Analyze Phase

When performing a Two Sample Proportion Test, the correct first step is to calculate the sample proportions for each supplier and determine the pooled proportion under the null hypothesis.

Here's why this is correct:

1. First, you calculate the individual sample proportions:
2. Supplier A: p̂₁ = 42/350 = 0.12 (12%)

3. Supplier B: p̂₂ = 58/420 = 0.138 (13.8%)

4. Then, under the null hypothesis (which assumes both populations have the same proportion), you calculate the pooled proportion:

5. p̂pooled = (42 + 58)/(350 + 420) = 100/770 = 0.13 (13%)

This pooled proportion is essential because it's used in the standard error calculation for the test statistic. The Two Sample Proportion Test assumes equal proportions under the null hypothesis, so pooling the data gives the best estimate of this common proportion.

The option suggesting unpooled standard errors is incorrect because the standard approach for hypothesis testing of two proportions uses the pooled proportion when calculating the standard error, not separate unpooled estimates.

The option mentioning chi-square distribution is incorrect because while chi-square can be used for comparing proportions, this question specifically asks about the Two Sample Proportion Test (z-test approach), and calculating the difference first before applying critical values reverses the proper sequence.

The option about verifying the confidence interval contains zero is incorrect because checking whether a confidence interval contains zero is done after the analysis to interpret results, not as a preliminary step before calculating the test statistic.

In One-Way ANOVA, the fundamental principle being tested is whether the variation between group means (between-group variance) is significantly larger than what would be expected from random chance alone (within-group variance or error variance).

The F-statistic is calculated as the ratio of between-group variance to within-group variance (MSB/MSW). When the F-statistic (4.76) exceeds the critical value (2.87), we reject the null hypothesis, which states that all group means are equal.

The correct answer captures this principle precisely: rejecting the null hypothesis demonstrates that the variance attributable to supplier differences (between-group variance) is statistically larger than what random sampling variation within groups would produce. This is the core logic of ANOVA - we're comparing systematic variation due to the factor (suppliers) against random error variation.

The second option incorrectly reverses the logic - ANOVA doesn't conclude that within-group variance is smaller than between-group variance; it concludes that between-group variance is larger than expected given the within-group variance.

The third option is incorrect because ANOVA doesn't test whether total variance equals pooled variance estimates - it specifically partitions and compares variance components.

The fourth option incorrectly suggests proportionality between variance and means, which is not the statistical inference being made. ANOVA tests whether differences exist, not whether variance is proportional to mean deviations.

The correct answer states that approximately 5 tests would be expected to incorrectly reject true null hypotheses, representing the cumulative effect of alpha risk across multiple comparisons.

Here's the mathematical reasoning:

• Alpha (α) = 0.05 represents the probability of making a Type I error (falsely rejecting a true null hypothesis) for each individual test
• When all null hypotheses are actually true, the expected number of false rejections = n × α = 100 × 0.05 = 5 tests
• This is the fundamental concept of Type I error rate and demonstrates why multiple testing problems exist in statistical analysis

Why the other options are incorrect:

• The second option incorrectly references beta risk (Type II error), which relates to failing to detect real differences when they exist. However, the question states all null hypotheses are TRUE, meaning there are no real differences to detect. Beta risk is irrelevant in this scenario.

• The third option incorrectly states 95 tests would accept false null hypotheses. This is flawed because if all null hypotheses are true, there are no false null hypotheses to accept. The 95% represents the probability of correctly retaining true null hypotheses, not accepting false ones.

• The fourth option incorrectly discusses identifying true differences and statistical power. Since all null hypotheses are true (no real differences exist), power is not applicable here. Power relates to detecting effects when they actually exist.

This question highlights the important Lean Six Sigma concept of understanding Type I error accumulation when conducting multiple hypothesis tests, which is why techniques like Bonferroni correction are sometimes applied.