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A sample proportion represents the fraction of items in a sample that exhibit a particular characteristic, which in this case is being defective. When calculating sample proportions, it's essential to consider the size of the sample and the number of observed occurrences of the characteristic. In the context of the assembly line exercise, for Line A, there were 18 defective items out of a sample of 100. This gives us the sample proportion for Line A, denoted as \( \hat{p}_A = \frac{18}{100} = 0.18 \). Similarly, for Line B, the sample proportion \( \hat{p}_B \) is calculated as \( \frac{12}{100} = 0.12 \).

• Sample size for Line A: 100
• Number of defectives for Line A: 18
• Sample size for Line B: 100
• Number of defectives for Line B: 12

By understanding these individual sample proportions, we can begin to compare the proportion of defectives between the two lines.

The standard error is crucial when determining the reliability of a sample statistic by quantifying the extent to which the sample results are expected to vary. In this exercise, we calculate the standard error of the difference in sample proportions between two assembly lines. The formula used is:\[ SE = \sqrt{ \frac{\hat{p}_A(1-\hat{p}_A)}{n_A} + \frac{\hat{p}_B(1-\hat{p}_B)}{n_B} } \]Where:

• \( \hat{p}_A \) and \( \hat{p}_B \) are the sample proportions for Line A and Line B, respectively.
• \( n_A \) and \( n_B \) are the sample sizes, each equal to 100 in this case.

The calculated standard error \( SE \) is approximately \( 0.0503 \). This value tells us about the typical deviation between their respective sample proportions, aiding in the formation of a confidence interval.

Hypothesis testing is a method of statistical inference used to decide if the evidence in a sample supports a particular hypothesis about a population parameter. In the exercise, the focus is on determining whether there is a significant difference in the proportions of defective items between the two lines. For this, the null hypothesis (\( H_0 \)) assumes no difference in proportion, meaning \( \hat{p}_A - \hat{p}_B = 0 \). The alternative hypothesis (\( H_1 \)) suggests there is a difference. The confidence interval approach in this scenario helps to visually test the hypothesis without explicitly calculating a test statistic. Since the confidence interval \(-0.0572 \text{ to } 0.1772\) includes zero, it implies no significant evidence to reject the null hypothesis, concluding there is no significant difference in the defect proportions.

To calculate the confidence interval, we need the critical value, which comes from the standard normal distribution table or a z-score table. It reflects how many standard deviations our sample statistic is away from the hypothesized population parameter under the null hypothesis.In this exercise, a 98% confidence level means that there's a 1% risk in each tail of the distribution. This leads us to a critical value of approximately \( z = 2.33 \).

• The critical value helps determine the margin of error for the confidence interval.
• Multiplied with the standard error, it extends around the sample mean to create the confidence interval.

By harnessing this critical value, we determine how confident we are that the true difference in proportions falls within the calculated interval.