FizzBuzz for Statistics

Inspired by a Reddit thread and the original FizzBuzz for programmers.

General

1. Explain p-value and confidence interval.

Experiment

1. Explain why, when applicable, the paired t-test is usually preferable over an unpaired t-test. (credit)

Regression

1. Interpreting interaction term: The model is \(y = \beta_0 + \beta_1 X_1 + \beta_2 X_2 + \beta_3 X_3 + \epsilon\). What is the change in \(y\) given 1 unit increase in \(X_1\)?

Probability

1. I fire two missiles at a ship at the same time. Each independently has a 60% chance of hitting and sinking it. I turn to yell at somebody, and when I turn back around, the ship is sinking. what’s the probability both missiles hit it? (credit)

2. Sum of two random variables

General (solution)

1. p-value and confidence interval

• p-value is the probability of getting observed, or more extreme, data given that the null hypothesis is true. Wrong: p-value is the probability that the null hypothesis is true.

• If we repeat an experiment many many times, we would expect that X% of the times the true parameter will fall within the X% confidence interval of that parameter.

Experiment (solution)

1. Paired t-test takes into account the paired structure, which reduces the variance of the difference. Consider two paired observations, whose outcomes are denoted \(X\) and \(Y\). We are interested in the difference \(Z = X - Y\).

\[\begin{align} Var(Z) &= Var(X - Y) = Var(X) + Var(Y) - 2Cov(X, Y) \\ \end{align}\]

Notice that if \(Cov(X, Y) > 0\), e.g. paired units tend to have similar outcomes, then \(Var(Z)\) is smaller.

Regression (solution)

1. Interpreting interaction term:

\[\frac{\partial}{\partial X_1} y = \beta_1 + \beta_3 X_2\]

Key point: the marginal effect of \(X_1\) on \(y\) depends on the value of \(X_2\)

Probability (solution)

1. After firing the two missiles, 4 disjoint events can happen:

• P(both hit) = 0.6 * 0.6 = 0.36
• P(only 1st hits) = 0.6 * 0.4 = 0.24
• P(only 2nd hits) = 0.4 * 0.6 = 0.24
• P(none hits) = 0.4 * 0.4 = 0.16

P(both hit | ship sinks) = P(both hit | at least one hit) = \(\frac{0.36}{0.36 + 0.24 + 0.24} = 0.43\)