Analyzing proportions (Part 2)

• Homework #6 has been posted (due Monday)
• Reading Assignment for Friday: Chapter 8, Fitting probability models to frequency data (QUIZ)

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The sampling distribution for the sample estimate of the proportion is a “scaled” binomial distribution.

\[ \hat{p} = \frac{\mathrm{Number \ of \ successes \ in \ sample}}{\mathrm{Total \ sample \ size}} = \frac{\hat{X}}{n} \]

Definition: The binomial distribution provides the probability distribution for the number of “successes” in a fixed number of independent trials, when the probability of success is the same in each trial.

Properties:

• Number of trials is fixed (\( n \))
• Probability of success in each trial (\( p \)) is the same in every trial
• Separate trials are independent

Definition: The binomial distribution provides the probability distribution for the number of “successes” in a fixed number of independent trials, when the probability of success is the same in each trial.

Properties:

• Number of trials is fixed (\( n \)) [sample size]
• Probability of success in each trial (\( p \)) is the same in every trial [proportion of successes in population]
• Separate trials are independent [random sample]

If we have \( n \) trials, and the probability of success in each trial is \( p \), we have \[ \mathrm{Pr[}X \mathrm{ \ successes]} = \left(\begin{array}{c}{n \\ X}\end{array}\right)p^{X}(1-p)^{n-X}, \] where \[ \left(\begin{array}{c}{n \\ X}\end{array}\right) = \frac{n!}{X!(n-X)!}, \] and \[ n! = n\times(n-1)\times(n-2)\cdots 2\times 1. \]

Why?

To figure out Pr[\( X \) successes], first ask

Question: “What are all different outcomes of \( X \) successes in \( n \) trials?”

Example: Suppose \( n=3 \) and \( X=2 \).

\[ 2 \ \mathrm{successes} = \{SSF, SFS, FSS\} \]

\[ \mathrm{Pr}[SSF] = \mathrm{Pr}[S]\times \mathrm{Pr}[S]\times \mathrm{Pr}[F] = p^2(1-p) \]

\[ \mathrm{Pr}[SFS] = \mathrm{Pr}[S]\times \mathrm{Pr}[F]\times \mathrm{Pr}[S] = p^2(1-p) \]

\[ \mathrm{Pr}[FSS] = \mathrm{Pr}[F]\times \mathrm{Pr}[S]\times \mathrm{Pr}[S] = p^2(1-p) \]

To figure out Pr[\( X \) successes], first ask

Question: “What are all different outcomes of \( X \) successes in \( n \) trials?”

Example: Suppose \( n=3 \) and \( X=2 \).

\[ 2 \ \mathrm{successes} = \{SSF, SFS, FSS\} \]

\[ \mathrm{Pr}[SSF] = \mathrm{Pr}[SFS] = \mathrm{Pr}[FSS] = p^2(1-p) = p^X(1-p)^{n-X} \]

How many ways are there to have 2 successes in 3 trials? 3 choose 2!!

\[ \mathrm{Pr[2 \ successes]} = \left(\begin{array}{c}{3 \\ 2}\end{array}\right)p^2(1-p) \]

Question: Given \( p=0.3 \) and \( n=20 \), what is Pr[6 successes]? (Write out using notation)

Answer: \[ \mathrm{Pr[}6 \mathrm{ \ successes]} = \left(\begin{array}{c}{20 \\ 6}\end{array}\right)0.3^{6}\times 0.7^{14}. \]

Using R and dbinom,

(ans <- dbinom(x=6, size=20, prob=0.3))

[1] 0.191639

Thus, Pr[6 successes] = 0.191639.

Let's plot the distribution:

barplot(pdist, 
        names.arg=0:10, 
        col="firebrick", 
        xlab="X (Number of successes)", 
        ylab="Probability")

Let's just look at a lower probability of success, say \( p=0.1 \):

pdist <- dbinom(0:10, 10, 0.1)

barplot(pdist, 
        names.arg=0:10, 
        col="firebrick", 
        xlab="X (Number of successes)", 
        ylab="Probability")

\[ \hat{p} = \frac{\mathrm{Number \ of \ successes \ in \ sample}}{\mathrm{Total \ sample \ size}} = \frac{1}{n}\hat{X} \]

• \( \hat{X} \) is a binomial random variable
• \( \hat{p} \) is thus a scaled binomial random variable
• Standard deviation of \( \hat{X} \) is \( \sigma_{\hat{X}} = \sqrt{np(1-p)} \)
• From rules for scaled random variables, the standard deviation of \( \hat{p} \) is thus
  \[ \sigma_{\hat{p}} = \frac{1}{n}\sqrt{np(1-p)} = \sqrt{\frac{p(1-p)}{n}} \]

This is the standard error for \( \hat{p} \)!!!

Definition: The binomial test uses data to test whether a population proportion (\( p \)) matches a null expectation (\( p_{0} \)) for the proportion.

Definition: The null hypothesis \( H_{0} \) and alternative hypothesis \( H_{A} \) for a binomial test are given by:

    \( H_{0} \): Relative frequency of successes in population is \( p_{0} \).
    \( H_{A} \): Relative frequency of successes in population is not \( p_{0} \).

Do people typically use a particular ear preferentially when listening to strangers? Marzoli and Tomassi (2009) had a researcher approach and speak to strangers in a noisy nightclub. An observer scored whether the person approached turned either the left or right ear toward the questioner. Of 25 participants, 19 turned the right ear toward the questioner and 6 offered the left ear. Is this evidence of population difference from 50% for each ear?

Discuss: What is the null and alternative hypotheses?

Answer:
\[ \begin{array}{ll} H_{0}\,: & p = 0.5 \\ H_{A}\,: & p \neq 0.5 \end{array} \]

Do people typically use a particular ear preferentially when listening to strangers? Marzoli and Tomassi (2009) had a researcher approach and speak to strangers in a noisy nightclub. An observer scored whether the person approached turned either the left or right ear toward the questioner. Of 25 participants, 19 turned the right ear toward the questioner and 6 offered the left ear. Is this evidence of population difference from 50% for each ear?

Discuss: What is the observed value of the test statistic?

Answer: Number of right ears is 19 (\( \hat{X}=19 \)).

Discuss: Under the null hypothesis, calculate the probability of getting exactly 19 right ears and six left ears.

(prob <- dbinom(x = 19, size = 25, prob = 0.5))

[1] 0.005277991

\[ \mathrm{Pr[19]} = \left(\begin{array}{c}{25 \\ 19}\end{array}\right)0.5^{19}0.5^{6} = 0.005278 \]

Discuss: Give the two-tailed \( P \)-value based on your previous answer.

(pval <- 2*sum(extreme_probs))

[1] 0.0146333

Discuss: State your conclusion.

Answer: Using a significance level of \( \alpha = 0.05 \), we reject \( H_{0} \) since \( P < 0.05 \). There is evidence that more people use the right ear than the left ear when listening to a stranger in the noisy nightclub.