When conducting a hypothesis test with a binomial distribution (sometimes called a Binomial Test), there are three ways to calculate the P-value (with additional variations possible).  The only exact calculation is to use the binomial probability distribution.  The other methods are approximations using the standardized normal distribution (when certain criteria have been achieved).  Of these two methods, one can use the sample counts or one can use the sample proportions.  Furthermore, it is possible in both of these approximating cases to apply a continuity correction to account for the use of a continuous distribution to approximate a discrete distribution.

This problem introduces the method to obtain an approximate P-value using the standard normal distribution as a reasonable approximation for the binomial distribution.  Instead of using the counts, this method uses the sample proportion, $\displaystyle \hat{{{p}}}=\frac{{k}}{{n}}$.  As shown below, this worked example DOES NOT demonstrate the continuity correction, and this demonstration uses Excel to obtain the answers.  VERY IMPORTANT:  This method will work if the minimum of $\displaystyle {n}{p}$ or $\displaystyle {n}{\left({1}-{p}\right)}$ is 10 or greater.  If this criteria is not achieved, then the normal approximation is not very accurate.

For this demonstration problem, we will test a hypothesis that the population proportion is not 45%.  We collect a random sample of size $\displaystyle {n}={400}$ and we obtain 153 successful observations.  While the sample proportion $\displaystyle \hat{{{p}}}=\frac{{153}}{{400}}={0.3825}$ is not equal to the hypothesized proportion $\displaystyle {p}={45}\%$, a hypothesis test is necessary to determine if this observation could reasonably happen by chance.

To start, we clearly construct the hypotheses for this problem.  Because there is no indication of which direction the difference from the hypothesized proportion might occur, this would suggest a two-tailed test (as can be seen in the choice of $\displaystyle {H}_{{a}}$):
      $\displaystyle {H}_{{o}}:{p}={0.45}$
      $\displaystyle {H}_{{a}}:{p}\ne{0.45}$

The distribution under examination is the binomial distribution for proportions.  As indicated above, this is a nearly normal distribution with a standard error of$\displaystyle \sigma=\sqrt{{\frac{{{p}\cdot{\left({1}-{p}\right)}}}{{n}}}}$.  With this information, we can calculate a z-score as the test statistic for this scenario using $\displaystyle \hat{{{p}}}$: $\displaystyle {z}=\frac{{\hat{{{p}}}-{p}}}{\sqrt{{\frac{{{p}\cdot{\left({1}-{p}\right)}}}{{n}}}}}=\frac{{{0.3825}-{0.45}}}{\sqrt{{\frac{{{0.45}\cdot{\left({1}-{0.45}\right)}}}{{400}}}}}=-{2.764}$ It can be argued that a continuity correction could be applied to $\displaystyle {k}$ before calculating $\displaystyle \hat{{{p}}}$, but for the moment we will overlook this (as the approximation obtained is very good).

Using the standard normal distribution, we can now calculate the P-value for this scenario:
      $\displaystyle {P}{\left({Z}\le-{2.764}\right)}={0.0028549}$
This can be obtained from Excel using the following formula:
=NORMSDIST(-2.764) However, this provides the probability estimate for values as extreme as our observed value or below.  In other words, this is the P-value for the left-tail of the distribution.  To obtain the two-tailed P-value, we would double this value.  The final P-value would be 0.0057.  Using a pre-selected significance level, this P-value can be used to decide whether to reject the null hypothesis or not.

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Exercise Problem
In a certain school district, it was observed that 31% of the students in the element schools were classified as only children (no siblings).  However, in the special program for talented and gifted children, 77 out of 207 students are only children.  The school district administrators want to know if the proportion of only children in the special program is significantly different from the proportion for the school district. Test at the $\displaystyle \alpha={0.05}$ level of significance.

What is the hypothesized population proportion for this test?
$\displaystyle {p}=$
(Report answer as a decimal accurate to 2 decimal places.  Do not report using the percent symbol.)

Based on the statement of this problem, how many tails would this hypothesis test have?

Choose the correct pair of hypotheses for this situation:

(A)	(B)	(C)
$\displaystyle {H}_{{0}}:{p}={0.31}$ $\displaystyle {H}_{{a}}:{p}<{0.31}$	$\displaystyle {H}_{{0}}:{p}={0.31}$ $\displaystyle {H}_{{a}}:{p}\ne{0.31}$	$\displaystyle {H}_{{0}}:{p}={0.31}$ $\displaystyle {H}_{{a}}:{p}>{0.31}$
(D)	(E)	(F)
$\displaystyle {H}_{{0}}:{p}={0.372}$ $\displaystyle {H}_{{a}}:{p}<{0.372}$	$\displaystyle {H}_{{0}}:{p}={0.372}$ $\displaystyle {H}_{{a}}:{p}\ne{0.372}$	$\displaystyle {H}_{{0}}:{p}={0.372}$ $\displaystyle {H}_{{a}}:{p}>{0.372}$

Using the normal approximation for the binomial distribution (without the continuity correction), was is the test statistic for this sample based on the sample proportion?
$\displaystyle {z}=$
(Report answer as a decimal accurate to 3 decimal places.)

You are now ready to calculate the P-value for this sample.
P-value =
(Report answer as a decimal accurate to 4 decimal places.)

This P-value (and test statistic) leads to a decision to...

As such, the final conclusion is that...