/*! 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 13 Skittles Statistics teacher Jaso... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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.

Short Answer

Expert verified
The hypotheses are each color has a 20% proportion. Expected counts are 12 per color. Critical values are 9.488 and 13.277. Example chi-square statistic is 1.00.

Step by step solution

01

Formulate the Hypotheses

To test the company's claim about the distribution of Skittles colors, we use the null hypothesis \(H_0\) and the alternative hypothesis \(H_a\). - \(H_0\): Each color of Skittle (lemon, lime, orange, strawberry, grape) has a population proportion of 0.20.- \(H_a\): Not all colors have a population proportion of 0.20.
02

Calculate Expected Counts

The expected count for each color in a bag of 60 Skittles, given the company's claim, is calculated as follows:- Total candies = 60- Proportion for each color = 0.20Expected count for each color: \[ 0.20 \times 60 = 12 \]
03

Determine Critical Values for Chi-Square

We use the chi-square distribution table to find the critical values for significance levels \(\alpha = 0.05\) and \(\alpha = 0.01\). - \(df = k - 1 = 5 - 1 = 4\) where \(k\) is the number of categories.- Critical value for \(\alpha = 0.05\): \(\chi^2_{0.05, 4} = 9.488\) - Critical value for \(\alpha = 0.01\): \(\chi^2_{0.01, 4} = 13.277\)
04

Create Observed Counts and Calculate Chi-Square Statistic

Create a set of observed counts such that the \(P\)-value is between 0.01 and 0.05. Here is an example:Observed counts: {[10, 13, 14, 11, 12]}Calculate the \(\chi^2\) statistic:\[ \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \]Where \(O_i\) are the observed counts and \(E_i\) are the expected counts (12 for each color).\[\chi^2 = \frac{(10-12)^2}{12} + \frac{(13-12)^2}{12} + \frac{(14-12)^2}{12} + \frac{(11-12)^2}{12} + \frac{(12-12)^2}{12} = \frac{4}{12} + \frac{1}{12} + \frac{4}{12} + \frac{1}{12} + \frac{0}{12} = 1.00\]This chi-square statistic should ideally be between the critical values found in Step 3 (i.e., create different observed counts if necessary) to get a \(P\)-value between 0.01 and 0.05.

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.

Statistical Hypotheses
When conducting a Chi-Square Test on Skittles flavors, statistical hypotheses play a crucial role.
Hypotheses are statements that we propose to evaluate using statistical data.

In this context, we use two types of hypotheses: the null hypothesis and the alternative hypothesis.
  • **Null Hypothesis ( \(H_0\) )**: This hypothesis states that each color of Skittles (lemon, lime, orange, strawberry, grape) makes up 20% of the total population of Skittles. Thus, mathematically expressed as: each color's proportion = 0.20.
  • **Alternative Hypothesis ( \(H_a\) )**: Contrary to the null hypothesis, it suggests that one or more Skittle colors do not have the claimed 20% distribution.
These hypotheses are critical as they set the framework for the Chi-Square Test and determine the direction of our test analysis. Success or failure in rejecting the null hypothesis will indicate whether Mars, Inc.'s claim about Skittles' color distribution is supported by the test data or not.
By clearly defining these hypotheses, statisticians can objectively evaluate if the observed sample data aligns with the expected distribution in the hypothesis.
Expected Counts
In a Chi-Square Test, expected counts are vital calculations that help us determine if observed data significantly deviates from what we would expect. To compute expected counts for the Skittles, we base our calculation on Mars, Inc.'s claim that each color is equally distributed at 20% in every packet.
  • **Total candies**: For a bag containing 60 Skittles, each color should theoretically represent 20% or one-fifth of the total count.
  • **Expected count per color**: Mathematically, we multiply the proportion by the total candies. Hence, the expected count becomes: \[0.20 \times 60 = 12\]
This expected count of 12 Skittles per color serves as a baseline for comparison with actual sample observations.
Deviations from this expected count help us determine the variance in the Chi-Square Test. It's important because this comparison tells us if any discrepancies are due to random chance or suggest a different distribution pattern among the Skittles colors.
By understanding and calculating the expected counts, we can delve deeper into assessing whether the claimed distribution aligns with our sample data.
Significance Level
The significance level, denoted by \(\alpha\), is a threshold used in hypothesis testing to determine statistical significance.
It helps us decide whether to accept the null hypothesis or not.
  • Common significance levels include **0.05** and **0.01**, representing a 5% and 1% risk of concluding that a difference exists when there is none.
  • These levels act as benchmarks; a result falling below \(\alpha\) suggests significant evidence against the null hypothesis.
In this Skittles exercise, finding the critical values at these significance levels helps identify the size of the Chi-Square statistic needed to find significant evidence against Mars, Inc.'s claim about color distribution.Using a chi-square distribution table:
  • For \(\alpha = 0.05\) with \(df = 4\), the critical value is approximately 9.488.
  • For \(\alpha = 0.01\), it is around 13.277.
If the calculated Chi-Square statistic exceeds these values, we reject the null hypothesis.
Understanding these levels allows researchers, like ourselves, to assess whether Mars, Inc.'s color distribution claims for Skittles are statistically plausible.
It's a key aspect of data analysis, with the final decision hinging on whether our test statistic falls within or outside these critical threshold values.

