chapter 9 ex 3rd edition
9-2 a) The type I error probability is the probability of rejecting the null hypothesis (H0: μ
= 12) when it is actually true. To find the type I error probability, we need to find the
critical value that defines the critical region. In this case, the critical region is defined as x̄
< 11.5 kilograms.
Since the sample size is small (n = 4), we can use the t-distribution instead of the normal
distribution to find the critical value. The critical value can be found using a t-distribution
table with degrees of freedom (df) equal to n-1 = 3 and a significance level of α.
Using the t-distribution table, we can find the critical t-value that corresponds to a onetailed test with a significance level of α and df = 3. Let's assume α is set at 0.05. The
critical t-value would be -2.365. This means that if the sample mean is less than -2.365
standard deviations below the null hypothesis mean (12), we reject H0.
The type I error probability is the probability of observing a sample mean less than 11.5
kilograms when the true mean is equal to 12 kilograms. We can use the t-distribution
formula to calculate the type I error probability:
P(x̄ < 11.5 | μ = 12) = P(t < -2.365 | μ = 12)
(b) The β (beta) is the probability of failing to reject the null hypothesis (H0) when the
alternative hypothesis (H1) is true. To find β, we need to find the probability of observing
a sample mean greater than or equal to 11.5 kilograms when the true mean is 11.25
kilograms.
P(x̄ ≥ 11.5 | μ = 11.25) = P(t ≥ -2.365 | μ = 11.25)
We can use the t-distribution formula to calculate β:
β = P(t ≥ -2.365 | μ = 11.25)
This value can be found using a t-distribution table with df = 3.
9-3 a) The type I error probability is the probability of rejecting the null hypothesis (H0: μ
= 12) when it is actually true. To find the type I error probability, we need to find the
critical value that defines the critical region. In this case, the critical region is defined as x̄
< 11.5 kilograms.
Since the sample size is larger (n = 16), we can use the normal distribution instead of the
t-distribution to find the critical value. The critical value can be found using a standard
normal table with a significance level of α.
Let's assume α is set at 0.05. The critical z-value would be -1.833. This means that if the
sample mean is less than -1.833 standard deviations below the null hypothesis mean (12),
we reject H0.
The type I error probability is the probability of observing a sample mean less than 11.5
kilograms when the true mean is equal to 12 kilograms. We can use the standard normal
distribution formula to calculate the type I error probability:
P(x̄ < 11.5 | μ = 12) = P(z < -1.833 | μ = 12)
(b) The β (beta) is the probability of failing to reject the null hypothesis (H0) when the
alternative hypothesis (H1) is true. To find β, we need to find the probability of observing
a sample mean greater than or equal to 11.5 kilograms when the true mean is 11.25
kilograms.
P(x̄ ≥ 11.5 | μ = 11.25) = P(z ≥ -1.833 | μ = 11.25)
We can use the standard normal distribution formula to calculate β:
β = P(z ≥ -1.833 | μ = 11.25)
This value can be found using a standard normal table.
9-4 In Exercise 9-2, the hypothesis test is to test H0: μ = 12 against H1: μ < 12 using a
sample size of n = 4. The type I error probability, alpha, is specified to be 0.01.
To find the boundary of the critical region, we need to find the critical value that
corresponds to alpha = 0.01. In this case, we are using the t-distribution since the sample
size is small (n = 4).
The critical value can be found using a t-distribution table with 3 degrees of freedom (df
= n - 1) and a significance level of α = 0.01. The critical t-value would be -3.182. This
means that if the sample mean is less than -3.182 standard deviations below the null
hypothesis mean (12), we reject H0.
The critical region can be defined as x̄ < 11.49 kilograms (12 + (-3.182) * (0.5 / sqrt(4))).
The boundary of the critical region is 11.49 kilograms.
(Hence, the boundary of the critical region for the test with type I error probability of 0.01
and a sample size of 4 is x-critical = 12.5825. Any sample mean less than 12.5825 would
result in rejection of the null hypothesis.)
