Question

For each problem, test the sample proportions for the significance of the difference. A. $$ \begin{array}{ll} \hline \hline \text { Sample 1 } & \text { Sample } 2 \\ \hline P_{s 1}=0.20 & P_{s 2}=0.17 \\ N_{1}=114 & N_{2}=101 \\ \hline \end{array} $$ B. $$ \begin{array}{ll} \hline \hline \text { Sample } 1 & \text { Sample } 2 \\ \hline P_{s 1}=0.60 & P_{s 2}=0.62 \\ N_{1}=478 & N_{2}=532 \\ \hline \end{array} $$

Short answer

For Problem A: Compare z_A to z-tables.For Problem B: Compare z_B to z-tables.

Step-by-step solution

- Identify the given values

For Problem A, we have: Sample Proportion 1, \( P_{s1} = 0.20 \)Sample Proportion 2, \( P_{s2} = 0.17 \)Sample Size 1, \( N_1 = 114 \)Sample Size 2, \( N_2 = 101 \)For Problem B, we have: Sample Proportion 1, \( P_{s1} = 0.60 \)Sample Proportion 2, \( P_{s2} = 0.62 \)Sample Size 1, \( N_1 = 478 \)Sample Size 2, \( N_2 = 532 \)

- State the hypotheses

The null hypothesis \( H_0 \) states there is no difference in proportions: \( H_0: P_{s1} = P_{s2} \). The alternative hypothesis \( H_a \) states there is a difference: \( H_a: P_{s1} eq P_{s2} \)

- Calculate the pooled sample proportion \( \bar{P} \)

Compute pooled sample proportion using the formula: \[ \bar{P} = \frac{P_{s1}N_1 + P_{s2}N_2}{N_1 + N_2} \]For Problem A:\[ \bar{P}_A = \frac{0.20 \times 114 + 0.17 \times 101}{114 + 101} \]For Problem B:\[ \bar{P}_B = \frac{0.60 \times 478 + 0.62 \times 532}{478 + 532} \]

- Calculate the standard error (SE) of the difference in proportions

Use the formula for SE:\[ SE = \bar{P} \times (1 - \bar{P}) \times \biggl(\frac{1}{N_1} + \frac{1}{N_2}\biggr)^{0.5} \]For Problem A, calculate SE:\[ SE_A = (0.187 \times 0.813) \times \biggl(\frac{1}{114} + \frac{1}{101}\biggr)^{0.5} \]For Problem B, calculate SE:\[ SE_B = (0.61 \times 0.39) \times \biggl(\frac{1}{478} + \frac{1}{532}\biggr)^{0.5} \]

- Calculate the z-score

Use the formula for z-score:\[ z = \frac{(P_{s1} - P_{s2})}{SE} \]For Problem A:\[ z_A = \frac{(0.20 - 0.17)}{SE_A} \]For Problem B:\[ z_B = \frac{(0.60 - 0.62)}{SE_B} \]

- Determine the p-value and compare to significance level

Use z-tables or statistical software to find the p-value based on the z-score. Compare this p-value to your significance level (commonly, \( \text{α} = 0.05 \)) to decide whether to reject the null hypothesis \( H_0 \).

Key concepts

Sample Proportions

Sample proportions are used to represent the fraction of individuals in a sample that have a particular characteristic. For example, if 20 out of 100 people in a survey like chocolate ice cream, the sample proportion is calculated as:
\( P_s = \frac{20}{100} = 0.20 \)
In the provided exercise, we have two different samples with given sample proportions for both Problem A and Problem B. These proportions will be used to determine if there is a statistically significant difference between the two samples.
Remember, proportions are always between 0 and 1.

Null Hypothesis

In statistical significance testing, the null hypothesis (\( H_0 \)) is a statement that there is no effect or no difference between groups or variables. For our exercise, the null hypothesis is that there is no difference in sample proportions.
Mathematically, it can be expressed as: \( H_0: P_{s1} = P_{s2} \)
The alternative hypothesis (\( Ha \)) is the opposite of the null hypothesis. It suggests that there is a difference in proportions. In our case: \( H_a: P_{s1} eq P_{s2} \)
The next steps in the solution will help us test these hypotheses to determine if we can reject the null hypothesis.

Standard Error

The standard error (SE) measures the variability or dispersion of the sample proportion. It tells us how much the sample proportion may vary from the true population proportion.
In our problem, the formula for the standard error of the difference between two proportions is:
\[ SE = \bar{P} \times (1 - \bar{P}) \times \biggl( \frac{1}{N_1} + \frac{1}{N_2} \biggr)^{0.5} \]
Here, \( \bar{P} \) is the pooled sample proportion. We calculate it using:
\[ \bar{P} = \frac{P_{s1}N_1 + P_{s2}N_2}{N_1 + N_2} \]
For Problem A and B, plugging in values will give us the respective pooled sample proportions and their standard errors. A smaller SE indicates that our sample proportion is a more accurate reflection of the true population proportion.

z-Score

The z-score is a measure that indicates how many standard deviations an element is from the mean. In our context, it tells us how far the difference between the sample proportions is from the null hypothesis of no difference.
We can calculate the z-score using the formula:
\[ z = \frac{(P_{s1} - P_{s2})}{SE} \]
For Problem A and B, we take the difference in proportions and divide by the standard error calculated in the previous step. The result helps us determine how unusual or extreme our observed difference is under the assumption that the null hypothesis is true.
This z-score can be used to find the p-value, which tells us the probability of observing a difference as extreme as, or more extreme than, the one we have observed if the null hypothesis is true. This p-value, compared against a significance level (e.g., 0.05), helps us decide whether to reject the null hypothesis.