Question

In: Statistics and Probability

Question 31 Suppose a researcher is interested in a new treatment plan for increasing CD4 cell...

Question 31

Suppose a researcher is interested in a new treatment plan for increasing CD4 cell count in immune-compromised patients with HIV. In order to investigate the new drug, the researcher gathers a group of 48 individuals with a mean CD4 count of 317 cells/mm3 after treatment with a standard deviation of 24.6. Test the hypothesis that the mean CD4 count has increased above the goal of 275 cells/mm3. Write out your hypotheses and correctly interpret your pvalue. (Use an alpha level of 0.01)

Ho: µ=275, Ha: µ<275

Z value -12.45, pvalue of <0.001; There is less than a 0.1% chance of finding this information given the null hypothesis were true.

Ho: µ=275, Ha: µ>275

Z value -11.83, pvalue of <0.0001; There is less than a .01% chance of finding this information given the null hypothesis were true.

Ho: µ=275, Ha: µ>275

Z value -10.31, pvalue of <0.05; There is less than a .01% chance of finding this information given the null hypothesis were false.

Question 32

A research program is concerned about the potential exposure to hazardous lead levels at a community playground and the potential for cognitive impairment among children in the surrounding area. Suppose the prevalence of cognitive impairment-related disorders in children in the community at large is only 7%. The researchers collect samples from 9 children and find that 3 children have reported cognitive-impairment related symptoms.

  1. What is the expected number of children that should report cognitive impairment related symptoms based on the information provided?
  2. What is the probability associated with the findings presented by the surveyors?

               A) 0.63~1

B) 0.0186

               A) 1.73~2

B) 0.2586

               A) 2.67~3

B) 0.5879

Question 33

A hospital decides to bring in a new surgical resident with a known high success rate for total knee replacements. The new resident has a current success rate of 0.96. Suppose on his first day, the resident is supposed to complete 4 total knee replacement operations in a row. Assume his probability of success does not change.

  1. What is the expected number of successes for the intern?
  2. What is the variance of the distribution?

               a) 4.56~5

b) 0.228

               a) 3.84~4

b) 0.392

               a) 5.68~6

b) 0.447

Question 34

Which of the following statistics is not necessary to know prior to performing a power calculation for a one-sample z test?

The detectable difference worth measuring

The false positive rate of the test

The population standard deviation

The sample size

Question 35

A major university has recently reported instituting a new program aimed at reducing binge drinking behaviors seen in on-campus residents after it was recently ranked as a top party school. After investing a considerable amount of funding into a particular program, representatives for the university have been charged with the task of reporting the progress after 4 years. Suppose the following table represents the number of drinks per week prior to and after attending the program for a SRS of students. Assume that a normal approximation is acceptable.

Student

Drinks/week Prior to Intervention

Drinks/week After intervention

A

16

4

B

18

5

C

17

3

D

19

3

E

21

4

F

20

2

A) Is there a significant amount of evidence to support the hypothesis that the university program has been successful at reducing the number of drinks/week? (Use an alpha=0.05)

B) Interpret the results above.

A) Ho: µ12 Ha: µ12

T statistic 12.526

Pvalue 0.005

B) We have sufficient evidence with 95% confidence that there is a significant difference in drinks consumption pre and post intervention.

A) Ho: µ12 Ha: µ12

T statistic 15.526

Pvalue 0.000

B) We have sufficient evidence with 95% confidence that there is a significant difference in drinks consumption pre and post intervention.

A) Ho: µ12 Ha: µ12

T statistic 21.256

Pvalue 0.0001

B) We have sufficient evidence with 95% confidence that there is a significant difference in drinks consumption pre and post intervention.

Question 36

Suppose a researcher is interested in the effectiveness of a new respiratory treatment drug aimed at increasing forced vital capacity (FVC) among children with respiratory illnesses. In this study, each patient treated with the study drug is frequency paired with another patient on a number of characteristics treated with a placebo to insure independence among samples. Assume the following table represents the FVC for each participant and the paired control.

Participant

Treated with new drug

FVC (%)

Participant

Treated with placebo

FVC (%)

A

91

A’

84

B

90

B’

89

C

87

C’

87

D

88

D’

84

E

82

E’

80

F

86

F’

81

A) What study type is this?

B) Carry out a paired sample t test to determine if the new therapy is significantly effective at increasing average FVC in children with respiratory illnesses. Write out the null and alternative hypotheses using an α=0.01. Interpret your results.

               A) Case-control

B) Ho: µ12 Ha: µ12

T is 2.244

Pvalue is 0.0001

Reject the Ho that there is no significant increase in FVC among patients treated in this study.

               A) Matched-Pairs

B) Ho: µ12 Ha: µ12

T is 2.939

Pvalue is 0.016

Reject the Ho that there is no significant increase in FVC among patients treated in this study.

               A) Matched-Pairs

B) Ho: µ12 Ha: µ12

T is 3.112

Pvalue is 0.05

Fail to reject the Ho that there is no significant increase in FVC among patients treated in this study.

Question 37

Which of the following would correctly result in a type II error for the Ho that the average age of children with Hodgkin’s lymphoma is 13 years old and the Ha that the average age of children with Hodgkin’s lymphoma is not 13 years old based on the following resulting confidence interval? (13.2, 17.8 years)

Reject the Ho based on the fact that the Ho value is not within the interval

Fail to reject the Ho based on the fact that the Ho value lies within the interval

Fail to reject the Ho based on the fact that the Ho value does not lie within the interval

Reject the Ho based on the fact that the Ho value lies within the interva

Question 38

The cumulative probability for the entire binomial distribution does not have to sum to 1.

True

False

Question 39

A university reported its annual findings about sexually transmitted HPV. The results proved to be quite disheartening, specifically pertaining the HPV rates observed throughout the year. Suppose 40% of women living on campus that year tested positive for HPV, and 6 female students were chosen at random for a new prevention program promoting safe sex practices.

  1. What is the probability of finding at least 2 females with HPV in the sample?
  2. What is the expected number of female students based on the sample?

               A) 0.881

B) 1.9

               A) 0.521

B) 3.2

               A) 0.767

B) 2.4

Question 40

Suppose the distribution of BMI (kg/m2) among a group of diabetic children varies normally with a mean of 26.8 and a σ=3.7.

A) What is the probability of randomly selecting an obese child (BMI>30) based on a sample of 15 children?

B) What is the probability of randomly selecting a “normal” child with a BMI between 18.5 and 24.9?

               A) 0.194

B) 0.291

               A) 0.227

B) 0.382

               A) 0.357

B) 0.179

Solutions

Expert Solution

Question 31

Here we have to test that

Where

n = 48

Sample mean =

Sample standard deviation = s = 24.6

Here n = 48 > 30

So we use here z test.

Test statistic:

z = 11.83                 (Round to 2 decimal)

Test statistic = z = 11.83

P value:

Test is one tailed (right tailed test).

alpha = level of significance = 0.01

P value = P(z > 11.83)

             = 1 - P(z < 11.83)

P(z < 11.83) from excel using function:

= NORMSDIST(11.83)

= 1

P value = 1 - P(z < 11.83) = 1 - 1 = 0

P value = 0

P value < 0.0001

Here p value < alpha

So we reject H0.

Conclusion: We can conclude that the mean CD4 count has increased above the goal of 275 cells/mm3.

Answer is

Ho: µ = 275, Ha: µ > 275

Z value = 11.83, p value of <0.0001; There is less than 0.01% chance of finding this information given the null hypothesis were true.


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