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Skittles庐 Statistics teacher Jason Mole sky contacted Mars, Inc., to ask about the color distribution for Skittles candies. Here is an excerpt from the response he received: 鈥淭he original flavor blend for the Skittles Bite Size Candies is lemon, green apple, 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.鈥

a. State appropriate hypotheses for a significance test of the company鈥檚 claim.

b. Find the expected counts for a random sample of 60 candies.

c. How large a 2test statistic would you need to have significant evidence against the company鈥檚 claim at the 伪=0.05 level? At the =0.01 level?

d. Create a set of observed counts for a random sample of 60 candies that gives a P-value between 0.01 and 0.05 Show the calculation of your chi-square test statistic.

Short Answer

Expert verified

Part (a)H0:plemon=0.20,plime=0.20,porange=0.20,pstrawberry=0.20,pgraph=0.20

Ha:Atleastoneofthepisisincorrect.

Part (b) Expected count for each flavor is 12

Part (c) If chi square statistic is greater than above critical values at specified level of significance then reject H0

Part (d)

Step by step solution

01

Part (a) Step 1: Given information

The flavor blend is 20% of each flavor.

Sample size is 60 candies.

02

Part (a) Step 2: Calculation

The null and alternative hypotheses:

H0:plemon=0.20,plime=0.20,porange=0.20,pstrawberry=0.20,pgrape=0.20

H0:Atleastoneofthepisisincorrect.

03

Part (b) Step 1: Calculation

To get the predicted count, multiply each frequency of flavor by 60As a result, the projected count is,

04

Part (c) Step 1: Calculation

The degrees of freedom =df=c-1

The df=5-1=4

For each degree of significance, various critical values will be used.

When significance level=a=0.05

2=9.49Usingexcelformula,=CHIINV(0.05,4)2=13.28Usingexcelformula,=CHIINV(0.01,4)

Reject H0 if the chi square statistic is greater than the crucial values at the selected level of significance.

05

Part (d) Step 1: Calculation

The frequency of the random sample is shown in the table below:

Using excel,

Therefore, thep-value is between 0.01 and 0.05

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Most popular questions from this chapter

Which of the following is the correct number of degrees of freedom for the chi-square test using these data?

a.4

b.8

c. 10

d.20

e.4876

Which of the following statements about chi-square distributions are true?

I. For all chi-square distributions, P(x20)=1

II. A chi-square distribution with fewer than 10degrees of freedom is roughly symmetric.

III. The more degrees of freedom a chi-square distribution has, the larger the mean of the distribution.

a. I only

b. II only

c. III only

d. I and III

e. I, II, and III

Is your random number generator working? Use your calculator鈥檚 RandInt function to generate 200 digits from 0 to 9 and store them in a list.

a. State appropriate hypotheses for a chi-square test for goodness of fit to determine whether your calculator鈥檚 random number generator gives each digit an equal chance of being generated.

b. Carry out a test at the =0.05 significance level. Hint: To obtain the observed

counts, make a histogram of the list containing the 200 random digits, and use the trace feature to see how many of each digit were generated. You may have to adjust your window to go from 0.5to9.5 with an increment of 1

c. Assuming that a student鈥檚 calculator is working properly, what is the probability that the student will make a Type I error in part (b)?

d. Suppose that 25 students in an AP庐 Statistics class independently do this exercise for homework and that all of their calculators are working properly. Find the probability that at least one of them makes a Type I error.

The manager of a high school cafeteria is planning to offer several new types of food for student lunches in the new 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 100students and asks them, 鈥淲hich type of food do you prefer: Ramen, tacos, pizza, or hamburgers?鈥 Here are her data:

The chi-square test statistic is

a. (1825)225+(2225)225+(3925)225+(2125)225

b. (2518)218+(2522)222+(2539)239+(2521)221

c. (1825)25+(2225)25+(3925)25+(2125)25

d. (1825)2100+(2225)2100+(3925)2100+(2125)2100

e. (0.180.25)20.25+(0.220.25)20.25+(0.390.25)20.25+(0.210.25)20.25

Testing a genetic model Biologists wish to cross pairs of tobacco plants having genetic makeup Gg, indicating that each plant has one dominant gene (G) and one recessive gene (g) for color. Each offspring plant will receive one gene for color from each parent. The Punnett square shows the possible combinations of genes received by the offspring.

The Punnett square suggests that the expected ratio of green (GG) to yellow-green (Gg) to albino (gg) tobacco plants should be 1:2:1. In other words, the biologists predict that 25%of the offspring will be green, 50%will be yellow-green, and 25%will be albino. To test their hypothesis about the distribution of offspring, the biologists mate 84randomly selected pairs of yellow-green parent plants. Of 84offspring, 23plants were green, 50were yellow-green, and 11 were albino. Do the data provide convincing evidence at the =0.01 level that the true distribution of offspring is different from what the biologists predict?

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