/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 81 \( \quad \wedge\) machine is sup... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

\( \quad \wedge\) 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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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’s 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 α , it represents the risk we are willing to take of making a Type I error—rejecting 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. ‌Choosing a level of significance involves balancing rigor in detecting effects and safety against false alarms.
‌ 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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In a study to estimate the proportion of residents in a certain city and its suburbs who favor the construction of a nuclear power plant, it is found that 63 of 100 urban residents favor the construction while only 59 of 125 suburban residents are in favor. Is there a significant difference between the proportion of urban and suburban residents who favor construction of the nuclear plant? Make use of a P-value.

Suppose that, in the past, \(40 \%\) of all adults favored capital punishment. Do we have reason to believe that the proportion of adults favoring capital punishment today has increased if, in a random sample of 15 adults, 8 favor capital punishment? Use a 0.05 level of significance.

A soft-drink machine at a steak house is regulated so that the amount of drink dispensed is approximately normally distributed with a mean of 200 milliliters and a standard deviation of 15 milliliters. The machine is checked periodically by taking a sample of 9 drinks and computing the average content. If \(x\) falls in the interval \(191

10.35 To find out whether a new serum will arrest leukemia, 9 mice, all with an advanced stage of the disease, are selected. Five mice receive the treatment and 4 do not. Survival times, in years, from the time the experiment commenced are as follows. $$\begin{array}{l|ccccc}\text { Treatment } & 2.1 & 5.3 & 1.4 & 4.6 & 0.9 \\\\\hline \text { No Treatment } & 1.9 & 0.5 & 2.8 & 3.1 &\end{array}$$ At the 0.05 level of significance can the serum be said to be effective? Assume the two distributions to be normally distributed with equal variances.

According to the published reports, practice under fatigued conditions distorts mechanisms which govern performance. An experiment was conducted using 15 college males who were trained to make a continuous horizontal right- to-left arm movement from a microswitch to a barrier, knocking over the barrier coincident with the arrival of a clock sweephand to the 6 o'clock position. The absolute value of the difference between the time, in milliseconds, that it took to knock over the barrier and the time for the sweephand to reach the 6 o'clock position \((500 \mathrm{msec})\) was recorded Each participant performed the task five times under prefatigue and postfatigue conditions, and the sums of the absolute differences for the five performances were recorded as follows: An increase in the mean absolute time differences when the task is performed under postfatigue conditions would support the claim that practice under fatigued conditions distorts mechanisms that govern performance. Assuming the populations to be normally distributed, test this claim.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.