To determine whether there is a statistically significant difference in the defect rates between the two factories, we can perform a two-proportion z-test. The null hypothesis in this case would be that the defect rates for both factories are equal, and the alternative hypothesis would be that the defect rates are different.

Step-by-Step Solution:

1. Define the Problem:

Let:

• $p_A$ be the defect rate for Factory A
• $p_B$ be the defect rate for Factory B

From the data:

• Factory A produced 1000 units, with 95 defective.
• Factory B produced 900 units, with 90 defective.

So, the defect rates are:

• $p_A = \frac{95}{1000} = 0.095$
• $p_B = \frac{90}{900} = 0.1$

2. Hypotheses:

• Null hypothesis (H₀): $p_A = p_B$ (there is no significant difference between the defect rates)
• Alternative hypothesis (H₁): $p_A \neq p_B$ (there is a significant difference between the defect rates)

3. Standard Error of the Difference in Proportions:

First, calculate the pooled proportion, since we are assuming the null hypothesis is true (i.e., that the defect rates are the same in both factories): $p_{\text{pooled}} = \frac{x_A + x_B}{n_A + n_B}$ where:

• $x_A = 95$, the number of defectives in Factory A
• $x_B = 90$, the number of defectives in Factory B
• $n_A = 1000$, the total units produced by Factory A
• $n_B = 900$, the total units produced by Factory B

$p_{\text{pooled}} = \frac{95 + 90}{1000 + 900} = \frac{185}{1900} = 0.09737$

Next, calculate the standard error (SE) of the difference in proportions: $SE = \sqrt{p_{\text{pooled}}(1 - p_{\text{pooled}})\left(\frac{1}{n_A} + \frac{1}{n_B}\right)}$ $SE = \sqrt{0.09737(1 - 0.09737)\left(\frac{1}{1000} + \frac{1}{900}\right)}$

$SE = \sqrt{0.09737 \times 0.90263 \times \left(\frac{1}{1000} + \frac{1}{900}\right)}$ $SE = \sqrt{0.08796 \times (0.001111 + 0.001111)}$ $SE = \sqrt{0.08796 \times 0.002222}$ $SE = \sqrt{0.000195}$ $SE \approx 0.01396$

4. Calculate the Z-Score:

The z-score is calculated as: $z = \frac{p_A - p_B}{SE}$ $z = \frac{0.095 - 0.1}{0.01396}$ $z = \frac{-0.005}{0.01396} \approx -0.358$

5. Determine the P-Value:

The p-value is the probability that the observed difference (or a more extreme difference) occurs under the null hypothesis. For a two-tailed test, we calculate the probability that the z-score falls beyond ±0.358.

Using a standard normal distribution table or a statistical calculator, the p-value for a z-score of -0.358 is approximately 0.72.

6. Conclusion:

• P-value = 0.72 (which is much greater than the significance level of 0.05).
• Since the p-value is greater than 0.05, we fail to reject the null hypothesis. This means that there is no statistically significant difference in the defect rates between the two factories at the 95% confidence level.

Summary:

• P-value = 0.72
• There is no significant difference in the defect rates between Factory A and Factory B.

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Would you like further details on any step, or would you like to explore similar examples?

Here are some related questions:

1. How does the sample size affect the p-value in a two-proportion z-test?
2. What assumptions are made in a two-proportion z-test?
3. How do you perform a two-proportion z-test for dependent samples?
4. What is the effect of using a one-tailed test versus a two-tailed test in hypothesis testing?
5. How do you calculate confidence intervals for the difference in proportions?

Tip: Always ensure you understand whether your data is independent or paired before choosing between a two-proportion or paired z-test.