Question

A poll of 2,142 randomly selected adults showed that 92​% of them own cell phones. The...

A poll of

2,142

randomly selected adults showed that

92​%

of them own cell phones. The technology display below results from a test of the claim that

91​%

of adults own cell phones. Use the normal distribution as an approximation to the binomial​ distribution, and assume a

0.05

significance level to complete parts​ (a) through​ (e).

Test of

p=0.91

vs

p≠0.91

Sample

X

N

Sample p

​95% CI

​Z-Value

​P-Value

1

1970

2,142

0.919701

​(0.908193​,0.931210​)

1.57

0.117

a. Is the test​ two-tailed, left-tailed, or​ right-tailed?

​Left-tailed test

Right tailed test

​Two-tailed test

b. What is the test​ statistic?

The test statistic is

nothing.

​(Round to two decimal places as​ needed.)

c. What is the​ P-value?

The​ P-value is

nothing.

​(Round to three decimal places as​ needed.)

d. What is the null hypothesis and what do you conclude about​ it?

Identify the null hypothesis.

A.

H0: p≠0.91

B.

H0: p>0.91

C.

H0: p<0.91

D.

H0: p=0.91

Choose the correct answer below.

A.

Fail to reject

the null hypothesis because the​ P-value is

greater than

the significance​ level,

α.

B.

Reject

the null hypothesis because the​ P-value is

greater than

the significance​ level,

α.

C.

Fail to reject

the null hypothesis because the​ P-value is

less than or equal to

the significance​ level,

α.

D.

Reject

the null hypothesis because the​ P-value is

less than or equal to

the significance​ level,

α.

e. What is the final​ conclusion?

A.There

is not

sufficient evidence to support the claim that

91​%

of adults own a cell phone.

B.There

is not

sufficient evidence to warrant rejection of the claim that

91​%

of adults own a cell phone.

C.There

is

sufficient evidence to support the claim that

91​%

of adults own a cell phone.

0 0
Add a comment Improve this question Transcribed image text
Answer #1

Using the result given the answers are a) Two tailed test b) Test statistic z=1.57 c) P-value 2*P(21.57) = 0.117 or 0.1164 d)

Add a comment
Know the answer?
Add Answer to:
A poll of 2,142 randomly selected adults showed that 92​% of them own cell phones. The...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • a poll of 2094 randomly selected adults showed that 94% of them own cell phones A...

    a poll of 2094 randomly selected adults showed that 94% of them own cell phones A poll of 2,094 randomly selected adults showed that 94% of them own cell phones. The technology display below results from a test of the claim that 92% of adults own cell phones. Use the normal distribution as an approximation to the binomial distribution, and assume a 0.01 significance level to complete parts (a) through (e). Test of p = 0.92 vs p*0.92 Samplex N...

  • A poll of 2,084 randomly selected adults showed that 94% of them own cell phones. The...

    A poll of 2,084 randomly selected adults showed that 94% of them own cell phones. The technology display below ret from a test of the claim that 92% of adults own cell phones. Use the normal distribution as an approximation to the bin distribution, and assume a 0.01 significance level to complete parts (a) through (e). Test of p = 0.92 vs p+0.92 Z-Value P-value Sample p 95% CI N Sample X 0.000 4.01 (0.930869,0.956847) 1 1967 2,084 0.943858 a....

  • A survey of 1,680 randomly selected adults showed that 549 of them have heard of a...

    A survey of 1,680 randomly selected adults showed that 549 of them have heard of a new electronic reader. The accompanying technology display results from a test of the claim that 37% of adults have heard of the new electronic reader. Use the normal distribution as an approximation to the binomial distribution, and assume a 0.01 significance level to complete parts a through e Sample proportion: 0.326786 Test statistic Critical z: P-Value z:-3.6687 ± 2.5758 0.0002 a. Is the test...

  • In a study of 807 randomly selected medical malpractice​ lawsuits, it was found that 497 of...

