/*! 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 1 Aw, nuts! A company claims that ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Aw, nuts! A company claims that each batch of its deluxe mixed nuts contains \(52 \%\) cashews, \(27 \%\) almonds, \(13 \%\) macadamia nuts, and \(8 \%\) brazil nuts. To test this claim, a quality-control inspector takes a random sample of 150 nuts from the latest batch. The one-way table below displays the sample data. $$ \begin{array}{lcccc} \hline \text { Nut: } & \text { Cashew } & \text { Almond } & \text { Macadamia } & \text { Brazil } \\ \text { Count: } & 83 & 29 & 20 & 18 \\ \hline \end{array} $$ (a) State appropriate hypotheses for performing a test of the company's claim. (b) Calculate the expected counts for each type of nut. Show your work.

Short Answer

Expert verified
Null hypothesis: The proportions are as claimed. Expected counts: Cashews=78, Almonds=40.5, Macadamia=19.5, Brazil=12.

Step by step solution

01

Understand the problem

The company claims specific proportions of nuts in each batch, and we need to test these proportions using the given sample data.
02

Formulate the hypotheses

We set the null hypothesis to match the company's claim: the nut proportions are as claimed — 52% cashews, 27% almonds, 13% macadamia, and 8% Brazil nuts. The alternative hypothesis states that the nut proportions in the sample do not match the company's claim.
03

Identify the sample size

Identify that the sample size, which is the total number of nuts sampled, is 150.
04

Calculate expected counts for Cashews

Calculate the expected count for cashews by multiplying the sample size by the claimed proportion: \[0.52 \times 150 = 78\].
05

Calculate expected counts for Almonds

Calculate the expected count for almonds by multiplying the sample size by the claimed proportion: \[0.27 \times 150 = 40.5\].
06

Calculate expected counts for Macadamia Nuts

Calculate the expected count for macadamia nuts by multiplying the sample size by the claimed proportion: \[0.13 \times 150 = 19.5\].
07

Calculate expected counts for Brazil Nuts

Calculate the expected count for Brazil nuts by multiplying the sample size by the claimed proportion: \[0.08 \times 150 = 12\].

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 statistical method that helps us decide if there is enough evidence in a sample to draw conclusions about a population. In this exercise, we are working with nut proportions. The company claims certain percentages for cashews, almonds, macadamia, and Brazil nuts. We use hypothesis testing to verify if these claims hold true for the sample taken.

To begin, we start with two main hypotheses:
  • Null Hypothesis ( H_0 ): The company's claimed nut proportions are correct. This means the sample matches the claimed percentages: 52% cashews, 27% almonds, 13% macadamia nuts, and 8% Brazil nuts.
  • Alternative Hypothesis ( H_a ): The proportions of nuts in the sample are not as the company claims. This suggests a difference in at least one of the nut types from the stated percentages.
Once these hypotheses are clear, we proceed with the statistical test, such as the Chi-square test, to compare observed counts with expected counts and determine if any discrepancies are statistically significant.
Expected Counts Calculation
When performing a Chi-square test, calculating expected counts is a critical step. These are the counts we would expect to see for each category if the null hypothesis is true.

For our exercise, the expected count for each type of nut is calculated by multiplying the total sample size, which is 150 nuts, by the claimed proportion.
  • Cashews: With a claimed proportion of 52%, the expected count is calculated as \(0.52 \times 150 = 78\).
  • Almonds: Similarly, with a proportion of 27%, we expect \(0.27 \times 150 = 40.5\).
  • Macadamia Nuts: For 13%, the expected count is \(0.13 \times 150 = 19.5\).
  • Brazil Nuts: Finally, with 8%, the expected number is \(0.08 \times 150 = 12\).
These expected counts form the basis for comparison against observed counts to assess the fit of the data to the claims.
Proportion Testing
Proportion testing examines whether the sample proportions match a specified distribution. It’s particularly useful when you have data in categories, like types of nuts, and want to see if they adhere to expected norms.

In our scenario, we compare the observed nut counts from the sample to the expected counts under the claimed distribution. The proportions given by the company (52% cashews, 27% almonds, 13% macadamia, and 8% Brazil nuts) serve as the expected proportions.

Using the Chi-square test for goodness of fit, we calculate the test statistic:
  • For each nut type, find the difference between observed and expected counts.
  • Square these differences, divide by the expected counts, and sum up all these values across nut types.
The resulting Chi-square statistic is then compared with a critical value from the Chi-square distribution table, considering the degrees of freedom (equal to the number of categories minus one), to decide whether the observed distribution significantly departs from the expected one.

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

Why men and women play sports Do men and women participate in sports for the same reasons? One goal for sports participants is social comparison the desire to win or to do better than other people. Another is mastery-the desire to improve one's skills or to try one's best. A study on why students participate in sports collected data from independent random samples of 67 male and 67 female undergraduates at a large university. \({ }^{13}\) Each student was classified into one of four categories based on his or her responses to a questionnaire about sports goals. The four categories were high social comparison-high mastery (HSC-HM), high social comparison-low mastery (HSC-L.M), low social comparison-high mastery (LSC-HM), and low social comparison-low mastery (LSC-L.M). One purpose of the study was to compare the goals of male and female students. Here are the data displayed in a two-way table: $$ \begin{array}{lcc} {\text { Gender }} \\ \text { Goal } & \text { Female } & \text { Male } \\ \text { HSC-HM } & 14 & 31 \\ \text { HSC-LM } & 7 & 18 \\ \text { LSC-HM } & 21 & 5 \\ \text { LSC-LM } & 25 & 13 \\ \hline \end{array} $$ (a) Calculate the conditional distribution (in proportions) of the reported sports goals for each gender. (b) Make an appropriate graph for comparing the conditional distributions in part (a). (c) Write a few sentences comparing the distributions of sports goals for male and female undergraduates.

