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To answer this question, we need to conduct a hypothesis test comparing the proportions of defective chips from two different plants.
Step 1: State the Hypotheses
We are given:
- [tex]\( p_A \)[/tex]: the true proportion of defective chips from plant [tex]\( A \)[/tex]
- [tex]\( p_B \)[/tex]: the true proportion of defective chips from plant [tex]\( B \)[/tex]
The hypotheses can be stated as:
- [tex]\( H_0: p_A = p_B \)[/tex]
- [tex]\( H_a: p_A > p_B \)[/tex]
Step 2: Collect Sample Data
From the problem:
- Sample size from plant [tex]\( A \)[/tex], [tex]\( n_A = 80 \)[/tex]
- Defective chips from plant [tex]\( A \)[/tex], [tex]\( x_A = 12 \)[/tex]
- Sample size from plant [tex]\( B \)[/tex], [tex]\( n_B = 90 \)[/tex]
- Defective chips from plant [tex]\( B \)[/tex], [tex]\( x_B = 10 \)[/tex]
Step 3: Calculate Sample Proportions
Calculate the sample proportions of defective chips for each plant:
[tex]\[ \hat{p}_A = \frac{x_A}{n_A} = \frac{12}{80} = 0.15 \][/tex]
[tex]\[ \hat{p}_B = \frac{x_B}{n_B} = \frac{10}{90} = 0.1111111111111111 \][/tex]
Step 4: Calculate the Pooled Proportion
The pooled proportion [tex]\(\hat{p}_{\text{pooled}}\)[/tex] is calculated as:
[tex]\[ \hat{p}_{\text{pooled}} = \frac{x_A + x_B}{n_A + n_B} = \frac{12 + 10}{80 + 90} = \frac{22}{170} = 0.12941176470588237 \][/tex]
Step 5: Calculate the Standard Error
The standard error (SE) for the difference in proportions is given by:
[tex]\[ SE = \sqrt{\hat{p}_{\text{pooled}} (1 - \hat{p}_{\text{pooled}}) \left(\frac{1}{n_A} + \frac{1}{n_B}\right)} = \sqrt{0.12941176470588237 \cdot 0.8705882352941176 \left(\frac{1}{80} + \frac{1}{90}\right)} = 0.05157645508324752 \][/tex]
Step 6: Calculate the Z-Score
We calculate the z-score using the sample proportions and the standard error:
[tex]\[ z = \frac{\hat{p}_A - \hat{p}_B}{SE} = \frac{0.15 - 0.1111111111111111}{0.05157645508324752} = 0.7540046873349452 \][/tex]
Step 7: Determine the P-Value
Since we are conducting a one-tailed test (to see if [tex]\( p_A \)[/tex] is greater than [tex]\( p_B \)[/tex]), we need the area to the right of the z-score on the normal distribution:
[tex]\[ p\text{-value} = 1 - \Phi(z) \][/tex]
Using the z-score of 0.7540046873349452:
The corresponding p-value is 0.22542320358226453.
Step 8: Make a Decision
Given that our p-value (0.22542320358226453) is greater than any common significance level (like 0.05 or 0.01), we do not have sufficient evidence to reject the null hypothesis.
Thus, the correct p-value for the given hypotheses is approximately 0.23.
Step 1: State the Hypotheses
We are given:
- [tex]\( p_A \)[/tex]: the true proportion of defective chips from plant [tex]\( A \)[/tex]
- [tex]\( p_B \)[/tex]: the true proportion of defective chips from plant [tex]\( B \)[/tex]
The hypotheses can be stated as:
- [tex]\( H_0: p_A = p_B \)[/tex]
- [tex]\( H_a: p_A > p_B \)[/tex]
Step 2: Collect Sample Data
From the problem:
- Sample size from plant [tex]\( A \)[/tex], [tex]\( n_A = 80 \)[/tex]
- Defective chips from plant [tex]\( A \)[/tex], [tex]\( x_A = 12 \)[/tex]
- Sample size from plant [tex]\( B \)[/tex], [tex]\( n_B = 90 \)[/tex]
- Defective chips from plant [tex]\( B \)[/tex], [tex]\( x_B = 10 \)[/tex]
Step 3: Calculate Sample Proportions
Calculate the sample proportions of defective chips for each plant:
[tex]\[ \hat{p}_A = \frac{x_A}{n_A} = \frac{12}{80} = 0.15 \][/tex]
[tex]\[ \hat{p}_B = \frac{x_B}{n_B} = \frac{10}{90} = 0.1111111111111111 \][/tex]
Step 4: Calculate the Pooled Proportion
The pooled proportion [tex]\(\hat{p}_{\text{pooled}}\)[/tex] is calculated as:
[tex]\[ \hat{p}_{\text{pooled}} = \frac{x_A + x_B}{n_A + n_B} = \frac{12 + 10}{80 + 90} = \frac{22}{170} = 0.12941176470588237 \][/tex]
Step 5: Calculate the Standard Error
The standard error (SE) for the difference in proportions is given by:
[tex]\[ SE = \sqrt{\hat{p}_{\text{pooled}} (1 - \hat{p}_{\text{pooled}}) \left(\frac{1}{n_A} + \frac{1}{n_B}\right)} = \sqrt{0.12941176470588237 \cdot 0.8705882352941176 \left(\frac{1}{80} + \frac{1}{90}\right)} = 0.05157645508324752 \][/tex]
Step 6: Calculate the Z-Score
We calculate the z-score using the sample proportions and the standard error:
[tex]\[ z = \frac{\hat{p}_A - \hat{p}_B}{SE} = \frac{0.15 - 0.1111111111111111}{0.05157645508324752} = 0.7540046873349452 \][/tex]
Step 7: Determine the P-Value
Since we are conducting a one-tailed test (to see if [tex]\( p_A \)[/tex] is greater than [tex]\( p_B \)[/tex]), we need the area to the right of the z-score on the normal distribution:
[tex]\[ p\text{-value} = 1 - \Phi(z) \][/tex]
Using the z-score of 0.7540046873349452:
The corresponding p-value is 0.22542320358226453.
Step 8: Make a Decision
Given that our p-value (0.22542320358226453) is greater than any common significance level (like 0.05 or 0.01), we do not have sufficient evidence to reject the null hypothesis.
Thus, the correct p-value for the given hypotheses is approximately 0.23.
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