    In a study of 807 randomly selected medical malpractice​ lawsuits, it was found that 497 of them were dropped or dismissed. Use a 0.01 significance level to test the claim that most medical malpractice lawsuits are dropped or dismissed. Which of the following is the hypothesis test to be​ conducted? A. H0: p>0.5 H1: p=0.5 B. H0: p=0.5 H1: p<0.5 C. H0: p≠0.5 H1: p=0.5 D. H0: p=0.5 H1: p≠0.5 E. H0: p<0.5 H1: p=0.5 F. H0: p=0.5 H1: p>0.5...

  • In a study of 784 randomly selected medical malpractice lawsuits, it was found that 47 of...

    In a study of 784 randomly selected medical malpractice lawsuits, it was found that 47 of them were recorded Use a 0.01 finance level to test them that most medical practice lawsuits are dropped or dismissed Which of the following is the hypothesis test to be conducted? In a study of 784 randomly selected medical malpractice lawsuits, it was found that 479 of them were dropped or dismissed. Use a 0.01 significance level to test the dropped or dismissed. Turu www...

  • 1. Claim Fewer than 97 % of adults have a cell phone

    1. Claim Fewer than 97 % of adults have a cell phone. In a reputable poll of 1038 adults, 89 % said that they have a cell phone. Find the value of the test statistic.2. The test statistic of z=1.38 is obtained when testing the claim that p>0.2a. Identify the hypothesis test as being two-tailed, left-tailed, or right-tailed.b. Find the P-value by the calculator or by the table.c. Using a significance level of α=0.05 should we reject H₀ or should...

  • Marketers believe that 93% of adults in California own a cell phone. A cell phone manufacturer...

    Marketers believe that 93% of adults in California own a cell phone. A cell phone manufacturer believes that number is actually lower. 220 Californians are surveyed, of which 90% report having cell phones. Use a 1% level of significance to test the manufacturer’s hypothesis. (6 points) a. H0 : b. Ha : c. What is the test statistic? d. What is the p-value? e. Indicate the correct decision, reason for it, and write an appropriate conclusion using complete sentences. i....

  • In a study of 814 randomly selected medical practice lawsuits, was found that 504 of them...

    In a study of 814 randomly selected medical practice lawsuits, was found that 504 of them were dropped or domised Use a 0.05 significance level to beat the claim that most medical malpractice outs are dropped or dismissed Which of the following is the hypothesis est to be conducted? OA HP05 OBHp0.5 Mp>05 Hp<0.5 OCH:05 OD H:05 HD-05 HD 05 OEM: D05 OF H 10.5 H:05 Hyp=0.5 What is the best? (Round love deal places as needed) What is the...

  • In a study of 706 randomly selected medical malpractice wuits, it was found that 520 of...

    In a study of 706 randomly selected medical malpractice wuits, it was found that 520 of them were dropped or dismissed. Use a 0.01 significance level to test the claim that most medical malpractice lawsuits are dropped or dismissed. Which of the following is the hypothesis test to be conducted? OA Hp0.5 OBHO: p=0.5 Hyp<0.5 Hy: p0.5 O C. Họ p = 0,5 OD HD 0.5 Hy: p=0.5 Hp 05 O E HO P<0.5 OF Hp0.5 H: p=0.5 Hp>05 What...

  • In a study of Brandomly selected medical malpractice lawsuits, it was found that 516 of them...

    In a study of Brandomly selected medical malpractice lawsuits, it was found that 516 of them were dropped or dismissed Use a 0.05 significance level to test the claim that most medical practice lawsuits are dropped or dismissed Which of the folowng is the hypothesis best to be conducted? OA HP-0.5 OB H:p>0.5 HD H05 OCH 0.5 OD. He: p0.5 He p05 Hp 05 OEHP+0.5 OF Hp 05 HDD Hp.05 What is the best statistic? Round wo decimal predet) What...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT