SOLUTION: ch 8 #11
A normal population has a mean of $79 and standard deviation of $20. You select random samples of nine.
a. Apply the central limit theorem to describe the sampling dis
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-> SOLUTION: ch 8 #11
A normal population has a mean of $79 and standard deviation of $20. You select random samples of nine.
a. Apply the central limit theorem to describe the sampling dis
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Question 1202884: ch 8 #11
A normal population has a mean of $79 and standard deviation of $20. You select random samples of nine.
a. Apply the central limit theorem to describe the sampling distribution of the sample mean with n = 9. With the small sample size, what condition is necessary to apply the central limit theorem?
b. What is the standard error of the sampling distribution of sample means? (Round your answer to 2 decimal places.)
c. What is the probability that a sample mean is greater than $82? (Round z-value to 2 decimal places and final answer to 4 decimal places.)
d. What is the probability that a sample mean is less than $75? (Round z-value to 2 decimal places and final answer to 4 decimal places.)
e. What is the probability that a sample mean is between $75 and $82? (Round z-value to 2 decimal places and final answer to 4 decimal places.)
f. What is the probability that the sampling error (x⎯− μ) would be $9 or more? That is, what is the probability that the estimate of the population mean is less than $70 or more than $88? (Round z-value to 2 decimal places and final answer to 4 decimal places.) Answer by Theo(13342) (Show Source):
if the population is normal, then the central limit theorem applies with sample sizes less than 30.
the reference measures another criteriia:
if the minimum of np or n * (1 - p) is > 5 in a proportion type study, then the central limit theorem can also apply.
a proportion type study is called dichotonous.
the particular problem statement we are working on is not dichotonomous, but the population is normal, so the central limit theorem applies, even if the sample size is less than 30, according to the reference.
based on that understanding, we'll assume the central limit theorem applies.
A normal population has a mean of $79 and standard deviation of $20. You select random samples of nine.
a. Apply the central limit theorem to describe the sampling distribution of the sample mean with n = 9. With the small sample size, what condition is necessary to apply the central limit theorem?
population must be normal.
b. What is the standard error of the sampling distribution of sample means? (Round your answer to 2 decimal places.)
standard error = standard deviation / square root of sample size = 20 / sqrt(9) = 6.67 rounded to 2 decimal places.
c. What is the probability that a sample mean is greater than $82? (Round z-value to 2 decimal places and final answer to 4 decimal places.)
z = (x - m) / s
z is the z-score
x is the sample mean
m is the population mean
s is the standard error.
formula becomes z = (82 - 79) / 6.67 = .45 rounded to 2 decimal places.
area to the right of that z-score is equal to .3264 rounded to 4 decimal places.
that's the probability of getting a sample mean greater than 82.
d. What is the probability that a sample mean is less than $75? (Round z-value to 2 decimal places and final answer to 4 decimal places.)
z-score formula becomes z = (75 - 79) / 6.67 = -.60 rounded to 2 decimal places.
area to the left of that z-score is equal to .2743 rounded to 4 decimal places.
that's the probability of getting a sample mean less than 75.
e. What is the probability that a sample mean is between $75 and $82? (Round z-value to 2 decimal places and final answer to 4 decimal places.)
z-score for sample mean of 75 is -.60
z-score for sample mean of 82 is .45
area to the left of z-score of -.60 = .2743
area to the left of z-score of .45 = .6736
area in between = .6736 minus .2743 = .3993
that's the probability of getting a sample mean between 75 and 82.
f. What is the probability that the sampling error (x⎯− μ) would be $9 or more? That is, what is the probability that the estimate of the population mean is less than $70 or more than $88? (Round z-value to 2 decimal places and final answer to 4 decimal places.)
sampling error means the same as margin of error i believe.
that would be (x - m).
you want to know the probability that the margin of error is 9 dollars or more.
your sample means would have to be between 79 - 9 = 70 and 79 + 9 = 88.
z-score for sample mean of 70 is equal to (70 - 79) / 6.67 = -1.35 rounded to 2 decimal places.
z-score for sample mean of 88 is equal to (88 - 79) / 6.67 = 1.35 rounded to 2 decimal places.
area to the left of z-score of -1.35 = .0885 rounded to 4 decimal places.
area to the left of z-score of 1.35 = .9115 rounded to 4 decimal places.
area in between = .9115 minus .0885 = .8230.
that's the probability of geeting a sample mean with a sampling error less than 9.
the probability of getting a sampling error greater than 9 would then be 1 - .8230 = .1770.
here's what the results look like graphically.
any discrepance between the values i calculated manually and the values that the calculator provides are due to rounding diferences only.
