8.5.1 The
weights, heights, yields, and seed number for 10 plants grown in an
experimental plot are given in the table. Each measurement is
approximately normally distributed. We wish to determine whether plants
in the plot differ from those outside. In each case, state one- and
two-tailed alternative hypotheses and find their significance. The
variance for weight is 9.0, and plants outside the plot have mean weight
10.0. ...
Get solution
8.5.2
The
weights, heights, yields, and seed number for 10 plants grown in an
experimental plot are given in the table. Each measurement is
approximately normally distributed. We wish to determine whether plants
in the plot differ from those outside. In each case, state one- and
two-tailed alternative hypotheses and find their significance. The
variance for height is 16.0, and plants outside the plot have mean
height 38.0....
Get solution
8.5.3
The
weights, heights, yields, and seed number for 10 plants grown in an
experimental plot are given in the table. Each measurement is
approximately normally distributed. We wish to determine whether plants
in the plot differ from those outside. In each case, state one- and
two-tailed alternative hypotheses and find their significance. The
variance for yield is 6.25, and plants outside the plot have mean yield
9.0. ...
Get solution
8.5.4
The
weights, heights, yields, and seed number for 10 plants grown in an
experimental plot are given in the table. Each measurement is
approximately normally distributed. We wish to determine whether plants
in the plot differ from those outside. In each case, state one- and
two-tailed alternative hypotheses and find their significance. The
variance for seed number is 25.0, and plants outside the plot have mean
seed number 15.0....
Get solution
8.5.6 Consider
again the weights, heights, yields, and seed number for 10 plants given
in Exercises 1–4. Find the sample variance for each and use the t
distribution to perform a two-tailed test of the hypothesis. Plants outside the plot have mean height 38.0.
Get solution
8.5.6 Consider
again the weights, heights, yields, and seed number for 10 plants given
in Exercises 1–4. Find the sample variance for each and use the t
distribution to perform a two-tailed test of the hypothesis. Plants outside the plot have mean height 38.0.
Get solution
8.5.7 Consider
again the weights, heights, yields, and seed number for 10 plants given
in Exercises 1–4. Find the sample variance for each and use the t
distribution to perform a two-tailed test of the hypothesis. Plants outside the plot have mean yield 9.0.
Get solution
8.5.8 Consider
again the weights, heights, yields, and seed number for 10 plants given
in Exercises 1–4. Find the sample variance for each and use the t
distribution to perform a two-tailed test of the hypothesis. Plants outside the plot have mean seed number 15.0.
Get solution
8.5.9 Find the smallest values of the sample mean for which the given hypothesis is rejected. Under
the conditions in Exercise 1, find the smallest weight that can reject
the null hypothesis that the mean weight is 10.0 with a one-tailed test
at the 0.01 level.
Get solution
8.5.10 Find the smallest values of the sample mean for which the given hypothesis is rejected. Under
the conditions in Exercise 2, find the smallest height that can reject
the null hypothesis that the mean height is 38.0 with a two-tailed test
at the 0.01 level.
Get solution
8.5.11 Find the smallest values of the sample mean for which the given hypothesis is rejected. Under
the conditions in Exercise 3, find the smallest yield that can reject
the null hypothesis that the mean yield is 9.0 with a two-tailed test at
the 0.001 level.
Get solution
8.5.12 Find the smallest values of the sample mean for which the given hypothesis is rejected. Under
the conditions in Exercise 4, find the smallest seed number that can
reject the null hypothesis that the mean seed number is 15.0 with a
one-tailed test at the 0.001 level.
Get solution
8.5.13 Find the power of the test assuming the given true mean. The true mean weight is 13.0 in Exercise 9.
Get solution
8.5.14 Find the power of the test assuming the given true mean. The true mean height is 43.0 in Exercise 10.
Get solution
8.5.15 Find the power of the test assuming the given true mean. The true mean yield is 11.0 in Exercise 11.
Get solution
8.5.16 Find the power of the test assuming the given true mean. The true mean seed height is 18.0 in Exercise 12.
Get solution
8.5.17
Consider
again plants with the null hypothesis that mean height is 39.0. Assume
that the standard deviation is known to be 3.2 cm. Show that a measured
sample mean of 40.0 is highly significant if the sample size is n = 88.
Get solution
8.5.18
Consider
again plants with the null hypothesis that mean height is 39.0. Assume
that the standard deviation is known to be 3.2 cm. Why is the power
with this sample size only 90% (as found in the text), rather than more
than 99%?
Get solution
8.5.19 Use the normal approximation to test the given hypothesis. A
coin is flipped 100 times and comes out heads 44 times. It is thought
that the coin is fair (has probability of heads is equal to 0.5). Do the
data provide evidence that the coin is unfair?
Get solution
8.5.20 Use the normal approximation to test the given hypothesis. Of
1000 people polled in one state, 320 favor the use of mathematics in
biology. The legislature has passed a bill mandating that at least 36%
of people must be in favor. Does the poll provide evidence that the
proportion is smaller than 0.36? Is the state in violation of the law?
