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Pranay Aryal
Choose from one of the options below and press 'submit' below to know the answer and press "show answer" to see the explanation. Use the up and down arrows to navigate between slides
Choose from one of the options below and press 'submit' below to know the answer and press "show answer" to see the explanation
Yes two is the answer
Continuation of problem 21. The investigator will estimate the success rate of the new hypnotherapy program with a 95% confidence interval. If she wants this confidence interval to have a half-width of 0.03, and she believes that this success rate will be somewhere around 0.20, how many subject must she enroll (ignore any previous sample size calculations for this study)?
Formula used \(n=\frac{z_{0.975}^2.\hat p(1-\hat p)}{d^2}\) \(\text{where} z_{0.975}=1.96 , \text{ and d= 0.03 and } \hat p =0.20\) \(=\frac{1.96^2 X 0.2 X 0.8}{0.03^2}\)
(1.96^2)*0.2*0.8/((0.03)^2)
## [1] 683
therefore b is the answer
A physical therapist wants to test the reaction time of a group of athletes using a standardized dexterity instrument. For each individual, the dexterity instrument calculates the cumulative reaction time, which is the sum of the reaction times for a series of stimuli provided by the dexterity instrument. For the general population, the mean cumulative reaction time is 14.2 seconds. The physical therapist would like to test if the mean reaction time of this group of athletes is different than 14.2 seconds. He has reason to believe that the mean reaction time in this group of athletes is 13.0 seconds, and that the standard deviation is s = 3.1 seconds. If he conducts the test at a significance level of .05 and wants power equal to 0.80, how many subjects should he enroll? (Note that the 80th percentile of the standard normal distribution is 0.84).
7
53
63
Not enough information has been given to calculate sample size
formula used \(n=[\frac{(Z_{1-\alpha/2}+Z_{1-\beta}).\sigma}{\mu_a-\mu_0}]^2\)
\(=[\frac{(Z_{0.975}+Z_{0.80}) X 3.1}{13.0-14.2}]^2\)
((1.96+0.84)*3.1/(13-14.2))^2
## [1] 52.32
b is the answer
After collecting data on the first 15 athletes in the study, the therapist calculates the sample standard deviation to be 2.5, which is lower than what he had originally suspected (3.1). If he recalculates his sample size based on this new estimated standard deviation, what will be the effect?
Sample size will stay the same as the originally calculated value
Sample size will increase from the originally calculated value
Sample size will decrease from the originally calculated value
There is not enough information to determine what will happen to sample size
I think '3' is the answer because according to the formula \(n=[\frac{(Z_{1-\alpha/2}+Z_{1-\beta}). \sigma}{\mu_a-\mu_0}]^2\), if sigma is low, sample size will be low
For 24 formula used \(n=[\frac{(Z_{1-\alpha/2}+Z_{1-\beta}).\sigma}{\mu_a-\mu_0}]^2\)
\(=[\frac{(Z_{0.975}+Z_{0.80}) X 3.1}{13.0-14.2}]^2\)
((1.96+0.84)*3.1/(13-14.2))^2
## [1] 52.32
b is the answer