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The Practice of Statistics for AP Examination

Daren Starnes, Josh Tabor

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 837 pages

ISBN: 9781319113339

Chapter 11: Q. 29 (page 753)

More candy The two-way table shows the results of the experiment
described in Exercise 27.

Red Survey Blue Survey
Control Survey
Total
Red Candy $13$
$5$
$8$
$26$
Blue Candy
$7$
$15$
$12$
$34$
Total $20$
$20$
$20$
$60$
a. State the appropriate null and alternative hypotheses.
b. Show the calculation for the expected count in the Red/Red cell. Then provide a
complete table of expected counts.
c. Calculate the value of the chi-square test statistic.

Short Answer

Expert verified

(a)Null hyptheses:The true distributions of color choice are not different for the three types of surveys.

Alternative hyptheses: The true distributions of color choice are different for the three types of surveys.

(b)The expected count in Red/Red cell = $8.67$

(c)The value of the chi-square test statistic = $6.62$

Step by step solution

01

Part (a) Step 1:Given Information

We have been given that,

	Red Survey	Blue Survey	Control Survey	Total
Red Candy	$13$	$5$	$8$	$26$
Blue Candy	$7$	$15$	$12$	$34$
Total	$20$	$20$	$20$	$60$

02

Part (a) Step 2:Explanation

Null hyptheses:The true distributions of color choice are not different for the three types of surveys.

Alternative hyptheses: The true distributions of color choice are different for the three types of surveys.

03

Part (b) Step 1:Given Information

We have been given that,

	Red Survey	Blue Survey	Control Survey	Total
Red Candy	$13$	$5$	$8$	$26$
Blue Candy	$7$	$15$	$12$	$34$
Total	$20$	$20$	$20$	$60$

04

Part (b) Step 2:Explanation

The expected count is calculated as: $\frac{R o w T o t a l * C o l u m n T o t a l}{G r a n d T o t a l}$

So, The expected count in Red/Red cell = $\frac{26 * 20}{60}$ = $8.67$ $8.67$

Table of expected counts :

	Red Survey	Blue Survey	Control Survey
Red Candy	$8.67$	$8.67$	$8.67$
Blue Candy	$11.33$	$11.33$	$11.33$

05

Part (c) Step 1:Given Information

We have been given that,

	Red Survey	Blue Survey	Control Survey	Total
Red Candy	$13$	$5$	$8$	$26$
Blue Candy	$7$	$15$	$12$	$34$
Total	$20$	$20$	$20$	$60$

06

Part (c) Step 2:Explanation

Here,

f_o=observed count

f_e=expected count

f_o	f_e	f_o-f_e	(f_o-f_e)²	(f_o-f_e)²/f_e
$13$	$8.67$	$4.33$	$18.74$	$2.16$
$7$	$11.33$	$- 4.33$	$18.74$	$1.65$
$5$	$8.67$	role="math" localid="1654163603493" $- 3.67$	$13.46$	$1.55$
$15$	$11.33$	$3.67$	$13.46$	$1.18$
$8$	$8.67$	$- 0.67$	$0.44$	$0.05$
$12$	$11.33$	$0.67$	$0.44$	$0.03$
				Chi-Square value: $6.62$

Calculated chi-square value : $6.62$

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Most popular questions from this chapter

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

Roulette Casinos are required to verify that their games operate as advertised. American roulette wheels have $38$ 蝉濒辞迟蝉鈥� $18$ red, $18$ black, and $2$ green. In one casino, managers record data from a random sample of $200$ spins of one of their American roulette wheels. The table displays the results.

a. State appropriate hypotheses for testing whether these data give convincing evidence that the distribution of outcomes on this wheel is not what it should be.

b. Calculate the expected count for each color.

c. Calculate the value of the chi-square test statistic.

Spinning heads? When a fair coin is flipped, we all know that the probability the coin lands on heads is $0.50$ However, what if a coin is spun? According to the article 鈥淓uro Coin Accused of Unfair Flipping鈥� in the New Scientist, two Polish math professors and their students spun a Belgian euro coin $250$ times. It landed heads $140$ times. One of the professors concluded that the coin was minted asymmetrically. A representative from the Belgian mint indicated the result was just chance. Assume that the conditions for inference are met.

a. Carry out a chi-square test for goodness of fit to test if heads and tails are equally likely when a euro coin is spun.

b. In Chapter $9$ Exercise $50$ you analyzed these data with a one-sample $z$ test for a proportion. The hypotheses were $H_{0} : p = 0.5$ and $H_{a} : p 鈮� 0.5$

where $p =$ the true proportion of heads. Calculate the $z$ statistic and P-value for this test. How do these values compare to the values from part (a)?

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 table displays the sample data.

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

b. Calculate the expected count for each type of nut.

c. Calculate the value of the chi-square test statistic.

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 $100$ students 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. $\frac{{(18 鈭� 25)}^{2}}{25} + \frac{{(22 鈭� 25)}^{2}}{25} + \frac{{(39 鈭� 25)}^{2}}{25} + \frac{{(21 鈭� 25)}^{2}}{25}$

b. $\frac{{(25 鈭� 18)}^{2}}{18} + \frac{{(25 鈭� 22)}^{2}}{22} + \frac{{(25 鈭� 39)}^{2}}{39} + \frac{{(25 鈭� 21)}^{2}}{21}$

c. $\frac{(18 鈭� 25)}{25} + \frac{(22 鈭� 25)}{25} + \frac{(39 鈭� 25)}{25} + \frac{(21 鈭� 25)}{25}$

d. $\frac{{(18 鈭� 25)}^{2}}{100} + \frac{{(22 鈭� 25)}^{2}}{100} + \frac{{(39 鈭� 25)}^{2}}{100} + \frac{{(21 鈭� 25)}^{2}}{100}$

e. $\frac{{(0.18 鈭� 0.25)}^{2}}{0.25} + \frac{{(0.22 鈭� 0.25)}^{2}}{0.25} + \frac{{(0.39 鈭� 0.25)}^{2}}{0.25} + \frac{{(0.21 鈭� 0.25)}^{2}}{0.25}$

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