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 1? reaction times will be greater, than 180 milliseconds? 1.6583 b. Suppose we have 17 different people take this reaction time test. What is the probability that the average of these 17 reaction times will be less than 192 milliseconds? c. Suppose we have 23 different people take this reaction time test. What is the probability that the average of these 23 reaction times will be less than 185 milliseconds? d. Would it be appropriate to use the normal probability app to compute the probability that a single reaction time is less than 185 milliseconds? (You have two attempts for this question.) No, because we can only compute the probability that reaction time is less than *or equal to 185 milliseconds. O Yes, because the Law of Large Numbers states that as a variable increases, it becomes more accurate. No, because reaction times are not normally distributed, and the normal probability app is only for computing probabilities associated with a normally distributed variable. O No, because you cannot compute a probability for a single event, only for the long run relative frequency of an infinite number of events. O Yes, because everything in statistics is an approximation and so it doesn't matter if our methodology makes sense. All that matters is that we use a method that produces a number of some sort. Yes, because the Central Limit Theorem makes everything become normally distributed. Yes, because converting a variable to a z-score makes that variable become normally distributed

Answer 1

Answer:

a)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    183      
u = mean =    190      
n = sample size =    9      
s = standard deviation =    20      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    -1.05      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   -1.05   ) =    0.853140944 [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 =    187      
x2 = upper bound =    192      
u = mean =    190      
n = sample size =    17      
s = standard deviation =    20      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -0.618465844      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    0.412310563      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.268134153      
P(z < z2) =    0.659944096      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.391809943   [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 =    185      
u = mean =    190      
n = sample size =    22      
s = standard deviation =    20      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    -1.17260394      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -1.17260394   ) =    0.120477334 [ANSWER]