A hat on a greek letter indicates an estimator. For example, when you see \(\hat \mu\) it means an estimator of the population mean \(\mu\), such as \(\bar{x}\).
the \(Z\) distribution is another name for the Standard Normal Distribution, \(N(0, 1)\)
One Sample Categorical Data (Approximate)
Normal Approximation (\(\mu\) and \(\sigma\) are known parameters)
One Sample Continuous Data (\(\mu\) known, \(\sigma\) unknown)
As always, feel free to work in pairs or small groups to accomplish these exercises!
Suppose that the true average IQ is 95. Using the lead-IQ dataset as a sample, perform a test to see if the children have an average IQ that is different from the true average. Also, create a 95% confidence interval for the mean IQ based on this data.
We will be using a new distribution here–the t-distribution. Open the functions document to learn about the functions we will be using for t-distribution calculations.
Define our null hypothesis to be \(H_0: \mu = 95\)
leadIQ <- read.delim('https://raw.githubusercontent.com/IowaBiostat/data-sets/main/lead-iq/lead-iq.txt')
mu <- 95
mu.hat <- mean(leadIQ$IQ)
s <- sd(leadIQ$IQ)
n <- length(leadIQ$IQ)
df <- n-1
t <- (mu.hat-mu)/(s/sqrt(n))
pval <- 2*pt(t,df)
pval[1] 0.002981458The result, \(p- = 0.0029815\), indicats strong evidence that the true average IQ of children in this dataset is not 95.
ci <- mu.hat + qt(c(.025,.975),n-1) * s/sqrt(n)
ci[1] 88.52022 93.64107This gives a confidence interval of (88.52, 93.641). We could also use the t.test function (as shown below) for this dataset, and it provides us both the p-value and the 95% confidence interval. This function works similarly to the ‘binom.test’ function.
t.test(leadIQ$IQ, mu=95)
One Sample t-test
data: leadIQ$IQ
t = -3.03, df = 123, p-value = 0.002981
alternative hypothesis: true mean is not equal to 95
95 percent confidence interval:
88.52022 93.64107
sample estimates:
mean of x
91.08065 Suppose that the current commonly used screening test for breast cancer has a sensitivity of 68%. A new screening test was used to test 200 breast cancer patients, in which 147 patients tested positive.
p <- 147/200
p + c(-1, 1) * qnorm(0.975) * sqrt((p*(1-p))/200)[1] 0.6738355 0.7961645Define \(H_0:0.68\)
p_hat <- 147/200
p0 <- 0.68
z <- (p_hat - p0) / (sqrt((p0*(1-p0))/200))
2*pnorm(z, lower.tail = F) # don't forget to make it two-sided![1] 0.09542845Hint: Use the binomial distribution where X = number of positive test results.
binom.test(x = 147, n = 200, p = 0.68)
Exact binomial test
data: 147 and 200
number of successes = 147, number of trials = 200, p-value = 0.1111
alternative hypothesis: true probability of success is not equal to 0.68
95 percent confidence interval:
0.6681299 0.7947609
sample estimates:
probability of success
0.735 A patient recently diagnosed with Alzheimer’s disease takes a cognitive abilities test. The population mean of this test is \(\mu = 52\) and the population standard deviation is \(\sigma = 5\). Assume the cognitive abilities test scores are normally distributed. Find the answers to the following questions with the Z distribution table, your calculators, or in R. Remember the Z table gives you the left-tailed probability.
First, standardize the values of 47 and 56 into \(z\) values, then use standard normal curve to find area between these \(z\) values.
mu <- 52
sd <- sqrt(25)
z1 <- (47 - mu) / sd
z2 <- (56 - mu) / sd
pnorm(z2) - pnorm(z1)[1] 0.6294893First, standardize 60 into a z value, then use standard normal curve to find area above this \(z\) value.
mu <- 52
sd <- sqrt(25)
n <- 9
z <- (60 - mu) / (sd / sqrt(n))
pnorm(z, lower.tail=FALSE)[1] 7.933282e-07First, standardize 43 into a z value, then use standard normal curve to find area below this \(z\) value.
mu <- 52
sd <- sqrt(25)
n <- 9
z <- (43 - mu) / (sd)
perc_alt_tx <- pnorm((z))
perc_alt_tx[1] 0.03593032Find z value and “unstandardize” back to the scale of the data.
z <- qnorm(.271, lower.tail=FALSE)
52 + z*5[1] 55.04896From part c, we found the probability that a patient would be considered for alternative treatment. We use this probability in the binomial distribution, Binom(\(n,\pi\)) where \(\pi = 0.0359303\) and \(n=25\).
Remember that using lower.tail=FALSE shades the binomial probability from the right and stops at the value greater than ours. That is \(P(X > 1)\)
pbinom(1,
25,
perc_alt_tx, # found in part c, saved to this variable
lower.tail = FALSE) # shading to the right[1] 0.2261473Wilson’s orchard’s pumpkins’ weights are known to follow a normal distribution with population mean \(\mu = 18 lbs.\) and standard deviation \(\sigma = 4 lbs\). Each year Wilson’s orchard randomly selects 4 pumpkins and measures the mean weight of the pumpkins.
\(\bar{X} \sim N(\mu, \frac{\sigma}{\sqrt{n}}) \sim N(18, \frac{4}{\sqrt{4}}=2)\)
Standardize 16 to find the \(z\) value. Use pnorm to find the probability.
z <- (16 - 18) / sqrt(4)
pnorm(z)[1] 0.1586553Standardize 21 to find the \(z\) value. Use pnorm to find the probability. Remember to either subtract from 1 or use lower.tail=FALSE to find the shaded area to the right.
z <- (21 - 18) / sqrt(4)
1 - pnorm(z)[1] 0.0668072First, standardize the values of 14 and 20 to be \(z\) values.
Second, use the normal distribution to find the probability of a sample mean being between 14 and 20.
Third, use this probability for \(\pi\) in the binomial(\(n, \pi\)) probability distribution. Make sure to correctly shade above 1 \(P(X \ge 2) = 1-P(X<2) = 1 - P(X\le1)\)1.
z1 <- (14 - 18) / sqrt(4)
z2 <- (20 - 18) / sqrt(4)
(p <- pnorm(z2) - pnorm(z1))[1] 0.8185946(ge2 <- pbinom(1, 5, p, lower.tail = FALSE))[1] 0.9953711If this doesn’t make sense, ask your instructor about this mathematical statement.↩︎