Suppose \(X \sim \text{Normal}(\mu, \sigma).\)

Define a new random variable \(Z\) by \(Z = \frac{X - \mu}{\sigma}.\)

Fact: \(Z\) also follows a Normal distribution. What are the mean (i.e., expected value) and variance of \(Z\)?

"Recall" that if \(X\) is a random variable and \(a\) is a number, then

\(E(aX) = a E(X)\)

\(E(X + a) = E(X) + a\)

\(\text{SD}(aX) = a^2 \text{SD}(X)\)

• The Apgar score gives a quick sense of a baby's physical health, and is used to determine whether a baby needs immediate medical care.
• It ranges from 0 (critical health problems) to 10 (no health problems).
• Let's try to estimate the proportion of babies in the population who have an Apgar score of 10 using a sample of \(n = 300\) babies.

babies <- mutate(babies, apgar_eq_10 = (apgar5 == 10))
head(babies[, c("gestation", "apgar5", "apgar_eq_10")])

## # A tibble: 6 x 3
##   gestation apgar5 apgar_eq_10
##       <int>  <int>       <lgl>
## 1        41      9       FALSE
## 2        47      6       FALSE
## 3        37      9       FALSE
## 4        35      9       FALSE
## 5        37     10        TRUE
## 6        35      9       FALSE

table(babies$apgar_eq_10)

## 
##  FALSE   TRUE 
## 236381  21648

table(babies$apgar_eq_10) / nrow(babies)

## 
##      FALSE       TRUE 
## 0.91610245 0.08389755

babies_sample <- sample_n(babies, size = 300)
table(babies_sample$apgar_eq_10) / nrow(babies_sample)

## 
##     FALSE      TRUE 
## 0.8866667 0.1133333

• Our estimate of the population proportion based on this sample is WRONG!

• Can we get a sense of how wrong it might be, using only the data in our sample?

• On Monday we said that if \(n\) is big enough, \[\widehat{p} \sim \text{Normal}\left(p, \sqrt{\frac{p(1-p)}{n}}\right)\]
• In this case, the population proportion is \(p = 0.084\), and \(n = 300\), so… \[\widehat{p} \sim \text{Normal}\left(0.084, 0.016\right)\]

\[\widehat{p} \sim \text{Normal}\left(0.084, 0.016\right)\]

• For about 68% of samples of size \(n\) we could take, the sample proportion \(\widehat{p}\) will be within \(\pm\) 1 standard deviation (\(\pm\) 0.016) of the population proportion \(p = 0.084\)

• For about 95% of samples of size \(n\) we could take, the sample proportion \(\widehat{p}\) will be within \(\pm\) 2 standard deviations (\(\pm\) 0.032) of the population proportion \(p = 0.084\)

• If \(\widehat{p}\) is within \(\pm\) 2 standard deviations of \(p\), then \(p\) is contained in the interval

\[[\widehat{p} - 2 \, \text{SD}(\widehat{p}), \widehat{p} + 2 \, \text{SD}(\widehat{p})]\]

• We are "95% Confident" that the population proportion \(p\) is in the interval [0.081, 0.145].
• For 95% of samples, an interval constructed this way contains \(p\).

• The 95% confidence interval from a couple of slides ago was

\[[\widehat{p} - 2 \, \text{SD}(\widehat{p}), \widehat{p} + 2 \, \text{SD}(\widehat{p})]\]

• But SD\((\widehat{p})\) depends on the (unknown) population parameter \(p\):

\[\text{SD}(\widehat{p}) = \sqrt{\frac{p(1-p)}{n}}\]

• The 95% confidence interval from a couple of slides ago was

\[[\widehat{p} - 2 \, \text{SD}(\widehat{p}), \widehat{p} + 2 \, \text{SD}(\widehat{p})]\]

• But SD\((\widehat{p})\) depends on the (unknown) population parameter \(p\):

\[\text{SD}(\widehat{p}) = \sqrt{\frac{p(1-p)}{n}}\]

• We can estimate SD\((\widehat{p})\) by plugging our estimate of \(p\) into this formula. An estimate of the standard deviation of a sampling distribution is called a standard error:

\[\text{SE}(\widehat{p}) = \sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}\]

• What if we want a 90% CI instead of a 95% CI?

