Section 11.3 Hypothesis Test for one proportion
In this section, you will consider the following options for null hypothesis and corresponding alternate hypothesis with respect to the unknown value p:
\begin{equation*}
H_0 : p = p_0 \\ H_a : p \neq p_0
\end{equation*}
or
\begin{equation*}
H_0 : p \le p_0 \\ H_a : p \gt p_0
\end{equation*}
or
\begin{equation*}
H_0 : p \ge p_0 \\ H_a : p \lt p_0.
\end{equation*}
For any given problem, we choose only one of these three options. The first is called a "two-tailed" test since the alternate hypothesis can not be equal if it is actually larger or smaller. The last two are called "one-tailed" tests since the alternate hypothesis only allows for being on one side. Note that some people will write all of these null hypothesis options using equality but the alternate hypothesis determines the number of tails.
From the Central Limit Theorem, you found that every interesting distribution eventually becomes approximately normal. This includes the binomial distribution with mean \(\mu = n p_0\) and variance \(\sigma^2 = n p_0 (1-p_0)\text{.}\) Hence, the z-statistic
\begin{equation*}
Z = \frac{X - n p_0}{\sqrt{n p_0(1-p_0)}} = \frac{p - p_0}{\sqrt{p_0(1-p_0)/n}}
\end{equation*}
is approximately standard normal and so probabilities on this statistic can be computed as needed using the normal distribution.
Let’s look at an example for this by considering:
\begin{equation*}
H_0: p = 0.20
\end{equation*}
vs
\begin{equation*}
H_a: p \ne 0.20.
\end{equation*}
Tests like this are called two-tailed since there are two ways to reject the null hypothesis: we find that p should be less than 0.2 or we find that p should be greater than 0.2.
To test our hypothesis, let’s now chose a significance level of \(\alpha = 0.05\) and take a sample. Presuming we actually do this, let’s assume that we find that out of n=100 sample values we get X = 27 successes. Hence, our actual test statistic is \(p = \frac{27}{100} = 0.27\text{.}\)
So, in this case we have
\begin{equation*}
\sigma = \sqrt{\frac{p_0(1-p_0)}{n}} = \sqrt{\frac{0.2 \cdot 0.8}{100}} = 0.04
\end{equation*}
and the z-statistic (using the normal distribution) for the sample statistic of p = 0.27 is
\begin{equation*}
z = \frac{0.27 - 0.2}{0.04} = 1.75.
\end{equation*}
Remember, the alternate hypothesis has two tails so to determine the P value we need to determine from the normal distribution
\begin{equation*}
P(Z \gt 1.75) + P(Z \lt -1.75)
\end{equation*}
and find that this has probability approximately 0.0392 + 0.0392 = 0.0784. However, this P value is greater than our significance level \(\alpha = 0.05\) so we cannot reject the null hypothesis at the 5 percent significance level. However, if we had chosen initially to use a 10 percent significance level then we would have rejected the null hypothesis and accepted the alternate.
An article in the Washington Post on March 16, 1993 stated that nearly 45 percent of all Americans have brown eyes. A random sample of \(n =
76\) C of I students found 28 with brown eyes.
We test
\(H_0: p = .45\)
\(H_a: p \neq .45\)
(a) What is the \(z\)-statistic for this test? \(\)
(b) What is the P-value of the test? \(\)
Next is a one-tailed test for p. Notice, in this case you will only compute z-score probability for one tail and not both tails.
Checkpoint 11.3.2. WebWork - One Tailed Test.
A noted psychic was tested for ESP. The psychic was presented with 200 cards face down and was asked to determine if the card was one of 5 symbols: a star, cross, circle, square, or three wavy lines. The psychic was correct in 49 cases. Let \(p\) represent the probability that the psychic correctly identifies the symbol on the card in a random trial. Assume the 200 trials can be treated as an SRS from the population of all guesses.
To see if there is evidence that the psychic is doing better than just guessing, we test
\(H_0: p = .2\)
\(H_a: p > .2\)
(a) What is the \(z\)-statistic for this test? \(\)
(b) What is the P-value of the test? \(\)
