Question
I need the answers of this quiz: Question 2: In a study of 13 children with cystic fibrosis who completed an exercise program, exercise endurance
I need the answers of this quiz:
Question 2:
In a study of 13 children with cystic fibrosis who completed an exercise program, exercise endurance (duration on an exercise test) improved by a mean of 0.77 minutes. The standard deviation for the change in endurance was 0.83 minutes. What is the theoretical distribution of the mean?
- unansweredT-distribution (with 12 degrees of freedom); mean=true mean; standard error= 0.23Standard normal distribution; mean=true mean; standard error= 0.23T-distribution with 12 degrees of freedom; mean=true mean; standard error= 0.07Standard normal distribution; mean=true mean; standard error= 0.07T-distribution (with 12 degrees of freedom); mean=true mean; standard error= 0.83Question 3:
Thirty heart disease patients are put on an exercise regimen. Twenty patients improve on the exercise stress test and ten dont improve or get worse. Calculate the 95% confidence interval for the true proportion of heart disease patients who improve their fitness using this particular exercise regimen. Recall that proportions are normally distributed with a standard error of
p(1p)n
(You may use the observed proportion to calculate the standard error.)
- unanswered66%66%-70%50%-84%56%-76%Question 4:
The following data were collected from a case-control study of breast cancer and fat intake:
case | control | |
High-fat diet | 10 | 5 |
Low-fat diet | 40 | 55 |
50 | 60 |
Statistical inferences for odds ratios are based on the natural log of the odds ratio, rather than the odds ratio itself (because the distribution for an odds ratio does not follow a normal distribution). The sampling distribution of the natural log of the odds ratio (lnOR) follows a normal distribution, with standard error
=1a+1b+1c+1d
(where a, b, c, and d are the cells in the 2x2 table).
Calculate the odds ratio for breast cancer (comparing high-fat diet to low-fat diet) from the 2x2 table above.
- unanswered2.750.402.522.401.00Question 5:
Take the natural log of the odds ratio that you calculated in question 4.
- unanswered15.62.750.51.010Question 6:
Calculate the standard error of the lnOR, according to the formula given above.
- unanswered15.60.5860.091.1721.01Question 7:
Calculate the 95% confidence interval for the lnOR.
- unanswered.24, 1.76-.01, 1.96-.14, 2.160, -2.16-.34, 3.16Question 8:
Convert the upper and lower confidence limits that you calculated in (e) back to odds ratios by exponentiating (i.e., calculate the 95% confidence interval for the OR).
- unanswered0.87, 8.661.58, 3.921.01, 9.650, 2.160.51, 6.56Question 9:
TRUE OR FALSE. The confidence interval of an odds ratio is symmetric.
- unansweredTrueFalseQuestion 10:
TRUE or FALSE. The odds ratio here is not statistically significant at the p=.05 level.
- unansweredTrueFalseQuestion 11:
A hypothetical HIV vaccine trial involving 20,000 participants10,000 in the vaccine group and 10,000 in the placebo grouphad the following results: 6.3 infections per 1000 in the vaccine group and 9.0 infections per 1000 in the placebo group.
I ran a computer simulation to predict possible outcomes of the trial if the null hypothesis is truethat is, if vaccinated and unvaccinated people are equally likely to contract HIV. I ran 1000 virtual trials of 20,000 people (10,000 per group) assuming that the vaccine is ineffective. Outcomes are expressed as excess infections in the placebo group. Here are the results of the 1000 virtual trials displayed as a histogram.
- unanswered
Question 12:
From the simulation, I learn that the statistic excess infections in the placebo group follows a normal distribution with a mean of 0 and a standard deviation (standard error) of 12.3. Use this information to calculate the two-sided p-value more precisely than in (11). Round to the nearest thousandth.
- unanswered
Question 13:
How would the above histogram change if I ran the simulation 10,000 times rather than 1000 times?
- unansweredIt would simply be more smooth.The p-value would get smaller.The standard error would get smaller.The distribution would be more similar to a T-distribution.Question 14:
The effect size in this hypothetical trial (2.7 fewer infections per 1000 vaccinated) is slightly smaller than the effect size found in the real 2009 HIV vaccine trial (2.8 fewer infections per 1000 vaccinated); so why is the p-value smaller?
- unansweredThe hypothetical trial has less variability.The sample size of the hypothetical trial is larger. .The real trial did not follow a normal distribution.The real trial used an intention to treat analysis.Question 15:
The Centers for Disease Control (CDC) collects yearly statistics on drinking behavior in the United States by surveying a random sample of U.S. adults. The following data display the percent of adults aged 18 years and over who had 5 or more drinks in 1 day at least once in the given year.
Year | Percent | 95% confidence interval |
1997 | 21.4 | 20.8-22.0 |
1998 | 20.2 | 19.6-20.8 |
1999 | 20.3 | 19.6-21.0 |
2000 | 19.2 | 18.6-19.9 |
2001 | 20.0 | 19.4-20.6 |
2002 | 19.9 | 19.2-20.5 |
2003 | 19.1 | 18.5-19.8 |
For which years (or year) above could you reject the null hypothesis that more than 20% of U.S. adults had 5 or more drinks in 1 day, at the .05 significance level?
There may be more than one correct answer.
- unanswered1997199819992000200120022003Step by Step Solution
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