Get solution
8.5.21
Consider the following data on 30 waiting times for 2 types of
events. ... Find the p-value associated with the null
hypothesis that the mean of type a is 1.0.
Get solution
8.5.22
Consider the following data on 30 waiting times for 2 types of events.
... Consider the following data on 30 waiting times for 2
types of events. Find the p-value associated with the null hypothesis
that the mean of type b is 1.0.
Get solution
8.5.23 Consider
again the data in Exercises 21 and 22. Each type has one or more
outliers that strongly affect the mean and standard deviation. Exclude
the outlier or outliers and recompute the p-value associated with the
null hypothesis that the mean is 1.0. What do you think of this
procedure if you were told that the data were generated from an
exponential distribution with mean 1.0? The outlier is extreme value 6.33 at time 16.
Get solution
8.5.24 Consider
again the data in Exercises 21 and 22. Each type has one or more
outliers that strongly affect the mean and standard deviation. Exclude
the outlier or outliers and recompute the p-value associated with the
null hypothesis that the mean is 1.0. What do you think of this
procedure if you were told that the data were generated from an
exponential distribution with mean 1.0? The outliers are the extreme values 4.16 and 4.83.
Get solution
8.5.25
A
chronic condition improves spontaneously in 45% of people. A new
medication is being tested to try to increase this percentage. Of 25 of
50 patients tested with the new medication, 30 improve. Is this
significant?
Get solution
8.5.26
A
chronic condition improves spontaneously in 45% of people. A new
medication is being tested to try to increase this percentage. Of 26 of
100 patients tested with the new medication, 60 improve. Is this
significantly better?
Get solution
8.5.27 A
chronic condition improves spontaneously in 45% of people. A new
medication is being tested to try to increase this percentage. Suppose
that the true fraction that improves with the medication is 0.6. What
is the power to detect this at the 0.05 level with a sample of 50
patients?
Get solution
8.5.28 A
chronic condition improves spontaneously in 45% of people. A new
medication is being tested to try to increase this percentage. Suppose
that the true fraction that improves with the medication is 0.6. What
is the power to detect this at the 0.05 level with a sample of 100
patients? How much greater is the power?
Get solution
8.5.29 A
company develops a new method to reduce error rates in the polymerase
chain reaction (PCR). With the old method, the number of errors in a
well-studied piece of DNA is known to have a Poisson distribution with
mean 35.0. Use a one-tailed test in each case. Find the
significance level if the DNA with the new method has only 27 errors.
Make sure to start by finding the normal approximation to the null
hypothesis.
Get solution
8.5.30 A
company develops a new method to reduce error rates in the polymerase
chain reaction (PCR). With the old method, the number of errors in a
well-studied piece of DNA is known to have a Poisson distribution with
mean 35.0. Use a one-tailed test in each case. Find the significance level if the DNA with the new method has only 23 errors.
Get solution
8.5.31
A
company develops a new method to reduce error rates in the polymerase
chain reaction (PCR). With the old method, the number of errors in a
well-studied piece of DNA is known to have a Poisson distribution with
mean 35.0. Use a one-tailed test in each case. What is the largest
number of errors that would reject the null hypothesis at the 0.05
level?
Get solution
8.5.32
A
company develops a new method to reduce error rates in the polymerase
chain reaction (PCR). With the old method, the number of errors in a
well-studied piece of DNA is known to have a Poisson distribution with
mean 35.0. Use a one-tailed test in each case. What is the largest
number of errors that would reject the null hypothesis at the 0.01
level?
Get solution
8.5.33 A
company develops a new method to reduce error rates in the polymerase
chain reaction (PCR). With the old method, the number of errors in a
well-studied piece of DNA is known to have a Poisson distribution with
mean 35.0. Use a one-tailed test in each case. Instead of
measuring only a single piece of DNA with the new method, 10 pieces are
measured and 300 errors are found. Does the new method reduce the number
of errors?
Get solution
8.5.34 A
company develops a new method to reduce error rates in the polymerase
chain reaction (PCR). With the old method, the number of errors in a
well-studied piece of DNA is known to have a Poisson distribution with
mean 35.0. Use a one-tailed test in each case. Instead of
measuring only a single piece of DNA with the new method, 20 pieces are
measured and 650 errors are found. Does the new method reduce the number
of errors?
Get solution
8.5.35
Use
the normal approximation to test the following hypotheses about growing
populations. In each case, habitat improvements are tried and the
population grows from 1 to 250 individuals in 50 years. Is there reason
to think that the habitat improvements helped? The population in
Section 7.8, Exercise 39, where per capita production is a random
variable with p.d.
f. g(x)=5.0 for 1.0 ≤ x ≤ 1.2.
Get solution
8.5.36
Use
the normal approximation to test the following hypotheses about growing
populations. In each case, habitat improvements are tried and the
population grows from 1 to 250 individuals in 50 years. Is there reason
to think that the habitat improvements helped? The population in
Section 7.8, Exercise 40, where per capita production is a random
variable with p.d.
f. g(x)= 1.25 for 0.7 ≤ x ≤ 1.5. Can you explain the difference from the result in the previous problem?
Get solution
