Question

Notation for the sample data given in exercise 1, consider the salk vaccine treatment group to be the first sample. Identify the values of \({{\bf{n}}_{\bf{1}}}{\bf{,}}{{\bf{\hat p}}_{\bf{1}}}{\bf{,}}{{\bf{\hat q}}_{\bf{1}}}{\bf{,}}{{\bf{n}}_{\bf{2}}}{\bf{,}}{{\bf{\hat p}}_{\bf{2}}}{\bf{,}}{{\bf{\hat q}}_{\bf{2}}}{\bf{,\bar p}}\) and \({\bf{\bar q}}\). Round all values so that they have six significant digits.

Short answer

The values of notations are as follows:

\(\begin{array}{c}{n_1} = 201,229\\{{\hat p}_1} = 0.000164\\{{\hat q}_1} = 0.999836\\{n_2} = 200745\end{array}\)

\(\begin{array}{c}{{\hat p}_2} = 0.000573\\{{\hat q}_2} = 0.999427\\\bar p = 0.000368\\\bar q = 0.999632\end{array}\)

Step-by-step solution

Step-1: Given information

The study is conducted on 401974 children divided into two groups:

Treatment: Of 201229, 33 developed polio.

Placebo: Of 200,745, 115 developed polio.

Step-2: Interpretation of notations in two proportions test

The general notations are expressed as,

\({n_1} = \)size of first sample

\({n_2} = \)size of second sample

\({\hat p_1} = \)sample proportion of success in first sample

\({\hat q_1} = \)complement of sample successes in second sample.

\({\hat p_2} = \)sample proportion of success in second sample

\({\hat q_2} = \)complement of sample successes in first sample

\(\bar p = \)pooled sample proportion

\(\bar q = 1 - \bar p\)

Step-3: Identify the values from the given information

Let the treatment group be defined as group 1 and Placebo as group 2.

Then,

\(\begin{array}{l}{{\bf{n}}_{\bf{1}}}{\bf{ = 201229}}\\{{\bf{x}}_{\bf{1}}}{\bf{ = 33}}\\{{\bf{n}}_{\bf{2}}}{\bf{ = 200745}}\\{{\bf{x}}_{\bf{2}}}{\bf{ = 115}}\end{array}\)

Step-4:  Compute measure of sample proportions

Sample proportions are calculated as,

\(\begin{array}{c}{{\hat p}_1} = \frac{{{x_1}}}{{{n_1}}}\\ = \frac{{33}}{{201229}}\\ = 0.000164\end{array}\)

\(\begin{array}{c}{{\hat q}_1} = 1 - {{\hat p}_1}\\ = 1 - 0.000164\\ = 0.999836\end{array}\)

Similarly,

\(\begin{array}{c}{{\hat p}_2} = \frac{{{x_2}}}{{{n_2}}}\\ = \frac{{115}}{{200745}}\\ = 0.000573\end{array}\)

\(\begin{array}{c}{{\hat q}_2} = 1 - {{\hat p}_2}\\ = 1 - 0.000573\\ = 0.999427\end{array}\)

Find sample pooled proportion

Now, the pooled sample proportion can be calculated as,

\(\begin{array}{c}\bar p = \frac{{{x_1} + {x_2}}}{{{n_1} + {n_2}}}\\ = \frac{{33 + 115}}{{201229 + 200745}}\\ = 0.000368\end{array}\)

Complement of pooled sample proportion can be calculated as,

\(\begin{array}{c}\bar q = 1 - \bar p\\ = 1 - 0.000368\\ = 0.999632\end{array}\)