In: Statistics and Probability
Question 13
Researchers estimate the prevalence of caesarian section surgeries among pregnant women is on the rise. Suppose the researchers estimate that the current prevalence is 0.42.
A) 2.01 B) 1.61 C) 2.25 |
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A) 1.72 B) 1.98 C) 2.01 |
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A) 1.31 B) 1.57 C) 1.92 |
Question 14
Researchers are concerned about the prevalence of hypertension among a population of elderly patients at a cardiovascular clinic aged 45-60. Suppose the prevalence of hypertension in the group is actually quite high at 38%. Assume that a sample of 14 participants is collected.
A) 6.12 B) 0.009 |
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A) 4.87 B) 0.265 |
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A) 5.32 B) 0.104 |
Question 15
Recent research has suggested that adolescent drinking behaviors have escalated. Assume national statistics presume that only 14% of adolescents aged 13-17 engage in alcohol consumption. A sample of 58 participants in this age range is collected and 13 participants report consuming alcohol on a regular basis.
A) Can researchers carry out their analysis assuming an approximation to the normal distribution is acceptable?
B) Calculate the probability of observing the results witnessed in the sample.
C) Determine the mean and standard deviation for the distribution under the assumptions made in part A.
A) Yes, npq>5 B) 0.1225 C) µ=0.21, σ=0.051 |
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A) No, npq<5 B) 0.2501 C) µ=0.19, σ=0.14 |
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A) Yes, npq>5 B) 0.0325 C) µ=0.14, σ=0.046 |
Question 16
A researcher is interested in looking at a new type of insulin that can treat diabetics with high fasting glucose levels. Suppose the researcher wants to take a sample of participants from the population of participants in the study in order to assess the progress by calculating their average resting glucose level (mg/dL). The researcher wants to be 95% confident using a population standard deviation of 27 mg/dL and hopes to have a margin of error at plus or minus 5 mg/dL. Calculate the sample size the researcher should get in order to answer his research question with all of the given information.
121.02~122 |
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112.02~113 |
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110.59~111 |
Question 17
Suppose the temperature that most foods can stay bacteria free in restaurants varies approximately according to a normal distribution with a mean of 31.3 degrees and a standard deviation of 2.8 degrees Fahrenheit. The Federal Department of Agriculture mandates that food inspectors make sure that all restaurants fine a location up to $1,500.00 if the temperature in cold food storage goes above the bottom 15% of that distribution for safety and insurance purposes. What temperature correlates to the bottom 15% of this distribution?
28.40 |
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32.51 |
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30.18 |
Question 18
Suppose a researcher is interested in the effectiveness of a new novel talk therapy technique in reducing overall depression as identified via score report on the geriatric depression scale (GDS). In order to carry out this hypothesis, the researcher gathers a SRS of participants in the program and performs the GDS test prior to, and after initiation of the new therapy technique. Assume the SRS score report presented below represents an approximately normal distribution.
Participant |
GDS Score (Out of 15) Prior to intervention |
GDS Score (Out of 15) Following intervention |
A |
12 |
8 |
B |
13 |
7 |
C |
12 |
7 |
D |
14 |
9 |
E |
11 |
6 |
F |
11 |
7 |
A) What type of study design is this?
B) Conduct a paired sample t test investigating the effectiveness of the new therapy technique with 95% confidence. Write out your null and alternative hypotheses, and interpret your pvalue correctly.
C) Construct a 95% confidence interval representing the average difference in score on the GDS. Provide an accurate interpretation of your interval.
A) Case-control test B) Ho: µ1=µ2 Ha: µ1>µ2 T statistic 17.227 pvalue 0.001 Fail to eject the Ho that the new talk therapy results in the same score on the depression index. C) (3.267, 3.231) 0 is no within the interval, which would indicate a significant result |
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A) Pre/post test B) Ho: µ1=µ2 Ha: µ1>µ2 T statistic 15.727 pvalue 0.000 Reject the Ho that the new talk therapy results in the same score on the depression index. C) (4.043, 5.623) 0 is no within the interval, which would indicate a significant result |
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A) Cohort B) Ho: µ1=µ2 Ha: µ1>µ2 T statistic 21.112 pvalue 0.05 Reject the Ho that the new talk therapy results in the same score on the depression index. C) (2.143, 6.145) 0 is no within the interval, which would indicate a significant result |
Question 19
A new vitamin supplementation program is intended to decrease average resting heart rate in individuals at risk for hypertension. Assume that a team of researchers are hopeful that resting heart rate in their population will get down to less than 68 bpm, in a population with a standard deviation of 2 bpm. In order to test this goal reduction, the team gathers a SRS of 273 participants in their program and calculates a sample average resting heart rate of 74 bpm.
A) Carry out a one sample Z test to determine if the team can conclude that the supplementation program is successful in meeting their goal reduction in resting heart rate. Use an α=0.05.
B) Construct a 95% Confidence interval about the sample mean, and interpret the result.
A) Ho: µ=68, Ha: µ<68 Z statistic is 49.57 pvalue >0.9999 Fail to reject the Ho, conclude that there is no such significant effect of the medication at reducing the heart rates below 68 bpm. B) (73.76, 74.24) |
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A) Ho: µ=68, Ha: µ>68 Z statistic is 51.75 pvalue >0.0001 Reject the Ho, conclude that there is no such significant effect of the medication at reducing the heart rates below 68 bpm. B) (68.42, 69.24) |
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A) Ho: µ=68, Ha: µ<68 Z statistic is 46.87 pvalue >0.9999 Fail to reject the Ho, conclude that there is no such significant effect of the medication at reducing the heart rates below 68 bpm. B) (71.22, 75.89) |
Question 20
Suppose a database contains population based statistics for a group of hypertensive potential participants in a new clinical trial. The systolic blood pressure of the population varies according to a normal distribution with mean 141 and a standard deviation of 8.4 mmHG.
A) 0.1665 B) 149.02 mmHG |
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A) 0.1112 B) 151.43 mmHG |
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A) 0.0222 B) 154.82 mmHG |
QUESTION 13
Researchers estimate that the current prevalence = p = 0.42.
Standard deviation =
A) For n = 7
Standard deviation =
= 1.31 (Round to 2 decimal)
For n = 7, Standard deviation = 1.31
B) For n = 10
Standard deviation =
= 1.56 (Round to 2 decimal)
For n = 10, Standard deviation = 1.56
C) For n = 15
Standard deviation =
= 1.91 (Round to 2 decimal)
For n = 15, Standard deviation = 1.91