11.92 Delayed Perinatal Stroke. In the article "Prothrombotic Factors in Children With Stroke or Porencephaly" (Pediatrics Journal, Vol. 116, Issue 2, pp. 447-453), J. Lynch et al. compared differences and similarities in children with arterial ischemic stroke and porencephaly. Three classification categories were used: perinatal stroke, delayed perinatal stroke, and childhood stroke. Of 59children, 25were diagnosed with delayed perinatal stroke. At the 5%significance level, do the data provide sufficient evidence to conclude that delayed perinatal stroke does not comprise one-third of the cases among the three categories?

Short Answer

Expert verified

At the 5% level, the test results are statistically significant. Yes, the data is sufficient to establish that delayed perinatal stroke does not affect one-third of the cases in any of the three groups.

Step by step solution

01

Given information

To find the data provide sufficient evidence to conclude that delayed perinatal stroke does not compromise one-third of the cases among the three categories at the significance level 5%.

02

Explanation

Since, the significance level is5%,n=59,p0=0.3,and x=25
Determine the value of np0 as:
np0=(59)(0.3)
=17.7
Then, find the value of n1-p0 as:
n(1-p0)=(59)(1-0.3)
Both np0and n1-p0are greater than 5. So one proportion ztest is appropriate to use.

The null hypothesis: H0:p0=0.3
The alternate hypothesis: Ha:p00.3

03

Explanation

Determine the sample proportion as:
p^=xn

=2529

=0.465
For z, state the expression as:
z=p^-p0P01-p0n
=0.424-0.30.3(1-0.3)59
=0.1240.06
=2.074
Also,
α=0.05
The test is two tailed.
04

Explanation

The critical value of zfrom the standard table for α=0.025 as:
z0.05=±1.96
In the rejection region, the test statistic falls . Therefore, the hypothesis H0 is rejected.
At the 5%level, the test results are statistically significant.
As a result, yes, the data is sufficient to establish that delayed perinatal stroke does not affect one-third of the cases in any of the three groups.

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Most popular questions from this chapter

Margin of error=0.02

Confidence level=95%

Educated guess=0.6

(a) Obtain a sample size that will ensure a margin of error of at most the one specified (provided of course that the observed value of the sample proportion is further from 0.5that of the educated guess.

(b). Compare your answer to the corresponding one and explain the reason for the difference, if any.

Government Surveillance. Gallup conducted a national survey of 1008American adults, asking "As you may know, as part of its efforts to investigate terrorism, a federal government agency obtained records from larger U.S. telephone and Internet companies in order to compile telephone call logs and Internet communications. Based on what you have heard or read about the program, would you say that you approve or disapprove of this government program?" Of those surveyed, 534said they disapprove.

a. Determine and interpret the sample proportion.

b. At the 5%significance level, do the data provide sufficient evidence to conclude that a majority (more than 50%) of American adults disapprove of this government surveillance program?

a. Determine the sample proportion.

b. Decide whether using the one-proportion z-test is appropriate.

c. If appropriate, use the one-proportion z-test to perform the specified hypothesis test.

x=35

n=50

H0:p=0.6

role="math" localid="1651304589496" Ha:p>0.6

α=0.05

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a. At the 5% significance level, do the data provide sufficient evidence to conclude that a majority (more than 50% ) of U.S. adults favored passage?
b. The headline on the website featuring the survey read, "In U.S., Slim Majority Supports Economic Stimulus Plan." In view of your result from part (a), discuss why the headline might be misleading.
c. How could the headline be made more precise?

Drinking Habits. In a nationwide survey, conducted by Pulse Opinion Research, LLC for Rasmussen Reports, 1000 American adults were asked, among other things, whether they drink alcoholic beverages at least once a week; 38% said "yes." Determine and interpret a 95% confidence interval for the proportion, p, of all American adults who drink alcoholic beverages at least once a week.

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