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

How to quit smoking It's hard for smokers to quit. Perhaps prescribing a drug to fight depression will work as well as the usual nicotine patch. Perhaps combining the patch and the drug will work better than either treatment alone. Here are data from a randomized, double-blind trial that compared four treatments. \({ }^{19} \mathrm{~A}\) "success" means that the subject did not smoke for a year following the beginning of the study. $$ \begin{array}{llcc} \hline \text { Group } & \text { Treatment } & \text { Subjects } & \text { Successes } \\ 1 & \text { Nicotine patch } & 244 & 40 \\ 2 & \text { Drug } & 244 & 74 \\ 3 & \text { Patch plus drug } & 245 & 87 \\ 4 & \text { Placebo } & 160 & 25 \\ \hline \end{array} $$ (a) Summarize these data in a two-way table. Then compare the success rates for the four treatments. (b) Explain in words what the null hypothesis \(H_{0}: p_{1}=\) \(p_{2}=p_{3}=p_{4}\) says about subjects' smoking habits. (c) Do the data provide convincing evidence of a difference in the effectiveness of the four treatments at the \(\alpha=0.05\) significance level?

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: $$ \begin{array}{lcccc} \hline \text { Type of Food: } & \text { Asian } & \text { Mexican } & \text { Pizza } & \text { Hamburgers } \\ \text { Count: } & 18 & 22 & 39 & 21 \\ \hline \end{array} $$ Which of the following is false? (a) A chi-square distribution with \(k\) degrees of freedom is more right-skewed than a chi-square distribution with \(k+1\) degrees of freedom. (b) A chi-square distribution never takes negative values. (c) The degrees of freedom for a chi-square test is determined by the sample size. (d) \(P\left(\chi^{2}>10\right)\) is greater when \(\mathrm{df}=k+1\) than when \(\mathrm{df}=k\) (e) The area under a chi-square density curve is always equal to \(1 .\)

Preventing strokes Aspirin prevents blood from clotting and so helps prevent strokes. The Second European Stroke Prevention Study asked whether adding another anticlotting drug named dipyridamole would be more effective for patients who had already had a stroke. Here are the data on strokes during the two years of the study: \(^{20}\). $$ \begin{array}{llcc} \hline \text { Group } & \text { Treatment } & \text { Number of patients } & \text { Number of strokes } \\ 1 & \text { Placebo } & 1649 & 250 \\ 2 & \text { Aspirin } & 1649 & 206 \\ 3 & \text { Dipyridamole } & 1654 & 211 \\ 4 & \text { Both } & 1650 & 157 \\ \hline \end{array} $$ (a) Summarize these data in a two-way table. Then compare the stroke rates for the four treatments. (b) Explain in words what the null hypothesis \(H_{0}: p_{1}=p_{2}=p_{3}=p_{4}\) says about the incidence of strokes. (c) Do the data provide convincing evidence of a difference in the effectiveness of the four treatments at the \(\alpha=0.05\) significance level?

How are schools doing? The nonprofit group Public Agenda conducted telephone interviews with three randomly selected groups of parents of high school children. There were 202 black parents, 202 Hispanic parents, and 201 white parents. One question asked was "Are the high schools in your state doing an excellent, good, fair, or poor job, or don't you know enough to say?" Here are the survey results: \({ }^{14}\) $$ \begin{array}{lccc} \hline & \begin{array}{c} \text { Black } \\ \text { parents } \end{array} & \begin{array}{c} \text { Hispanic } \\ \text { parents } \end{array} & \begin{array}{c} \text { White } \\ \text { parents } \end{array} \\ \text { Excellent } & 12 & 34 & 22 \\ \text { Good } & 69 & 55 & 81 \\ \text { Fair } & 75 & 61 & 60 \\ \text { Poor } & 24 & 24 & 24 \\ \text { Don't know } & 22 & 28 & 14 \\ \text { Total } & 202 & 202 & 201 \\ \hline \end{array} $$ (a) Calculate the conditional distribution (in proportions) of responses for each group of parents. (b) Make an appropriate graph for comparing the conditional distributions in part (a). (c) Write a few sentences comparing the distributions of responses for the three groups of parents.

No chi-square A school's principal wants to know if students spend about the same amount of time on homework each night of the week. She asks a random sample of 50 students to keep track of their homework time for a week. The following table displays the average amount of time (in minutes) students reported per night: $$ \begin{array}{lccccccc} \hline \text { Night: } & \text { Sunday } & \text { Monday } & \text { Tuesday } & \text { Wednesday } & \text { Thursday } & \text { Friday } & \text { Saturday } \\ \text { Average } & 130 & 108 & 115 & 104 & 99 & 37 & 62 \\ \text { time: } & & & & & & & \\ \hline \end{array} $$ Explain carefully why it would not be appropriate to perform a chi-square test for goodness of fit using these data.

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.