Aplia: Student Question S. Using z-scores with the distribution of sample means

Question

Aplia: Student Question S. Using z-scores with the distribution of sample means The Stroop task is a psychological test of mental flexibility. The task takes advantage of the fact that we read words more quickly and automatically than we can name colors. If a word is printed or displayed in a color different from the color it actually names, we will say the word more readily than we can name the color it is written in. For example, if the word "green" is written in blue ink, we will say the word "green" more readily than we can name the color as blue. A professor of cognitive psychology is interested in the reaction time on the Stroop task among college students. She measures the reaction time for 36 randomly selected students. The professor knows that the distribution of scores is normal, but she does not know that the true average reaction time on the Stroop task among college students is 758.7 msecs with a standard deviation of 78.20 msecs The expected value of the mean of the 36 randomly selected students, M, is mean and/or standard deviation just given to calculate the expected value of M.) . (Hint: Use the population . (Hint: Use the population mean and/or standard deviation just given to The standard error of M is calculate the standard error.) The DataView tool that follows displays a data set consisting of 200 potential samples (each sample has 36 observations). Sample Variables2 Observations 200 Statistics for 200 Random Samples [n = 36) drawn from a normal distribution R was used to generate the samples values 200 200 Variable Type Form Missing Sample Means Qvantitative Numeric Numeric Sample SD Quantitative

Explanation / Answer

When sample size (n) = 36
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
a.
expected mean ( u ) = 758.7
b.
standard error ( se )= 78.2/ Sqrt ( 36 ) = 13.033
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When sample size (n) = 33
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
a.
expected mean ( u ) = 758.7
b.
standard error ( se )= 78.2/ Sqrt ( 33 ) = 13.6129
c.
P(X > 758.7) = (758.7-758.7)/78.2/ Sqrt ( 33 )
= 0/13.613= 0
= P ( Z >0) From Standard Normal Table
= 0.5

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