Question

Sometimes experiments involving success or failure responses are run in a paired or before/after manner. Suppose that before a major policy speech by a political candidate, n individuals are selected and asked whether \((S)\)or not (F) they favor the candidate. Then after the speech the same n people are asked the same question. The responses can be entered in a table as follows:

Before	After
		S	F
	S	\({{\bf{X}}_{\bf{1}}}\)	\({{\bf{X}}_{\bf{2}}}\)
	F	\({{\bf{X}}_{\bf{3}}}\)	\({{\bf{X}}_{\bf{4}}}\)

Where\({{\bf{x}}_{\bf{1}}}{\bf{ + }}{{\bf{x}}_{\bf{2}}}{\bf{ + }}{{\bf{x}}_{\bf{3}}}{\bf{ + }}{{\bf{x}}_{\bf{4}}}{\bf{ = n}}\). Let\({{\bf{p}}_{\bf{1}}}{\bf{,}}{{\bf{p}}_{\bf{2}}}{\bf{,}}{{\bf{p}}_{\bf{3}}}\), and \({p_4}\)denote the four cell probabilities, so that \({p_1} = P(S\) before and S after), and so on. We wish to test the hypothesis that the true proportion of supporters (S) after the speech has not increased against the alternative that it has increased.

a. State the two hypotheses of interest in terms of\({p_1},{p_2}\),\({p_3}\), and \({p_4}\).

b. Construct an estimator for the after/before difference in success probabilities

c. When n is large, it can be shown that the rv \(\left( {{{\bf{X}}_{\bf{i}}}{\bf{ - }}{{\bf{X}}_{\bf{j}}}} \right){\bf{/n}}\) has approximately a normal distribution with variance given by\(\left[ {{{\bf{p}}_{\bf{i}}}{\bf{ + }}{{\bf{p}}_{\bf{j}}}{\bf{ - }}{{\left( {{{\bf{p}}_{\bf{i}}}{\bf{ - }}{{\bf{p}}_{\bf{j}}}} \right)}^{\bf{2}}}} \right]{\bf{/n}}\). Use this to construct a test statistic with approximately a standard normal distribution when \({H_0}\)is true (the result is called McNemar's test).

d. If\({{\bf{x}}_{\bf{1}}}{\bf{ = 350,}}\;\;\;{{\bf{x}}_{\bf{2}}}{\bf{ = 150,}}\;\;\;{{\bf{x}}_{\bf{3}}}{\bf{ = 200}}\), and\({x_4} = 300\), what do you conclude?

Short answer

a. The hypotheses of interest are \({H_0}:{p_3} = {p_2}\)versus \({H_a}:{p_3} > {p_2};\)

b. \(\frac{{{X_3} - {X_2}}}{n};\)

\(c.Z = \frac{{{X_3} - {X_2}}}{{\sqrt {{X_3} + {X_2}} }};\)

d. Reject null hypothesis.

Step-by-step solution

Step 1: State the two hypotheses of interest in terms of\({{\bf{p}}_{\bf{1}}}{\bf{,}}{{\bf{p}}_{\bf{2}}}\), \({{\bf{p}}_{\bf{3}}}\), and \({{\bf{p}}_{\bf{4}}}\).

(a):

The probability that after the speech people favor the major is the sum of \({p_1}\)and \({p_3}\)- both are the corresponding probabilities to cells 1 and 3. The probability that before the speech people do not favor the major is the sum of \({p_1}\)and \({p_2}\)both are the corresponding probabilities to cells 1 and 2. This is because the first column represents success for "after" and the first row represents the row for "before" success.

From\({p_1}\)+\({p_3}\)=\({p_1}\)+\({p_2}\), which would be the null hypothesis,

The hypotheses of interest are \({H_0}:{p_3} = {p_2}\) versus \({H_a}:{p_3} > {p_2}\)

Step 2:Construct an estimator for the after/before difference in success probabilities

(b):

Every cell in the table represents number of people for corresponding described event, this indicates that using that data you could estimate the probabilities. There are total of n individuals, and\({x_1} + {x_2} + {x_3} + {x_4} = n\). From this, the estimate of \({p_1} + {p_3} - \left( {{p_1} + {p_2}} \right)\) is simply

\(\frac{{{x_1} + {x_3} - \left( {{x_1} + {x_2}} \right)}}{n} = \frac{{{x_3} - {x_2}}}{n}\)

W hich is obviously the after/before difference in success probabilities estimate. \(\sigma = \sqrt {\frac{{{p_3} + {p_2}}}{n}} ,\)However, the estimator requires random variables and it is\(\frac{{{X_3} - {X_2}}}{n}\)

Step 3:to construct a test statistic with approximately a standard normal distribution

(c):

Assume that \({H_0}\)is true, thus\({p_3} = {p_2}\). Based on the given information, the variance of the given estimator in (b) is

\({\mathop{\rm Var}\nolimits} \frac{{{X_3} - {X_2}}}{n} = \frac{{{p_2} + {p_3} - {{\left( {{p_2} - {p_3}} \right)}^2}}}{n}\)

which is, for \({p_3} = {p_2}\)

\({\mathop{\rm Var}\nolimits} \frac{{{X_3} - {X_2}}}{n} = \frac{{{p_3} + {p_2}}}{n}.\)

The standard deviation is which needs to be estimated with\(\hat \sigma = \sqrt {\frac{{{{\hat p}_3} + {{\hat p}_2}}}{n}} \)

Since the expected value is obviously zero, and random variable

\(\frac{{{X_3} - {X_2}}}{n}\)has approximately normal distribution with parameter mean 0 and standard deviation\(\sigma \), you can construct the Z statistic as

\(\begin{array}{c}Z = \frac{{\frac{{{X_3} - {X_2}}}{n}}}{{\sqrt {\frac{{{{\hat p}_3} + {{\hat p}_2}}}{n}} }}\\ = \frac{{{X_3} - {X_2}}}{{\sqrt {{X_3} + {X_2}} }}\end{array}\)

which has standard normal distribution.

Step 4:To conclude for the final proof

(d):

the hypotheses of interest are \({H_0}:{p_3} = {p_2}\) versus\({H_a}:{p_3} > {p_2}\), and the value of the Z test statistic is

\(\begin{array}{c}z = \frac{{{X_3} - {X_2}}}{{\sqrt {{X_3} + {X_2}} }}\\ = \frac{{200 - 150}}{{\sqrt {200 + 150} }}\\ = 2.68\end{array}\)

The test is one sided, upper, thus the P value is the area under the standard normal curve to the right of z. Therefore, the P value is

\(\begin{array}{c}P = P(Z > 2.68)\\ = 1 - P(Z \le 2.68)\\ = 1 - \Phi (2.68)\\ = 0.0037\end{array}\)

where the \(\Phi ( \cdot )\)value can be obtained from the table in the appendix. Depending on significance level, the conclusion might change. At significance level 0.01 or 0.05 which are usually used, because

\(P = 0.0037 < 0.01 < 0.05\)we reject the null hypothesis

However, you might not reject in for other significance levels.