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Chapter 10: Problem 81

\( \quad \wedge\) machine is supposed to mix peanuts, hazelnuts, cashews, and pecans in the ratio 5: 2: 2: 1 . A can containing 500 of these mixed nuts was found to have 269 peanuts, 112 hazelnuts, 74 cashews, and 45 pecans. At the 0.05 level of significance, test the hypothesis that the machine is mixing the nuts in the ratio \(5: 2 ; 2: 1\).

Short Answer

Expert verified
The nut-mixing machine is not mixing the nuts in the ratio of 5:2:2:1 at the 0.05 level of significance, as the Chi-square test statistic (10.22) is greater than the critical contribution (7.815), which leads to the conclusion that there is significant evidence to reject the null hypothesis.

Step by step solution

01

Determine the Observed Frequencies

The given problem provides the observed frequencies for each type of nut: peanuts, 269; hazelnuts, 112; cashews, 74; pecans, 45.
02

Calculate the Expected Frequencies

In order to calculate the expected frequencies, determine the portion of the 500 based on the expected ratios. For peanuts (50% of 500), hazelnuts (20% of 500), cashews (20% of 500), and pecans (10% of 500), this would result in expected frequencies of: peanuts, 250; hazelnuts, 100; cashews, 100; pecans, 50.
03

Compute the Chi-square Test Statistic

The Chi-square test statistic is calculated with the formula \(\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}\), where \(O_i\) are the observed frequencies and \(E_i\) are the expected frequencies. Here, it would be \(\chi^2 = \sum \frac{(269 - 250)^2}{250} + \frac{(112 - 100)^2}{100} + \frac{(74 - 100)^2}{100} + \frac{(45 - 50)^2}{50} = 1.52 + 1.44 + 6.76 + 0.5 = 10.22
04

Determine the Critical Chi-square Value

For a 0.05 level of significance and 3 degrees of freedom (4 categories of nuts minus 1), we can lookup in a Chi-square distribution table for the critical value, which we find to be 7.815.
05

Making the decision for the Null Hypothesis

The Chi-square test statistic 10.22 is greater than the critical contribution 7.815. Therefore, at the 0.05 level of significance, we reject the null hypothesis which suggests that the machine is mixing the nuts at a ratio of 5:2:2:1.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a crucial part of statistics that allows us to make inferences or decisions about a population based on sample data. In our nut mixing example, we want to see if the machine actually mixes nuts in the expected ratio of 5:2:2:1.
  • The **null hypothesis** ( H_0 ) states that there is no difference between the observed and expected mixes. H_0: The machine mixes the nuts in the ratio 5:2:2:1.
  • The **alternative hypothesis** ( H_a ) proposes that the mix is not according to the expected ratio. H_a: The machine does not mix nuts in the ratio 5:2:2:1.

We use statistical tests, like the Chi-square test, to determine whether to reject the null hypothesis or not. Choosing to reject or not depends on the comparison between the critical value and the computed test statistic. Ensuring our inference is both rigorous and controlled by statistical measures is the aim of hypothesis testing.
Expected Frequencies
Expected frequencies are calculated based on the hypothesis that is being tested. In the context of the nut mixing problem, they represent the quantities we would anticipate if the machine were functioning correctly, according to the proposed ratio of 5:2:2:1.
To calculate expected frequencies:
  • **Peanuts:** 50% of 500 = 250
  • **Hazelnuts:** 20% of 500 = 100
  • **Cashews:** 20% of 500 = 100
  • **Pecans:** 10% of 500 = 50

Each expected frequency derives from the total number, 500, and adheres strictly to the proportions outlined in the hypothesis. These frequencies are later compared to the observed frequencies using the Chi-square test to see if they align with the machine's output.
Observed Frequencies
Observed frequencies are what we see directly in the data collected from our experiment or observation. These are the actual counts of the nuts provided by the machine. In this exercise, students will understand how real-world outputs are compared against theoretical or expected values.
For this problem:
  • **Peanuts:** Observed = 269
  • **Hazelnuts:** Observed = 112
  • **Cashews:** Observed = 74
  • **Pecans:** Observed = 45
The differences between observed and expected frequencies will be scrutinized in the Chi-square test. Observed frequencies are original data points, key in assessing if there鈥檚 a deviation from what was anticipated.
Level of Significance
The level of significance plays a critical role in hypothesis testing as it defines the probability threshold for assessing evidence against the null hypothesis. Often denoted as 伪 , it represents the risk we are willing to take of making a Type I error鈥攔ejecting a true null hypothesis.
In this nut mixing analysis, we use a 0.05 level of significance. This suggests that there is a 5% risk of asserting there is an issue with the nut mix, when there may not be. 鈥孋hoosing a level of significance involves balancing rigor in detecting effects and safety against false alarms.
鈥� Once this level is set, it is utilized to find the critical value in a Chi-square distribution table, helping guide the decision whether to reject the null hypothesis based on the test statistic calculated.

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Most popular questions from this chapter

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