8.3.1 Consider the following data on 20 plants. ... ... Find the following for the given measurement.
a. The sample mean.
b. The sample median.
c. The trimmed means ... The weight W
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8.3.2 Consider the following data on 20 plants. ... ... Find the following for the given measurement.
a. The sample mean.
b. The sample median.
c. The trimmed means ... The height H
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8.3.3 Consider the following data on 20 plants. ... ... Find the following for the given measurement.
a. The sample mean.
b. The sample median.
c. The trimmed means ... The yield Y
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8.3.4 Consider the following data on 20 plants. ... ... Find the following for the given measurement.
a. The sample mean.
b. The sample median.
c. The trimmed means ... The seed number S
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8.3.5 Find the sample variance, the sample standard deviation, and the standard error for the given measurement. The weight W in Exercise 1.
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8.3.6 Find the sample variance, the sample standard deviation, and the standard error for the given measurement. The height H in Exercise 2.
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8.3.7 Find the sample variance, the sample standard deviation, and the standard error for the given measurement. The yield Y in Exercise 3.
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8.3.8 Find the sample variance, the sample standard deviation, and the standard error for the given measurement. The seed number S in Exercise 4.
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8.3.9 Find
how many measurements lie (a) less than one sample standard deviation
from the sample mean and (b) more than two sample standard deviations
from the sample mean for the given measurement. Which behave more or
less like the normal distribution? The weight W in Exercise 1.
Get solution
8.3.10 Find
how many measurements lie (a) less than one sample standard deviation
from the sample mean and (b) more than two sample standard deviations
from the sample mean for the given measurement. Which behave more or
less like the normal distribution? The height H in Exercise 2.
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8.3.11 Find
how many measurements lie (a) less than one sample standard deviation
from the sample mean and (b) more than two sample standard deviations
from the sample mean for the given measurement. Which behave more or
less like the normal distribution? The yield Y in Exercise 3.
Get solution
8.3.12 Find
how many measurements lie (a) less than one sample standard deviation
from the sample mean and (b) more than two sample standard deviations
from the sample mean for the given measurement. Which behave more or
less like the normal distribution? The seed number S in Exercise 4.
Get solution
8.3.13 Find the 95% confidence limits around the sample mean for the given measurement, assuming that the sample variance s is a good estimate of the true variance σ . The weight W in Exercises 1 and 5.
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8.3.14 Find the 95% confidence limits around the sample mean for the given measurement, assuming that the sample variance s is a good estimate of the true variance σ . The height H in Exercises 2 and 6.
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8.3.15 Find the 95% confidence limits around the sample mean for the given measurement, assuming that the sample variance s is a good estimate of the true variance σ . The yield Y in Exercises 3 and 7.
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8.3.16 Find the 95% confidence limits around the sample mean for the given measurement, assuming that the sample variance s is a good estimate of the true variance σ . The seed number S in Exercises 4 and 8.
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8.3.17 Find the 95% confidence limits around the sample mean for the given measurement, without assuming that the sample variance s is a good estimate of the true variance σ (thus using the t distribution). The weight W in Exercises 1 and 5.
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8.3.18 Find the 95% confidence limits around the sample mean for the given measurement, without assuming that the sample variance s is a good estimate of the true variance σ (thus using the t distribution). The height H in Exercises 2 and 6.
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8.3.19 Find the 95% confidence limits around the sample mean for the given measurement, without assuming that the sample variance s is a good estimate of the true variance σ (thus using the t distribution). The yield Y in Exercises 3 and 7.
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8.3.20 Find the 95% confidence limits around the sample mean for the given measurement, without assuming that the sample variance s is a good estimate of the true variance σ (thus using the t distribution). The seed number S in Exercises 4 and 8.
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8.3.21 Find the given confidence limits around the sample mean for the given measurement, assuming that the sample variance s is a good estimate of the true variance σ . 98% confidence limits around the weight W in Exercise 1.
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8.3.22 Find the given confidence limits around the sample mean for the given measurement, assuming that the sample variance s is a good estimate of the true variance σ . 90% confidence limits around the height H in Exercise 2.
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8.3.23 Find the given confidence limits around the sample mean for the given measurement, assuming that the sample variance s is a good estimate of the true variance σ . 99.8% confidence limits around the yield Y in Exercise 3.
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8.3.24 Find the given confidence limits around the sample mean for the given measurement, assuming that the sample variance s is a good estimate of the true variance σ . 99.9% confidence limits around the seed number S in Exercise 4.
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8.3.25 Find the normal approximation to the following. The average of 30 numbers chosen from the exponential p.d.f.g(x ) = ...
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8.3.27 Find
95% confidence intervals in the following cases, assuming that the
standard deviations are known to match those in the earlier problem.
Does the confidence interval include the true mean? A sample mean of 0.4 is found in Exercise 25.
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8.3.27 Find
95% confidence intervals in the following cases, assuming that the
standard deviations are known to match those in the earlier problem.
Does the confidence interval include the true mean? A sample mean of 0.4 is found in Exercise 25.
Get solution
8.3.28 Find
95% confidence intervals in the following cases, assuming that the
standard deviations are known to match those in the earlier problem.
Does the confidence interval include the true mean? A sample mean of 0.7 is found in Exercise 26.
Get solution
8.3.29 Use the normal approximation to find 95% confidence limits around the estimated proportion ... in the following cases. A coin is flipped 100 times and comes out heads 44 times.
