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In an experiment to compare the quality of four different brands of reel-to- reel recording tape, five \(2400-\mathrm{ft}\) reels of each brand (A-D) were selected and the number of flaws in each reel was determined. \(\begin{array}{lrrrrr}\text { A: } & 10 & 5 & 12 & 14 & 8 \\ \text { B: } & 14 & 12 & 17 & 9 & 8 \\ \text { C: } & 13 & 18 & 10 & 15 & 18 \\ \text { D: } & 17 & 16 & 12 & 22 & 14\end{array}\) It is believed that the number of flaws has approximately a Poisson distribution for each brand. Analyze the data at level \(.01\) to see whether the expected number of flaws per reel is the same for each brand.

Short Answer

Expert verified
Reject H0; expected flaws differ across brands.

Step by step solution

01

State the Hypotheses

We need to determine if the expected number of flaws per reel is the same for each brand. Let's state the null hypothesis (H0) and alternative hypothesis (H1):- H0: \( \lambda_A = \lambda_B = \lambda_C = \lambda_D \) (the expected number of flaws, \( \lambda \), is the same for all brands).- H1: At least one brand has a different expected number of flaws.
02

Compute the Mean Number of Flaws for Each Brand

We need to calculate the sample mean (\( \bar{x} \)) for each brand, as this will help us estimate the rate parameter \( \lambda \) for the Poisson distribution.- Brand A: \(\bar{x}_A = \frac{10 + 5 + 12 + 14 + 8}{5} = 9.8 \)- Brand B: \(\bar{x}_B = \frac{14 + 12 + 17 + 9 + 8}{5} = 12 \)- Brand C: \(\bar{x}_C = \frac{13 + 18 + 10 + 15 + 18}{5} = 14.8 \)- Brand D: \(\bar{x}_D = \frac{17 + 16 + 12 + 22 + 14}{5} = 16.2 \)
03

Choose the Appropriate Statistical Test

Since we are dealing with means from a Poisson distribution with the same sample size for each treatment, we can use a one-way ANOVA to test the hypothesis that the means are equal across the groups. This will help us see if there are significant differences in the means at the \( \alpha = 0.01 \) level.
04

Compute the ANOVA Test Statistic

Calculate the between-group variability (SSB) and the within-group variability (SSW).1. Total Mean: \(\bar{x}_{total} = \frac{9.8 + 12 + 14.8 + 16.2}{4} = 13.2 \)2. SSB: \( SSB = 5[(9.8 - 13.2)^2 + (12 - 13.2)^2 + (14.8 - 13.2)^2 + (16.2 - 13.2)^2] = 138.0 \)3. SSW: Calculate each brand's deviations and sum of squares within their group. \( SSW = 4[(10-9.8)^2 + (5-9.8)^2 + (12-9.8)^2 + (14-9.8)^2 + (8-9.8)^2] + 4[(14-12)^2 + (12-12)^2 + ...] + ... = 128.4 \)
05

Calculate the F-statistic and Critical Value

The F-statistic is calculated as:\[ F = \frac{SSB / (k-1)}{SSW / (N-k)} = \frac{138 / 3}{128.4 / 16} \]Calculate:\[ F = \frac{46}{8.025} \approx 5.73 \]Where \(k\) is the number of groups (4) and \(N\) is total observations (20).Now, with \(3, 16\) degrees of freedom at \( \alpha = 0.01 \), find the critical F-value from F-distribution tables for 3 and 16 degrees of freedom, which is approximately \( 4.07 \).
06

Make a Decision

Since the calculated F-statistic \(5.73\) is greater than the critical F-value \(4.07\), we reject the null hypothesis.
07

Conclusion

There is sufficient evidence at the \( 0.01 \) significance level to suggest that the expected number of flaws per reel is different for at least one of the brands A, B, C, or D.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Poisson distribution
A Poisson distribution is a probability distribution that models the number of events occurring within a fixed interval of time or space. These events are assumed to occur with a constant mean rate, and each event happens independently of the time since the last event. In simpler terms, it's a way to predict the number of times an event will happen over a specified time period. For example, in this exercise, we're looking at the expected number of flaws on a reel of tape.
  • Rate Parameter (\( \lambda \)): This is the average number of events (flaws) in a given time period or space. In our case, each brand has a Poisson-distributed number of flaws.
  • Characteristics: The Poisson distribution is defined for events that occur rarely over a specified domain, where each event is independent.
To determine the best use of the Poisson distribution, you should check if your data matches these conditions: the occurrences should be random, independent, and the average rate should be fixed throughout your measurement.
hypothesis testing
Hypothesis testing is a statistical method used to make decisions about a population based on a sample. It's about testing assumptions and is widely used to determine if there is enough evidence to reject a stated hypothesis, called the null hypothesis (H0).
In this exercise, hypothesis testing is used to compare if the expected number of flaws in recording tapes from different brands are statistically the same.
  • Null Hypothesis (H0): Assumes no difference in the expected number of flaws across all brands. In math terms, \( \lambda_A = \lambda_B = \lambda_C = \lambda_D \).
  • Alternative Hypothesis (H1): At least one brand differs in its expected number of flaws.
Hypothesis testing guides us in deciding whether the observed differences in sample data are due to random chance or if they are significant enough to suggest a difference in the whole population. In essence, it's a great tool to produce reliable conclusions backed by data.
statistical significance
Statistical significance is a concept used to determine if the results of an experiment or study are likely to be true and not due to chance. It’s a measure of how much evidence we have against a null hypothesis. In our exercise, the statistical test aims to identify significant differences in the mean number of flaws among different brands of tapes.
  • Significance Level (\(\alpha\)): The threshold for determining if an observed effect is significant. In this exercise, it's set to 0.01, which indicates a 1% chance of believing there is an effect when there actually is none.
  • P-value: The probability that the observed data (or more extreme data) would occur if the null hypothesis were true. If the p-value is less than \(\alpha\), the results are statistically significant.
In our case, since the F-statistic was larger than the critical F-value, the test results are statistically significant. This means we have enough evidence to reject the null hypothesis and accept that there is a notable difference in the number of flaws among the brands.
F-distribution
The F-distribution is a continuous probability distribution often used in the context of ANOVA (Analysis of Variance). It's useful for comparing variances across different groups or samples to see if they come from the same population.
In this exercise, the F-distribution helps us determine if the variance between groups (different brands) is significantly greater than the variance within groups (individual reels from the same brand). The F-statistic is calculated by dividing the variance between the group means by the variance within the groups.
  • Degrees of Freedom: Important components in the F-distribution, degrees of freedom are linked to the number of groups and the size of each group. They are used to determine the critical value from an F-table.
  • Critical F-value: A value obtained from an F-distribution table that you compare your calculated F-statistic against. If the calculated F-statistic is greater than this critical value, you reject the null hypothesis.
By using the F-distribution in this analysis, we effectively assess if the differences in the observed data between the reel tapes of different brands are statistically significant. Understanding this distribution is crucial for deciding whether the differences observed are likely due to random variability or hold genuine significance.

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Most popular questions from this chapter

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