Reaction time studies are studies in which participants receive a stimulus and t
ID: 3126635 • Letter: R
Question
Reaction time studies are studies in which participants receive a stimulus and the amount of time it takes for them to react is measured. In one simple type of reaction time study, each participant holds a clicker button and stares at a screen. When the participant sees a part of the screen light up, he or she clicks the button as quickly as possible. The researcher then records how much time elapsed between when the screen lit up and when the participant clicked the button.
Suppose that, in these tests, the distribution of reaction times is skewed slightly to the right. Suppose also that mean reaction time is 190 milliseconds, and the standard deviation for reaction times is 20 milliseconds (for the purposes of this problem, you can treat the mean and standard deviation as population parameters). Use this information to answer the following questions, and round your answers to four decimal places.
a. Suppose we have 11 different people take this reaction time test. What is the probability that the average of these 11 reaction times will be greater than 180 milliseconds?
b. Suppose we have 15 different people take this reaction time test. What is the probability that the average of these 15 reaction times will be between 189 and 193 milliseconds?
c. Suppose we have 25 different people take this reaction time test. What is the probability that the average of these 25 reaction times will be less than 189 milliseconds?
Explanation / Answer
a)
We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as
x = critical value = 180
u = mean = 190
n = sample size = 11
s = standard deviation = 20
Thus,
z = (x - u) * sqrt(n) / s = -1.658312395
Thus, using a table/technology, the right tailed area of this is
P(z > -1.658312395 ) = 0.951372786 [ANSWER]
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b)
We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as
x1 = lower bound = 189
x2 = upper bound = 193
u = mean = 190
n = sample size = 15
s = standard deviation = 20
Thus, the two z scores are
z1 = lower z score = (x1 - u) * sqrt(n) / s = -0.193649167
z2 = upper z score = (x2 - u) * sqrt(n) / s = 0.580947502
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.423225298
P(z < z2) = 0.719362082
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.296136783 [ANSWER]
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c)
We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as
x = critical value = 189
u = mean = 190
n = sample size = 25
s = standard deviation = 20
Thus,
z = (x - u) * sqrt(n) / s = -0.25
Thus, using a table/technology, the left tailed area of this is
P(z < -0.25 ) = 0.401293674 [ANSWER]
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