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8.3.31 The
95% confidence limits with the t distribution are wider than those with
the normal distribution. For approximately how many degrees of freedom
do they match the given confidence limits with the normal distribution?
How large a sample does this correspond to? The 99% confidence limits with the normal distribution.
Get solution
8.3.31 The
95% confidence limits with the t distribution are wider than those with
the normal distribution. For approximately how many degrees of freedom
do they match the given confidence limits with the normal distribution?
How large a sample does this correspond to? The 99% confidence limits with the normal distribution.
Get solution
8.3.32 The
95% confidence limits with the t distribution are wider than those with
the normal distribution. For approximately how many degrees of freedom
do they match the given confidence limits with the normal distribution?
How large a sample does this correspond to? The 98% confidence limits with the normal distribution.
Get solution
8.3.33 In simple cases, we can see why using a denominator of n in
the equation for the sample variance produces a biased estimate.
Suppose a population consists of half 0 s and half 1 s. Use the
variance of a Bernoulli distribution with p = 0.5 to find the exact variance of each measurement.
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8.3.34 In simple cases, we can see why using a denominator of n in
the equation for the sample variance produces a biased estimate.
Suppose a population consists of half 0 s and half 1 s. Find all
possible samples of size 2 and their associated probabilities.
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8.3.35 In simple cases, we can see why using a denominator of n in
the equation for the sample variance produces a biased estimate.
Suppose a population consists of half 0 s and half 1 s. Find the mean
squared deviation from the mean for each and average them to find the
expected sample variance.
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8.3.36 In simple cases, we can see why using a denominator of n in
the equation for the sample variance produces a biased estimate.
Suppose a population consists of half 0 s and half 1 s. Compare with
the true answer and show that using a denominator of n − 1 rather than n would give the right answer. Try to explain the bias.
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8.3.37 Consider
the following data on immigration into four populations over 20 yr
(based on the probabilities in Section 7.8, Exercises 33–36.) For each,
find the sample mean and the sample standard deviation. Compare them
with the mathematical mean and standard deviation found in the earlier
problem. ... Population a (see Section 7.8, Exercise 33).
Get solution
8.3.38 Consider
the following data on immigration into four populations over 20 yr
(based on the probabilities in Section 7.8, Exercises 33–36.) For each,
find the sample mean and the sample standard deviation. Compare them
with the mathematical mean and standard deviation found in the earlier
problem. ... Population b (see Section 7.8, Exercise 34).
Get solution
8.3.39
Consider
the following data on immigration into four populations over 20 yr
(based on the probabilities in Section 7.8, Exercises 33–36.) For each,
find the sample mean and the sample standard deviation. Compare them
with the mathematical mean and standard deviation found in the earlier
problem. ... Population c (see Section 7.8, Exercise 35). Why
are the mean and variance so different from the mathematical
expectations?
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8.3.40 Consider
the following data on immigration into four populations over 20 yr
(based on the probabilities in Section 7.8, Exercises 33–36.) For each,
find the sample mean and the sample standard deviation. Compare them
with the mathematical mean and standard deviation found in the earlier
problem. ... Population d (see Section 7.8, Exercise 36).
Get solution
8.3.41 Find
the 95% confidence intervals around the mean number of immigrants using
both the true variance and the sample variance. Does the true mean lie
within the confidence limits? Population a
Get solution
8.3.42 Find
the 95% confidence intervals around the mean number of immigrants using
both the true variance and the sample variance. Does the true mean lie
within the confidence limits? Population b
Get solution
8.3.43 Find
the 95% confidence intervals around the mean number of immigrants using
both the true variance and the sample variance. Does the true mean lie
within the confidence limits? Population c
Get solution
8.3.44 Find
the 95% confidence intervals around the mean number of immigrants using
both the true variance and the sample variance. Does the true mean lie
within the confidence limits? Population d
Get solution
8.3.45
Several plants are crossed, producing the following proportions. Find
99% confidence limits around the fraction of tall plants in each case.
Of 50 offspring, 35 are tall.
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8.3.46
Several plants are crossed, producing the following proportions. Find
99% confidence limits around the fraction of tall plants in each case.
Of 500 offspring, 350 are tall.
Get solution
8.3.47
Several plants are crossed, producing the following proportions. Find
99% confidence limits around the fraction of tall plants in each case.
Of 100 offspring, 52 are tall.
Get solution
8.3.48
Several plants are crossed, producing the following proportions. Find
99% confidence limits around the fraction of tall plants in each case.
Of 200 offspring, 13 are tall.
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8.3.49 Mutations
are counted in a large section of the genome. This count finds 14
mutations in one set of 1 million base pairs. Then 30 different sets of 1
million base pairs are measured, and the average number of mutations
per million is found to be 13.5. Use the normal approximation to
the Poisson distribution to estimate 95% confidence limits around the
true mean number of mutations per million base pairs and compare with
Section 8.2, Exercise 43.
Get solution
8.3.50 Mutations
are counted in a large section of the genome. This count finds 14
mutations in one set of 1 million base pairs. Then 30 different sets of 1
million base pairs are measured, and the average number of mutations
per million is found to be 13.5. Thirty different sets of 1
million base pairs are measured, and the average number of mutations per
million is found to be 13.5. Estimate the standard deviation, and find
the standard error of the mean and the 99% confidence limits.
Get solution
8.3.51 Use
the Monte Carlo method to estimate the confidence limits in Exercises
41 and 42 . How close are your results to the exact answer?
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