Questions: A city council is trying to determine if library B, with extended hours, has a higher proportion of patrons under the age of 18 than library A, which closes at 6:00 pm. A random sample of patrons from both libraries are checked. The results of the data collected are shown below. Library A Library B --- --- Successes 56 Successes 97 Observations 150 Observations 200 Proportion 0.373 Proportion 0.485 Confidence level 95% z-score 2.091 p-value 0.0183

A city council is trying to determine if library B, with extended hours, has a higher proportion of patrons under the age of 18 than library A, which closes at 6:00 pm. A random sample of patrons from both libraries are checked. The results of the data collected are shown below.

Library A Library B
--- ---
**Successes** 56 **Successes** 97
**Observations** 150 **Observations** 200
**Proportion** 0.373 **Proportion** 0.485
**Confidence level** 95%
**z-score** 2.091
**p-value** 0.0183

Transcript text: A city council is trying to determine if library $B$, with extended hours, has a higher proportion of patrons under the age of 18 than library $A$, which closes at $6: 00 \mathrm{pm}$. A random sample of patrons from both libraries are checked. The results of the data collected are shown below. \begin{tabular}{|l|r|r|r|} \hline \multicolumn{2}{|c|}{ Library A } & \multicolumn{2}{c|}{ Library B } \\ \hline Successes & 56 & Successes & 97 \\ \hline Observations & 150 & Observations & 200 \\ \hline Proportion & 0.373 & Proportion & 0.485 \\ \hline & & & \\ \hline Confidence level & $95 \%$ & & \\ \hline z-score & 2.091 & & \\ \hline p-value & 0.0183 & & \\ \hline \end{tabular}

Solution

Solution Steps

Step 1: Identify the given data

Library A:
- Successes ($x_A$): 56
- Observations ($n_A$): 150
- Proportion ($\hat{p}_A$): 0.373
Library B:
- Successes ($x_B$): 97
- Observations ($n_B$): 200
- Proportion ($\hat{p}_B$): 0.485
Confidence level: 95%
Z-score: 2.091
p-value: 0.0183

Step 2: Calculate the sample proportions

Sample proportion for Library A ($\hat{p}_A$): \[ \hat{p}_A = \frac{x_A}{n_A} = \frac{56}{150} = 0.373 \]
Sample proportion for Library B ($\hat{p}_B$): \[ \hat{p}_B = \frac{x_B}{n_B} = \frac{97}{200} = 0.485 \]

Step 3: Calculate the standard error of the difference in proportions

Pooled proportion ($\hat{p}$): \[ \hat{p} = \frac{x_A + x_B}{n_A + n_B} = \frac{56 + 97}{150 + 200} = \frac{153}{350} = 0.437 \]
Standard error (SE): \[ SE = \sqrt{\hat{p}(1 - \hat{p}) \left( \frac{1}{n_A} + \frac{1}{n_B} \right)} = \sqrt{0.437 \times 0.563 \left( \frac{1}{150} + \frac{1}{200} \right)} \] \[ SE = \sqrt{0.437 \times 0.563 \left( 0.00667 + 0.005 \right)} = \sqrt{0.437 \times 0.563 \times 0.01167} = \sqrt{0.00287} = 0.0536 \]