Question

A survey of 1,570 randomly selected adults showed that 541 of them have heard of a new electronic reader. The accompanying technology display results from a test of the claim that 34% 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). a. Is the test two-tailed, left-tailed, or right-tailed? Right tailed test Left-tailed test Two-tailed test b. What is the test statistic? zequals=nothing (Round to two decimal places as needed.) c. What is the P-value? P-value equals=nothing (Round to four decimal places as needed.) d. What is the null hypothesis and what do you conclude about it? Identify the null hypothesis. A. Upper H 0 : p less than 0.34H0: p<0.34 B. Upper H 0 : p not equals 0.34H0: p≠0.34 C. Upper H 0 : p greater than 0.34H0: p>0.34 D. Upper H 0 : p equals 0.34H0: p=0.34 Choose the correct answer below. A. Reject the null hypothesis because the P-value is less than or equal toless than or equal to the significance level, alphaα. B. Fail to reject the null hypothesis because the P-value is greater than the significance level, alphaα. C. Fail to reject the null hypothesis because the P-value is less than or equal to the significance level, alphaα. D. Reject the null hypothesis because the P-value is greater than the significance level, alphaα. e. What is the final conclusion? A.There is not sufficient evidence to warrant rejection of the claim that 34% of adults have heard of the new electronic reader. B.There is sufficient evidence to support the claim that 34% of adults have heard of the new electronic reader. C.There is not sufficient evidence to support the claim that 34% of adults have heard of the new electronic reader. D.There is sufficient evidence to warrant rejection of the claim that 34% of adults have heard of the new electronic reader. |

Answer #1

A poll of
2 comma 1172,117
randomly selected adults showed that
9191%
of them own cell phones. The technology display below results
from a test of the claim that
9393%
of adults own cell phones. Use the normal distribution as an
approximation to the binomial distribution, and assume a
0.010.01
significance level to complete parts (a) through (e).
Test of
pequals=0.930.93
vs
pnot equals≠0.930.93
Sample
X
N
Sample p
95% CI
Z-Value
P-Value
1
19241924
2 comma 1172,117
0.9088330.908833
(0.8927190.892719,0.9249480.924948)...

A poll of 2,133 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
pequals
0.92vs
pnot equals
0.92
Sample
X
N
Sample p
95% CI
Z-Value
P-Value
1
1996
2 comma 133
0.935771
(0.922098,0.949444
)...

A poll of 2,061 randomly selected adults showed that 97% of them
own cell phones. The technology display below results from a test
of the claim that 94% 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).
Show your work. ***Correct answers are in BOLD (how do we
get those answers)
Test of p=0.94 vs p≠0.94
X= 1989, N= 2061, Sample...

In a study of 809 randomly selected medical malpractice
lawsuits, it was found that 471 of them were dropped or dismissed.
Use a 0.05 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. Upper H 0 : p
less than 0.5 Upper H 1 : p equals 0.5 B. Upper H 0 : p equals 0.5
Upper H 1 : p greater...

In a study of
824
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 claim that most medical
malpractice lawsuits are dropped or dismissed.
Which of the following is the hypothesis test to be
conducted?
A.
Upper H 0 : p not equals 0.5H0: p≠0.5
Upper H 1 : p equals 0.5H1: p=0.5
B.
Upper H 0 : p less than 0.5H0: p<0.5
Upper H...

A certain drug is used to treat asthma. In a clinical trial of
the drug, 26 of 285 treated subjects experienced headaches (based
on data from the manufacturer). The accompanying calculator
display shows results from a test of the claim that less than 11%
of treated subjects experienced headaches. Use the normal
distribution as an approximation to the binomial distribution and
assume a 0.01 significance level to complete parts (a) through
(e) below.
A. Is the test two-tailed, left-tailed, or...

In a recent poll, 801 adults were asked to identify their
favorite seat when they fly, and 476 of them chose a window seat.
Use a 0.05 significance level to test the claim that the majority
of adults prefer window seats when they fly. Identify the null
hypothesis, alternative hypothesis, test statistic, P-value,
conclusion about the null hypothesis, and final conclusion that
addresses the original claim. Use the P-value method and the
normal distribution as an approximation to the binomial...

In a recent court case it was found that during a period of 11
years 893 people were selected for grand jury duty and 38% of them
were from the same ethnicity. Among the people eligible for grand
jury duty, 78.9% were of this ethnicity. Use a 0.01 significance
level to test the claim that the selection process is biased
against allowing this ethnicity to sit on the grand jury. Identify
the null hypothesis, alternative hypothesis, test statistic,
P-value, conclusion...

An organization surveyed 1 comma 100 adults and asked, "Are you
a total abstainer from, or do you on occasion consume, alcoholic
beverages?" Of the 1 comma 100 adults surveyed, 362 indicated that
they were total abstainers. Sixty years later, the same question
was asked of 1 comma 100 adults and 410 indicated that they were
total abstainers. Has the proportion of adults who totally abstain
from alcohol changed? Use the alpha equals 0.05 level of
significance. Determine the null...

A genetic experiment involving peas yielded one sample
of offspring consisting of 416 green peas and 165 yellow peas. Use
a 0.01 significance level to test the claim that under the same
circumstances, 27% of offspring peas will be yellow. Identify the
null hypothesis, alternative hypothesis, test statistic,
P-value, conclusion about the null hypothesis, and final
conclusion that addresses the original claim. Use the P-value
method and the normal distribution as an approximation to the
binomial distribution.
What are the...

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