Question

Question 6 An investigator is looking at the relationship between periodontal disease and the onset of...

Question 6

An investigator is looking at the relationship between periodontal disease and the onset of hypertension. Suppose the investigator decides to look at height as a variable that may confound the relationship between exposure and case status in his study. After collecting information about the height of each participant, he assembles a database that approximates height with the normal distribution.

A) Assuming the mean of the distribution is 176.2 cm with a standard deviation of 17.5 cm, what are the highest and lowest 1% of heights in the population?

B) Assume the investigator is now looking at the population of Dutch men in his study, and that the mean height there is 184.8 cm with a standard deviation of 16.1 cm, answer the previous question.

               A) Highest=209.99 cm, Lowest=133.33

B) Highest=210.15 cm, Lowest=141.26

               A) Highest=216.91 cm, Lowest=135.49

B) Highest=222.25 cm, Lowest=147.35

               A) Highest=219.89 cm, Lowest=129.46

B) Highest=211.57 cm, Lowest=142.87

Question 7

Age is a variable that is almost always included in any prospective study. Suppose in one particular investigation, the mean age µ=37.5 years and the σ=6.2 years.

A) What is the probability of selecting an individual that is 39 years of age or older?

B) What is the probability of selecting someone between 32 and 36 years of age?

               A) 0.40442

B) 0.2169

               A) 0.5748

B) 0.1897

               A) 0.6987

B) 0.1524

Question 8

Suppose a professor for organic chemistry keeps a record of the final grades reported to the academic institution each semester. Assume the distribution of all final grades varies according to a normal distribution with mean 73.5% and standard deviation 9.8%.

A) What score corresponds to the highest 1% of the distribution?

B) What score corresponds to the lowest 1% of the distribution?

               A) 99.8%

B) 53.2%

               A) 96.3%

               B) 50.7

               A) 89.1%

B) 48.5%

Question 9

A researcher is interested in the effectiveness of a new program at reducing the average systolic blood pressure in a population with a new anti-hypertensive medication. The researcher wants to test the hypothesis that the mean difference in systolic blood pressure is greater than 40 mmHG. Which of the following represents the correct null and alternative hypotheses for the study?

Ho: µ1 - µ2 = 0, Ha: µ1 - µ2 = 40

Ho: µ1 - µ2 = 40 Ha: µ1 - µ2 ≠ 40

Ho: µD = 40 Ha: µD >40

Ho: µD = 40 Ha: µD ≠40

Question 10

The following table illustrates 8 different community playgrounds and their corresponding number of children with acute injuries occurring on the premises reported by their parents in one year.

Playground

Total Injuries (x10)

Sydney Hampton

2.7

Roger B Smith Memorial

4.8

Holy Trinity Baptist

10.4

Hoffman Triangle

11.8

Pleasant Ridge

6.4

Oak Woods

2.4

Olympic Gardens

1.6

Oak Park

0.7

  1. How many possible samples of size 3 can we select from the community parks?

56

45

52

Question 11

Suppose a researcher is interested in the relationship between average hours of sleep per night and a number of health outcomes. Suppose the following table represents a SRS suitable for a normal approximation from participants in the study.

Patient

Hours of Sleep

A

6.4

B

5.8

C

6.7

D

9.2

E

4.6

F

5.7

A) Construct a 95% confidence interval for the average hours of sleep.

B) Provide a correct interpretation of the interval.

               A) Ci (5.16, 7.64)

B) We are 95% confidence that the true mean hours of sleep per night in the population lies within the interval

               A) Ci (4.76, .24)

B) We are 95% confidence that the true mean hours of sleep per night in the population lies within the interval

               A) Ci (8.61, 9.64)

B) We are 95% confidence that the true mean hours of sleep per night in the population lies within the interval

Question 12

Write out the null and alternative hypotheses for the following hypothetical proposal. Carry out a one-sample Z test to determine significance at the α=0.05 level.

PROPOSAL

Recent scientific literature has suggested a link between exposure to the endocrine disruptor Bisphenol A (BPA) found ubiquitously in many plastic materials, and precocious puberty in adolescents. The transmission is maternal to child during the last weeks of gestation. Scientists can trace the presence of BPA in children for years after birth, and believe that it may be the cause for early onset of puberty in young girls. Suppose a researcher collects a SRS of 92 girls with high exposure rates of BPA and follows them through puberty. The sample average age of puberty was 9.7 years. Assume the national average age for onset of puberty is 11.2 years with a population standard deviation of 0.6 years. Does the researcher have sufficient evidence at the alpha level of 0.05 to determine if the exposure to BPA has led to a decreased age of puberty in girls?

Ho: µ=11.2, Ha: µ>11.2

Z stat=44.81 pvalue of <0.001 Fail to reject the Ho that the average onset of puberty is 11.2 years of age among the females exposed to BPA levels in the sample. Conclude that the girls exposed to BPA have not reached puberty at a significantly lower age.

Ho: µ=11.2, Ha: µ<11.2

Z stat=23.98 pvalue of <0.0001 Reject the Ho that the average onset of puberty is 11.2 years of age among the females exposed to BPA levels in the sample. Conclude that the girls exposed to BPA have reached puberty at a significantly lower age.

Ho: µ=11.2, Ha: µ<11.2

Z stat=30.21 pvalue of <0.005 Reject the Ho that the average onset of puberty is 11.2 years of age among the females exposed to BPA levels in the sample. Conclude that the girls exposed to BPA have reached puberty at a significantly lower age.

Homework Answers

Answer #1

Mean= 176.2 and S.D= 17.5

Z score corresponding to 0.01 is -2.33

Z= X-176.2/17.5

-2.33*17.5=X-176.2

X= 135.49cm

Z score corresponding to 0.99 is

Z= x-mesn/sigma

X= Z*sigma+mean

X= 2.33*17.5+176.2

X= 216.91 cm

Now for DUTCH PEOPLE

FOR LOWEST Z SCORE CORRESPONDING TO 0.01 IS -2.33

X= -2.33*16.1+184.8

X= -37.51+184.8

X= 147.35cm

For Highest

X= 2.33*16.1+184.8

X=222.25 cm

OPTION B

NOTE: AS PER THE GUIDELINES I HAVE DONE THE FIRST QUESTION PLEASE REPOST THE REST. THANK YOU.

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