Question 1202891: 8-6
Suppose your statistics instructor gave six examinations during the semester. You received the following exam scores (percent correct): 89, 64, 88, 89, 91, and 75. The instructor decided to randomly select two exam scores, compute their mean, and use this score to determine your final course grade.
e. Compute the mean of the sample means and the standard error of the sample means. (Round your answers to 2 decimal places.)
f. Applying the central limit theorem, if the instructor randomly samples two of your exam scores to compute an average as your final course grade, what is the probability your final course grade will be less than 82.67? What is the probability that your final course grade will be more than 82.67? (Round your answers to 1 decimal places.)
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the mean of the population is equal to82.66667.
the standard deviation of the population is equal to 9.87702.
the standard error of the sample means is equal to 9.87702 / sqrt(2) = 6.98411.
e. Compute the mean of the sample means and the standard error of the sample means. (Round your answers to 2 decimal places.)
the mean of the sample means will be equal to the population mean after an infinite number of samples of size 2.
that makes it eqal to 82.67.
the standard error of the sample means of size 2 will be equal to the standard deviation of the population divided by the square root of 2 which makes it equal to 6.98.
f. Applying the central limit theorem, if the instructor randomly samples two of your exam scores to compute an average as your final course grade, what is the probability your final course grade will be less than 82.67? What is the probability that your final course grade will be more than 82.67? (Round your answers to 1 decimal places.)
the z-score formula is z = (x - m) / s
z i the z-score
x is the sample mean
m is the population mean
s is the standard error of the sample means.
you get z = (82.67 - 82.67) / 6.98 = 0
the area to the left of that z-score is .5
the area to the right of that z-score is .5
the probability you will get a sample mean less than 82.67 is therefore .5 and the probability you will get a sample mean greter than 82.67 is therefore also .5.
keep in mind that the central limit theorem usually deals with sample sizes of 30 or more.
a sample size of 2 is not considered large enough to get an accurate representation, although infinite selections of samples of size 2 in a population of size 6 may be enough to justify the theorem.
i don't know enough to justify or refute the central limit theorem in tha case, although simulations by david lane do involve sample of size 2 as well.
here's a simulation from david lane tha uses cample sizes of 2 with 100,000 different samples of size 2 chosen randomly.
the mean of 100,000 sample means with samples of size 2 is very close to the population mean as shown in the demo.
the simulation calculator can be found at https://onlinestatbook.com/stat_sim/sampling_dist/
here's a refernce on cenral limit theorem by david lane.
https://sphweb.bumc.bu.edu/otlt/mph-modules/bs/bs704_probability/BS704_Probability12.html
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