Exercises 59 to 60 refer to the following setting. For their final project, a group of AP \(^{\otimes}\) Statistics students investigated the following question: "Will changing the rating scale on a survey affect how people answer the question?" To find out, the group took an SRS of 50 students from an alphabetical roster of the school's just over 1000 students. The first 22 students chosen were asked to rate the cafeteria food on a scale of 1 (terrible) to 5 (excellent). The remaining 28 students were asked to rate the cafeteria food on a scale of 0 (terrible) to 4 (excellent). Here are the data: $$ \begin{array}{lcccrc} &{1 \text { to 5 scale }} \\ \text { Rating } & 1 & 2 & 3 & 4 & 5 \\ \text { Frequency } & 2 & 3 & 1 & 13 & 3 \\ \hline & {0 \text { to 4 scale }} \\ \text { Rating } & 0 & 1 & 2 & 3 & 4 \\ \text { Frequency } & 0 & 0 & 2 & 18 & 8 \\ \hline \end{array} $$ Average ratings (1.3,10.2) The students decided to compare the average ratings of the cafeteria food on the two scales. (a) Find the mean and standard deviation of the ratings for the students who were given the 1 -to- 5 scale. (b) For the students who were given the 0 -to- 4 scale, the ratings have a mean of 3.21 and a standard deviation of \(0.568 .\) Since the scales differ by one point, the group decided to add 1 to each of these ratings. What are the mean and standard deviation of the adjusted ratings? (c) Would it be appropriate to compare the means from parts (a) and (b) using a two-sample \(t\) test? Justify your answer.

Skittles Statistics teacher Jason Molesky contacted Mars, Inc., to ask about the color distribution for Skittles candies. Here is an excerpt from the response he received: "The original flavor blend for the SKITTLES BITE SIZE CANDIES is lemon, lime, orange, strawberry and grape. They were chosen as a result of consumer preference tests we conducted. The flavor blend is 20 percent of each flavor." (a) State appropriate hypotheses for a significance test of the company's claim. (b) Find the expected counts for a bag of Skittles with 60 candies. (c) How large a \(\chi^{2}\) statistic would you need to have significant evidence against the company's claim at the \(\alpha=0.05\) level? At the \(\alpha=0.01\) level? (d) Create a set of observed counts for a bag with 60 candies that gives a \(P\) -value between 0.01 and \(0.05 .\) Show the calculation of your chi-square statistic.

Multiple choice: Select the best answer for Exercises 19 to 22 Exercises 19 to 21 refer to the following setting. The manager of a high school cafeteria is planning to offer several new types of food for student lunches in the following school year. She wants to know if each type of food will be equally popular so she can start ordering supplies and making other plans. To find out, she selects a random sample of 100 students and asks them, "Which type of food do you prefer: Asian food, Mexican food, pizza, or hamburgers?" Here are her data: $$ \begin{array}{lcccc} \hline \text { Type of Food: } & \text { Asian } & \text { Mexican } & \text { Pizza } & \text { Hamburgers } \\ \text { Count: } & 18 & 22 & 39 & 21 \\ \hline \end{array} $$ An appropriate null hypothesis to test whether the food choices are equally popular is (a) \(H_{0}: \mu=25,\) where \(\mu=\) the mean number of students that prefer each type of food. (b) \(H_{0}: p=0.25,\) where \(p=\) the proportion of all students who prefer Asian food. (c) \(H_{0}: n_{A}=n_{M}=n_{P}=n_{H}=25,\) where \(n_{A}\) is the number of students in the school who would choose Asian food, and so on. (d) \(H_{0}: p_{A}=p_{M}=p_{P}=p_{H}=0.25,\) where \(p_{A}\) is the proportion of students in the school who would choose Asian food, and so on. (e) \(\quad H_{0}: \hat{p}_{\mathrm{A}}=\hat{p}_{M}=\hat{p}_{P}=\hat{p}_{H}=0.25,\) where \(\hat{p}_{\mathrm{A}}\) is the pro- portion of students in the sample who chose Asian food, and so on.

Sorry, no chi-square How do U.S. residents who travel overseas for leisure differ from those who travel for business? The following is the breakdown by occupation: \({ }^{17}\) $$ \begin{array}{lcc} \hline \text { 0ccupation } & \text { Leisure } & \text { Business } \\ \text { Professional/technical } & \text { travelers (\%) } & \text { travelers (\%) } \\ \text { Manager/executive } & 36 & 39 \\ \text { Retired } & 23 & 48 \\ \text { Student } & 14 & 3 \\ \text { 0ther } & 7 & 3 \\ \text { Total } & 20 & 7 \\ \hline \end{array} $$ Explain why we can't use a chi-square test to learn whether these two distributions differ significantly.

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.