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A teacher has two large containers filled with blue, red, and green beads and claims the proportion of red beads is the same for both containers. The students believe the proportions are different. Each student shakes the first container, selects 50 beads, counts the number of red beads, and returns the beads to the container. The student repeats this process for the second container. One student's samples contained 10 red beads from the first container and 16 red beads from the second container. Let [tex][tex]$p_1$[/tex][/tex] be the true proportion of red beads in container 1 and [tex][tex]$p_2$[/tex][/tex] be the true proportion of red beads in container 2. Which of the following is the correct standardized test statistic and [tex][tex]$P$[/tex][/tex]-value for the hypotheses [tex][tex]$H_0: p_1 - p_2 = 0$[/tex][/tex] and [tex][tex]$H_A: p_1 - p_2 \neq 0$[/tex][/tex]?

A. [tex]z = \frac{0.20 - 0.32}{\sqrt{\frac{(0.26)(0.74)}{100}}}, P[/tex]-value [tex]= 0.0855[/tex]

B. [tex]z = \frac{0.20 - 0.32}{\sqrt{\frac{(0.20)(0.80)}{50} + \frac{(0.32)(0.68)}{50}}}, P[/tex]-value [tex]= 0.171[/tex]

C. [tex]z = \frac{0.20 - 0.32}{\sqrt{(0.26)(0.74)}}, P[/tex]-value [tex]= 0.171[/tex]

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Answer :

GinnyAnswer
Let's solve this problem by following these step-by-step:

### Step 1: Calculate sample proportions
We need to determine the sample proportions from the given data:
- For the first container:
- Number of red beads sampled = 10
- Total number of beads sampled = 50
- Sample proportion ([tex]\( p1_{hat} \)[/tex]) = [tex]\( \frac{10}{50} = 0.2 \)[/tex]

- For the second container:
- Number of red beads sampled = 16
- Total number of beads sampled = 50
- Sample proportion ([tex]\( p2_{hat} \)[/tex]) = [tex]\( \frac{16}{50} = 0.32 \)[/tex]

### Step 2: Determine the combined proportion
The combined proportion ([tex]\( p_{combined} \)[/tex]) of red beads is calculated by combining the red beads and total beads from both samples:
[tex]\[ p_{combined} = \frac{10 + 16}{50 + 50} = \frac{26}{100} = 0.26 \][/tex]

### Step 3: Calculate the standard error
The standard error (SE) of the difference between the sample proportions is given by:
[tex]\[ SE = \sqrt{ p_{combined} (1 - p_{combined}) \left(\frac{1}{n1} + \frac{1}{n2}\right) } \][/tex]
where:
- [tex]\( p_{combined} = 0.26 \)[/tex]
- [tex]\( n1 = 50 \)[/tex]
- [tex]\( n2 = 50 \)[/tex]

Let's substitute the values:
[tex]\[ SE = \sqrt{ 0.26 \times (1 - 0.26) \left(\frac{1}{50} + \frac{1}{50}\right) } \][/tex]
[tex]\[ SE = \sqrt{ 0.26 \times 0.74 \left(\frac{2}{50}\right) } \][/tex]
[tex]\[ SE = \sqrt{ 0.26 \times 0.74 \times 0.04 } \][/tex]
[tex]\[ SE \approx 0.0877 \][/tex]

### Step 4: Calculate the z-score
The z-score is calculated as:
[tex]\[ z = \frac{p1_{hat} - p2_{hat}}{SE} \][/tex]
Substituting the values:
[tex]\[ z = \frac{0.2 - 0.32}{0.0877} \][/tex]
[tex]\[ z \approx -1.368 \][/tex]

### Step 5: Calculate the p-value
The p-value is the probability that the observed difference between the sample proportions is at least as extreme as the difference observed, under the null hypothesis. Given the z-score calculated, the p-value (considering a two-tailed test) can be found using a z-table:
[tex]\[ \text{p-value} = 2 \left(1 - \Phi(|z|)\right) \][/tex]
Where [tex]\(\Phi(z)\)[/tex] represents the cumulative distribution function of the normal distribution.

For [tex]\( z \approx -1.368 \)[/tex]:
[tex]\[ \text{p-value} \approx 2 ( 1 - 0.0857) = 2 \times 0.9143 \approx 0.171 \][/tex]

### Conclusion
The correct standardized test statistic and p-value for testing the hypotheses [tex]\( H_0: p1 - p2 = 0 \)[/tex] and [tex]\( H_A: p1 \neq p2 \)[/tex] are:
[tex]\[ z = \frac{0.20 - 0.32}{\sqrt{\frac{(0.26)(0.74)}{100}}}, \][/tex]
[tex]\[ \text{p-value} = 0.171 \][/tex]

Hence, the correct option is:
[tex]\[ z = \frac{0.20 - 0.32}{\sqrt{\frac{(0.20)(0.80)}{50} + \frac{(0.32)(0.68)}{50}}}, \text{ p-value } = 0.171 \][/tex]

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