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Chapter 11: Q. 2.2 (page 684)

Let鈥檚 continue our analysis of Joey鈥檚 sample of M&M鈥橲 Peanut Chocolate Candies from the previous Check Your Understanding (page 681).

23% each of blue and orange, 15% each of green and yellow, and 12% each of red and brown. Joey bought a bag of Peanut Chocolate Candies and counted the colors of the candies in his sample: 12 blue, 7 orange, 13 green, 4 yellow, 8 red, and 2 brown.

2. Sketch a graph like Figure 11.4 that shows the P-value.

Short Answer

Expert verified

The p-value can be visualized (area shaded in blue) as follows:

Step by step solution

01

Given

23% each of blue and orange, 15% each of green and yellow, and 12% each of red and brown. Joey bought a bag of Peanut Chocolate Candies and counted the colors of the candies in his sample: 12 blue, 7 orange, 13 green, 4 yellow, 8 red, and 2 brown.

02

Formulation

In order to calculate the p value we need to calculate the chi-square statistics.

The formula for chi-square statistics is:

饾搂2=鈭�iOi-Ei2Ei

Upon putting the value in above formula and simplifying we get,

饾搂2=11.1516

03

The p-value

The p-value is depicted in the following figure (shaded in light blue) as follows:

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Most popular questions from this chapter

Regulating guns The National Gun Policy Survey asked a random sample of adults, 鈥淒o you think there should be a law that would ban possession of handguns except for the police and other authorized persons?鈥� Here are the responses, broken down by the respondent鈥檚 level of education:

(a) How do opinions about banning handgun ownership seem to be related to the level of education? Make an appropriate graph to display this relationship. Describe what you see.

(b) Determine whether or not the sample provides convincing evidence that education level and opinion about a handgun ban are independent in the adult population

Calculate the expected count for each color, assuming that the company鈥檚 claim is true. Show your work.

Software gives test statistic 蠂2=69.8and P-value close to 0 . The correct interpretation of this result is

(a) the probability of getting a random sample of 4877teens that yields a value of 蠂2of 69.8or larger is basically 0.

(b) the probability of getting a random sample of 4877teens that yields a value of 蠂2of 69.8or larger if H0is true is basically 0.

(c) the probability of making a Type I error is basically 0.

(d) the probability of making a Type II error is basically 0.

(e) it's very unlikely that these data are true.

Biologists wish to mate pairs of fruit flies having genetic makeup RrCc, indicating that each has one dominant gene (R) and one recessive gene (r) for eye color, along with one dominant (C) and one recessive (c) gene for wing type. Each offspring will receive one gene for each of the two traits from each parent. The following Punnett square shows the possible combinations of genes received by the offspring:

Any offspring receiving an R gene will have red eyes, and any offspring receiving a C gene will have straight wings. So based on this Punnett square, the biologists predict a ratio of 9 red-eyed, straight-winged (x):3 red-eyed, curly-winged (y):3 white-eyed, straight-winged (z):1 white-eyed, curly-winged (w) offspring. To test their hypothesis about the distribution of offspring, the biologists mate a random sample of pairs of fruit flies. Of 200 offspring, 99 had red eyes and straight wings, 42 had red eyes and curly wings, 49 had white eyes and straight wings, and 10 had white eyes and curly wings. Do these data differ significantly from what the biologists have predicted? Carry out a test at the A  0.01 significance level

No chi-square A school鈥檚 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 50students 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:

Explain carefully why it would not be appropriate to perform a chi-square goodness-of-铿乼 test using these data.
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