6.5.2 Discrete Tests - GCEALevelMaths9709

In this topic we will learn how to:

formulate hypotheses and carry out a hypothesis test in the context of a single observation from a population which has a binomial or poisson distribution using direct evaluation of probabilities

The steps for formulating a hypothesis test under discrete conditions are very similar to those under normal conditions:
$\textbf{Step 1}$ Define the random variable
$\textbf{Step 2}$ State the distribution of the random variable
$\textbf{Step 3}$ State null and alternative hypotheses
$\textbf{Step 4}$ Define the distribution assuming $H_{0}$ is true
$\textbf{Step 5}$ State the rejection rule $\textbf{Step 6}$ Substitute the test statistic into $H_{1}$ and evaluate the probability $\textbf{Step 7}$ Use the rejection rule to determine whether to reject or accept $H_{0}$ $\textbf{Step 8}$ Conclude in context

Let’s look at some past paper questions.

1. Arvind uses a fair $6$ -sided die to play a game. He believes he has a system to predict the score when the die is thrown. Before each throw of the die, he writes down what he thinks the score will be. He claims that he can write the correct score more often than he would if he were just guessing. His friend Laxmi tests his claim by asking him to write down the score before each of $15$ throws of the die. Arvind writes down the correct score on exactly $5$ out of $15$ throws. Test Arvind’s claim at the $10\%$ significance level. (9709/61/M/J/22 number 2)

Define the random variable,
$\textmd{X - r.v, number of correct scores that Arvind gets}$ State the distribution of $X$ ,
$X \sim B(15, p)$ State null and alternative hypotheses,
$H_{0}: p = \frac{1}{6}$ $H_{1}: p > \frac{1}{6}$ Note: We’re testing for a definite increase, since we want to test whether Arvind gets the scores correct more often with his system, hence upper-tail test.

Assuming $H_{0}$ is true,
$X \sim B\left(15, \frac{1}{6}\right)$ For an upper-tail test, at the $10\%$ significance level, state the rejection rule,
$\textmd{Reject }H_{0} \textmd{ if }P(X\ \textcolor{#2192FF}{\ge}\ c)\ \textcolor{#0f0}{<}\ 0.100$ Note: The sign inside the probability statement is based upon the alternative hypothesis. If it is an upper tail test we use $\ge$ , lower-tail we use $\le$ . We always check whether that probability is less than the siginificance level. $c$ represents the test statistic.
The test statistic is $5$ . Substitute the test statistic into the rejection rule,
$P(X \ge 5) = 1 - P(X < 5)$

$P(X \ge 5) = 1 - \left[^{15}C_{4} \times \left(\frac{1}{6}\right)^{4} \times \left(\frac{5}{6}\right)^{11} + ^{15}C_{3} \times \left(\frac{1}{6}\right)^{3} \times \left(\frac{5}{6}\right)^{12} + ^{15}C_{2} \times \left(\frac{1}{6}\right)^{2} \times \left(\frac{5}{6}\right)^{13} + ^{15}C_{1} \times \left(\frac{1}{6}\right) \times \left(\frac{5}{6}\right)^{14} + \left(\frac{5}{6}\right)^{15}\right]$

$P(X \ge 5) = 0.0898$ Compare $0.0898$ to the rejection rule,
$\textmd{Reject }H_{0} \textmd{ if } P(X \ge c) < 0.100$ $0.0898 < 0.100$ $\textmd{Reject }H_{0}\textmd{ in favour of }H_{1}$ Conclude in context,
There is evidence at the $10\%$ significance level, that Arvind has a system to predict the score when the die is thrown.

2. The number of absences per week by workers at a factory has the distribution $Po(2.1)$ . Following a change in working conditions, the management wished to test whether the mean number of absences has decreased. They found that, in a randomly chosen $3$ -week period, they were exactly $2$ absences. Carry out the test at the $10\%$ significance level. (9709/63/O/N/20 number 5)

Define the random variable,
$\textmd{X - r.v, number of absences per 3-week period}$ State the distribution of $X$ ,
$X \sim Po(\lambda)$ State null and alternative hypotheses,
$H_{0}: \lambda = 6.3$ $H_{1}: \lambda < 6.3$ Note: We’re testing for a definite decrease, hence lower-tail test.

Assuming $H_{0}$ is true,
$X \sim Po(6.3)$ For an upper-tail test, at the $10\%$ significance level, state the rejection rule,
$\textmd{Reject }H_{0}\textmd{ if }P(X \le c) < 0.100$ The test statistic is $5$ . Substitute the test statistic into the rejection rule,
$P(X \le 2) = e^{-6.3}\left(\frac{6.3^{2}}{2!} + 6.3 + 1\right)$ $P(X \le 2) = 0.0498$ Compare $0.0498$ to the rejection rule,
$\textmd{Reject }H_{0} \textmd{ if }P(X \le c) < 0.100$ $0.0498 < 0.100$ $\textmd{Reject }H_{0}\textmd{ in favour of }H_{1}$ Conclude in context,
There is evidence at the $10\%$ significance level, that the mean number of absences has decreased.