• We need to know: 90% of sample means will be within how many standard deviations of the population mean?

• This is called the critical value, and denoted by \(z^*\)

• Our new CI formula: \([\widehat{p} - z^* \text{SE}(\widehat{p}), \widehat{p} + z^* \text{SE}(\widehat{p})]\)

• For a 90% CI, the critical value is the 95th percentile of a Normal(0, 1) distribution:

qnorm(0.95, mean = 0, sd = 1)

## [1] 1.644854

• More generally: for a \((1 - \alpha) \times 100\)% CI, the critical value is the (1 - )th quantile of a Normal(0, 1) distribution:
  • \(\alpha = 0.1\) -> 90% CI. 1 - 0.05 = 0.95th quantile.
  • \(\alpha = 0.05\) -> 95% CI. 1 - 0.025 = 0.975th quantile.
  • \(\alpha = 0.01\) -> 99% CI. 1 - 0.005 = 0.995th quantile.

• \(\widehat{p} \sim \text{Normal}(p, \text{SD}(\widehat{p}))\)

• For a 90% CI, we need the total area to the left of \(p + z^* \text{SD}(\widehat{p})\) to be 0.95, in a Normal(\(p\), SD(\(\widehat{p}\))) distribution.

• \(\widehat{p} \sim \text{Normal}(p, \text{SD}(\widehat{p}))\)

• For a 90% CI, we need the total area to the left of \(p + z^* \text{SD}(\widehat{p})\) to be 0.95, in a Normal(\(p\), SD(\(\widehat{p}\))) distribution.

• Let's define \(Z = \frac{\widehat{p} - p}{\text{SD}(\widehat{p})}\). Then \(Z \sim \text{Normal}(0, 1)\) (see warmup)

• For a 90% CI, area to the left of \(p + z^* \text{SD}(\widehat{p})\) is 0.95.

• Define \(Z = \frac{\widehat{p} - p}{\text{SD}(\widehat{p})}\). Then \(Z \sim \text{Normal}(0, 1)\)

• Area to the left of \(\frac{[p + z^* \text{SD}(\widehat{p})] - p}{\text{SD}(\widehat{p})} = z^*\) is 0.95.

• CI formula: \([\widehat{p} - z^* \text{SE}(\widehat{p}), \widehat{p} + z^* \text{SE}(\widehat{p})]\)
• Standard Error of \(\widehat{p}\): \(\text{SE}(\widehat{p}) = \sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}\)
• Critical Value: \(z^*\) is the 97.5th percentile of a standard normal distribution if we want a 95% CI
  • Use qnorm function in R
• Margin of Error: \(z^* \text{SE}(\widehat{p})\) (how much we add and subtract from the point estimate \(\widehat{p}\))
• Interpretation: In repeated sampling, a confidence interval constructed using this procedure contains the population parameter for 95% of samples (or whatever your confidence level is).

• Two outcomes (that are relevant to this analysis)
• Same probability of success
• People/items in our sample are independent
  • Think about how data were collected/if there is a connection between units
  • 10% Condition: Sample size less than 10% of population size?
• Sample size large enough to use normal approximation to the sampling distribution:
  • \(np \geq 10\) and \(n(1 - p) \geq 10\)
  • … but we don't actually know \(p\)!
  • Check that there are at least 10 "successes" and 10 "failures" in the data set.

table(babies_sample$apgar_eq_10) / nrow(babies_sample)

## 
##     FALSE      TRUE 
## 0.8866667 0.1133333

p_hat <- 0.1133333
se_p_hat <- sqrt(p_hat * (1 - p_hat) / 300)
z_star <- qnorm(0.975, mean = 0, sd = 1)
p_hat - z_star * se_p_hat

## [1] 0.07746206

p_hat + z_star * se_p_hat

## [1] 0.1492045

library(mosaic)
confint(binom.test(
  babies_sample$apgar_eq_10,
  conf.level = 0.95,
  ci.method = "wald",
  success = TRUE))

##   probability of success      lower     upper level
## 1              0.1133333 0.07746209 0.1492046  0.95