard errors using the formula below.

SE for a Difference in Means (Independent Samples)

\[ \begin{align*} se(\bar{x}_1-\bar{x}_2)=&\sqrt{se(\bar{x}_1)^2+se(\bar{x}_2)^2}\\ =& \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}} \end{align*} \]

\[ 33410 \pm 2\times \sqrt{\frac{11586.90^2}{13}+\frac{9069.678^2}{24}} \]

33410 + 2*sqrt(11586.90^2/13+9069.678^2/24)

[1] 40827.51

33410 - 2*sqrt(11586.90^2/13+9069.678^2/24)

[1] 25992.49

What do we notice about this interval for the population, or true, mean difference?

• The upper limit and the lower limit are both positive - likely \( \mu_{bhb} \) > \( \mu_{pm} \)
• This interval does not contain zero
• If zero is not contained in the interval, and zero is not a plausible value for the true mean difference
• If the lower limit for the interval is negative and the upper limit is positive – then zero is plausible.

So, we have a meaningful finding in the face of uncertainty

The same logic applies for proportions (with sufficiently large samples), the main complication is the SE

• 13 of the 33 seeds (39.4%) of seeds germinated in Blockhouse Bay soil and 24 of the 33 seeds (72.7%) of seeds germinated in potting mix ie. \[ \hat{p}_{bhb}=0.394,~~~~\hat{p}_{pm}=0.727 \]
• Germination rates for Blockhouse Bay look lower than potting mix, but sampling uncertainty exists

The approach is similar

• we want to estimate \( p_{bhb}-p_{pm} \).
• We can find the standard errors for each estimate separately but we need to combine them.

We can combine these standard errors using the formula below.

SE for a Difference between Proportions (Independent Samples)

\[ \begin{align*} se(\hat{p}_1-\hat{p}_2)=&\sqrt{se(\hat{p}_1)^2+se(\hat{p}_2)^2}\\ =& \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1}+\frac{\hat{p}_2(1-\hat{p}_2)}{n_2}} \end{align*} \]

What do we notice about this interval for the population, or true, difference between proportions?

• The upper and lower limits are both negative — it's likely \( p_{bhb} \) is smaller than \( p_{pm} \)
• This interval does not contain zero
• If zero is not contained in the interval, the zero is not a plausible value for the true difference i.e. a difference is likely

What do we notice about this interval for the population, or true, difference between proportions??

• The two standard error interval for \( p_{bhb}-p_{pm} \) was [-0.563,-0.103].
• This means we can be fairly confident that germination rates in potting mix are between 10% and 56% better, on average, than those in Blockhouse Bay.

We don't know the true SE

• When we sample data from a Normal (\( Z \)) distribution and we know the population standard deviation \( \sigma \), the sample mean is exactly Normally distributed about the true population mean, \( \mu \)
• When we don't know the population SD, we introduce a new source of variability – we use the sample SD of our data (\( s_x \)) instead of a known population SD (\( \sigma \))
• When the population SD is unknown the \( t \)-distribution is used instead

• is indexed by a parameter called the degrees of freedom (\( df \)); \( df \) is determined by the sample size (\( df=n-1 \))
• is symmetrical about zero and has a similar shape to the Normal distribution
• becomes more and more like the Normal as \( n \) (and thus \( df \)) increases; a \( t \)-distribution with small \( n \) (& \( df \)) has fatter tails and a flatter top compared with the Normal distribution
• with \( df=\infty \) and the Normal (0,1) distribution are two ways of describing the same distribution

• For large samples, the sample mean falls within about two standard errors of the population mean for approximately 95% of samples taken
• For small samples, the distribution of \( T \) is quite different to the Normal (\( Z \)).
• For small samples, we must build intervals which are more than two-standard errors to capture the population mean most of the time

• For instance, if we have a sample size of 7, the sample mean only falls within \( \pm 2 \) standard errors about 91% of the time (even when the data is exactly Normal) ie. 90.8% of the \( t \)-distribution falls between 2 standard errors when \( df=n-1=7-1=6 \) \newline \[ pr(-2 \leq T \leq 2)=0.908 \]
• If we want to ensure intervals for parameters contain the true parameter value a set proportion of times, we often need to replace \( \pm 2 \) with a bigger number