We are going to look at data from a study that examined medical records of elderly patients for calcium and inorganic phosphorus blood levels. One of the purposes of the study was to determine if different laboratories yield substantially different results for calcium measurements. There are six labs—each denoted by a number (1 – 6). All conditions of ANOVA were met.

C) Assuming it was appropriate to reject the null hypothesis (which may or may not be correct) of this test, what is the appropriate conclusion?

Question

We are going to look at data from a study that examined medical records of elderly patients for calcium and inorganic phosphorus blood levels. One of the purposes of the study was to determine if different laboratories yield substantially different results for calcium measurements.  There are six labs—each denoted by a number (1 – 6).  All conditions of ANOVA were met.

C) Assuming it was appropriate to reject the null hypothesis (which may or may not be correct) of this test, what is the appropriate conclusion?

BlackTom AI · Accepted Answer

Rejecting the null hypothesis in a one-way ANOVA implies that not all group means are equal, so at least one laboratory yields a different average calcium measurement.

1) Relevant concepts
- One-way ANOVA compares k group means (here k = 6 labs) using between-group and within-group variability.
- Null hypothesis H0: μ1 = μ2 = ... = μ6 (all lab means equal). Alternative Ha: at least one μi differs.
- Test statistic: F = MSB / MSW, where MSB = SSB/(k−1) and MSW = SSW/(N−k). If F is large and p-value small (below α), reject H0.
- Conditions met (stated) justify using the ANOVA inference.

2) Step-by-step reasoning
- Compute each lab’s sample mean and the overall mean.
- Compute SSB = Σ ni (mean_i − overall_mean)^2 (measures variation between lab means).
- Compute SSW = Σ Σ (xij − mean_i)^2 (measures variation within labs).
- Degrees of freedom: df_between = k−1 = 5; df_within = N−k.
- Compute MSB = SSB / 5 and MSW = SSW / (N−6).
- Compute F = MSB / MSW and corresponding p-value.
- Given the problem statement “Assuming it was appropriate to reject the null hypothesis,” the calculated F must have produced a p-value < α, so we reject H0.

3) Calculation template (intermediate steps)
- Example symbolic steps: SSB → divide by 5 → MSB; SSW → divide by N−6 → MSW; F = MSB/MSW → compare to F_{5, N−6, α} or use p-value.
- Because rejection is assumed, F exceeded the critical value or p < α.

4) Why the correct answer is correct
- Rejecting H0 in this ANOVA means there is statistical evidence that the lab performing the bloodwork affects the true average calcium level (i.e., at least one lab mean differs). ANOVA does not tell which labs differ or the direction; additional post-hoc pairwise comparisons would be needed to identify specific differences.

5) Why the other option is incorrect
- “There is not evidence...” corresponds to failing to reject H0. That contradicts the given assumption that the null was rejected, so it is not correct.

6) Conclusion
- Therefore, the appropriate conclusion is: “There is evidence to show that which lab does the bloodwork affects the true average calcium amounts.” This matches the provided answer.

类似问题

Using the same dataset, Flaye now decides to investigate whether concert attendance differs by the day of the concert (Friday, Saturday, or Sunday). Assuming all assumptions are met, what statistical test would you suggest they run?

更多留学生实用工具

智能学习助手

风格化写作助手

论文查重助手

文献引用助手

课堂转译助手

课堂笔记助手

Quiz搜索助手

学校历年真题

智能刷题助手

智能匹配练习

希望你的学习变得更简单