/*! 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} Q 16. Spinning heads? When a fair coin... [FREE SOLUTION] | 91影视

91影视

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 H0:p=0.5 and Ha:p0.5

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)?

Short Answer

Expert verified

Part (a) When a euro coin is spun, there is no persuasive evidence that heads and tails are not equally likely.

Part (b) When a euro coin is spun, there is no persuasive evidence that heads and tails are not equally likely.

Step by step solution

01

 Part (a) Step 1: Given information

Sample size =n=25

Level of significance ==0.05

02

Part (a) Step 2: Concept

Test statistic: 2=(OE)2E

03

Part (a) Step 3: Calculation

The null and alternative hypotheses:

H0:p1=p2=12=0.5Ha:Atleastoneofthepisisincorrect.

Expected values can be found as,

E1=np1=2500.5=125E2=np2=2500.5=125

Therefore, test statistic is,

2=(140125)2125+(110125)2125=3.6

P-value using excel formula, =CHIDIST(3.6,1)

P-value =0.0578

Decision: P-value>0.05,failtorejectH0

When a euro coin is spun, there is no persuasive evidence that heads and tails are not equally likely.

04

Part (b) Step 1: Calculation

The null and alternative hypotheses:

H0:p=0.5Ha:p0.5

Using excel,

Decision: P-value>0.05,failtorejectH0

When a euro coin is spun, there is no persuasive evidence that heads and tails are not equally likely.

Here, z2=(1.90)2=3.6=2

Also, P-value is same for both tests.

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 91影视!

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

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.

鈥淲ill changing the rating scale on a survey affect how people answer the question?鈥 To find out, the group took an SRS of 50students from an alphabetical roster of the school鈥檚 just over 1000students. The first 22students chosen were asked to rate the cafeteria food on a scale of 1(terrible) to 5(excellent). The remaining 28students were asked to rate the cafeteria food on a scale of 0(terrible) to 4(excellent). Here are the data:

a. Was this an observational study or an experiment? Justify your answer.

b. Explain why it would not be appropriate to perform a chi-square test in this setting.

The two-way table shows the results of the experiment described in

Exercise 28.

Hatching Statuswater temperature:coldwater temperature:neutral
water temperature:hot
Total
Yes16
38
75
129
No11
18
29
58
Total
27
56
104
187

a. State the appropriate null and alternative hypotheses.

b. Show the calculation for the expected count in the Cold/Yes cell. Then provide a

complete table of expected counts.

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

All current-carrying wires produce electromagnetic (EM) radiation, including the electrical wiring running into, through, and out of our homes. High-frequency EM radiation is thought to be a cause of cancer. The lower frequencies associated with household current are generally assumed to be harmless. To investigate the relationship between current configuration and type of cancer, researchers visited the addresses of a random sample of children who had died of some form of cancer (leukemia, lymphoma, or some other type) and classified the wiring configuration outside the dwelling as either a high-current configuration (HCC) or a low-current configuration (LCC). Here are the data:

Computer software was used to analyze the data. The output included the value X2=0.435

Which of the following is the appropriate degrees of freedom for the X2test?

a. 1

b. 2

c. 3

d. 4

e. 5

More candy The two-way table shows the results of the experiment

described in Exercise 27.


Red SurveyBlue Survey
Control Survey
Total
Red Candy13
5
8
26
Blue Candy
7
15
12
34
Total20
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.

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.