Chapter 7. Hypothesis Testing: Hypothesis Test for a Population Proportion
Small-sample Proportion Test: Test Statistic and p-value
Binomial Test for a Population Proportion: Test Statistic
When the sample size is small (#n\lt 30#), the Central Limit Theorem no longer applies.
In such cases, however, the number of successes #X# can be used as the test statistic.
Under the assumption that the null hypothesis #H_0# is true, #X# has the binomial distribution with parameters #n# and #\pi_0#. That is, #X \sim B(n, \pi_0)#.
Calculating the p-value of a Binomial Test for a Population Proportion with Statistical Software
Suppose you observe #X=k#.
The calculation of the #p#-value of a binomial test for #\pi# is dependent on the direction of the test and can be performed using either Excel or R.
To calculate the #p#-value of a binomial test for #\pi# in Excel, make use of one of the following commands:
\[\begin{array}{llll}
\phantom{0}\text{Direction}&\phantom{0000}H_0&\phantom{0000}H_a&\phantom{0000000000}\text{Excel Command}\\
\hline
\text{Left-tailed}&H_0:\pi \geq \pi_0&H_a:\pi \lt \pi_0&=\text{BINOM.DIST}(k, n, \pi_0,1)\\
\text{Right-tailed}&H_0:\pi \leq \pi_0&H_a:\pi \gt \pi_0&=1\text{ - }\text{BINOM.DIST}(k \text{ - }1,n,\pi_0 ,1)\\
\text{Two-tailed}&H_0:\pi = \pi_0&H_a:\pi \neq \pi_0&\text{twice the smaller of the two probabilities above}\\
\end{array}\]
To calculate the #p#-value of a binomial test for #\pi# in R, make use of one of the following commands:
\[\begin{array}{llll}
\phantom{0}\text{Direction}&\phantom{0000}H_0&\phantom{0000}H_a&\phantom{000000}\text{R Command}\\
\hline
\text{Left-tailed}&H_0:\pi \geq \pi_0&H_a:\pi \lt \pi_0&\text{pbinom}(k, n, \pi_0, \text{TRUE})\\
\text{Right-tailed}&H_0:\pi \leq \pi_0&H_a:\pi \gt \pi_0&\text{pbinom}(k \text{ - }1,n,\pi_0 , \text{FALSE})\\
\text{Two-tailed}&H_0:\pi = \pi_0&H_a:\pi \neq \pi_0&\text{twice the smaller of the two probabilities above}\\
\end{array}\]
If #p \leq \alpha#, reject #H_0# and conclude #H_a#. Otherwise, do not reject #H_0#.
This small-sample approach can also be used for large samples, as the #p#-values will be approximately the same for either approach.
A gambling commissioner is suspicious of this claim and thinks the true chances of winning are lower than #22\%#. The commissioner plans on using a statistical test to test her suspicion.
In #20# independent trials, the commissioner wins #1# time.
Calculate the #p#-value of the test and make a decision regarding #H_0: \pi \geq 0.22#. Round your answer to #4# decimal places. Use the significance level #\alpha = 0.07#.
On the basis of this #p#-value, #H_0# should be rejected, because #\,p# #\lt# #\alpha#.
There are a number of different ways we can calculate the #p#-value of the test. Click on one of the panels to toggle a specific solution.
Let #X# denote the number of observed wins out of #20#, then #X=1#.
A sample size of #20# is not considered large enough for the Central Limit Theorem to apply. This means that we will need to use #X# as the test statistic.
Under the null hypothesis #H_0#, #X# is binomially distributed with parameters #n# and #\pi_0#. That is:
\[X \sim B(20, 0.22)\]
Assuming #X \sim B(20, 0.22)#, the #p#-value of a left-tailed binomial test for #\pi# can be computed with the following Excel command:
\[\begin{array}{c}
= \text{BINOM.DIST}(X,n,\pi_0 ,1)\\
\downarrow\\
= \text{BINOM.DIST}(1, 20, 0.22 ,1)
\end{array}\]
This gives:
\[p =0.0461\]
Since #\,p# #\lt# #\alpha#, #H_0: \pi = 0.22# should be rejected.
Let #X# denote the number of observed wins out of #20#, then #X=1#.
A sample size of #20# is not considered large enough for the Central Limit Theorem to apply. This means that we will need to use #X# as the test statistic.
Under the null hypothesis #H_0#, #X# is binomially distributed with parameters #n# and #\pi_0#. That is:
\[X \sim B(20, 0.22)\]
Assuming #X \sim B(20, 0.22)#, the #p#-value of a left-tailed binomial test for #\pi# can be computed with the following R command:
\[\begin{array}{c}
\text{pbinom}(X,n,\pi_0 ,\text{TRUE})\\
\downarrow\\
\text{pbinom}(1, 20, 0.22 ,\text{TRUE})
\end{array}\]
This gives:
\[p =0.0461\]
Since #\,p# #\lt# #\alpha#, #H_0: \pi = 0.22# should be rejected.
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