9-5
In Exercise 9-2, the hypothesis test is to test H0: μ = 12 against H1: μ < 12 using a sample
size of n = 4. The type I error probability, alpha, is specified to be 0.05.
To find the boundary of the critical region, we need to find the critical value that
corresponds to alpha = 0.05. In this case, we are using the t-distribution since the sample
size is small (n = 4).
The critical value can be found using a t-distribution table with 3 degrees of freedom (df
= n - 1) and a significance level of α = 0.05. The critical t-value would be -1.833. This
means that if the sample mean is less than -1.833 standard deviations below the null
hypothesis mean (12), we reject H0.
The critical region can be defined as x̄ < 11.93 kilograms (12 + (-1.833) * (0.5 / sqrt(4))).
The boundary of the critical region is 11.93 kilograms.
9-5
To find the boundary of the critical region when the type I error probability (alpha) is 0.05,
we need to find the critical value for a one-tailed, lower-tailed test with a significance
level of 0.05. This can be done using a t-distribution table with 3 degrees of freedom
(sample size - 1).
Since the hypothesis is H0: mu = 12 against H1: mu < 12, we are performing a lower-tailed
test. So, we need to find the critical value such that the area under the t-distribution to
the left of that value is 0.05.
From the t-distribution table, the critical value for a one-tailed, lower-tailed test with a
significance level of 0.05 and 3 degrees of freedom is -2.492.
Therefore, the boundary of the critical region is given by xbar < -2.492 * s/sqrt(n). Using
the sample information from Exercise 9-2 (n = 4, s = 0.5, and xbar = the sample mean),
we can calculate the critical region boundary as follows:
xbar < -2.492 0.5 / sqrt(4) = -2.492 0.5 / 2 = -0.622
So, if the sample mean (xbar) is less than -0.622, we will reject the null hypothesis H0:
mu = 12 and conclude that the mean elongation is less than 12 kilograms.
เช็คตัวเลขใหม่ทุกอัน
9-6 a) The type I error probability α is the probability of rejecting the null hypothesis
when it is actually true. To find α, we need to find the probability that the sample mean
falls outside of the acceptance region when the true population mean is 100.
Since the population is approximately normally distributed with mean 100 and standard
deviation 2, the standard error of the mean is 2/sqrt(9) = 0.447.
Using a z-score, we can find the proportion of the normal distribution outside of the
acceptance region:
z = (98.5 - 100) / 0.447 = -2.206 z = (101.5 - 100) / 0.447 = 2.206
Using a standard normal table, we find that the area to the left of -2.206 is 0.0132, and
the area to the right of 2.206 is 0.0132. Therefore, α = 0.0132 + 0.0132 = 0.0264.
b) To find β for the case where the true mean heat evolved is 103, we need to find the
probability that the sample mean falls in the acceptance region when the true population
mean is 103.
Using the same standard error of the mean (0.447), we can find the z-score:
z = (103 - 100) / 0.447 = 6.729
Using a standard normal table, we find that the area to the left of 6.729 is 1, which means
that the sample mean will always be in the acceptance region when the true population
mean is 103, so β = 0.
c) To find β for the case where the true mean heat evolved is 105, we need to find the
probability that the sample mean falls in the acceptance region when the true population
mean is 105.
Using the same standard error of the mean (0.447), we can find the z-score:
z = (105 - 100) / 0.447 = 11.831
Using a standard normal table, we find that the area to the left of 11.831 is 1, which means
that the sample mean will always be in the acceptance region when the true population
mean is 105, so β = 0.
This value of β is smaller than the one found in part (b) above because the difference
between the true population mean (105) and the hypothesized mean (100) is larger than
the difference between the true population mean (103) and the hypothesized mean (100).
When the difference between the true population mean and the hypothesized mean is
larger, the sample mean is less likely to fall in the acceptance region, so β is smaller.
9-7 For the given hypothesis test, H0: μ = 100 versus H1: μ ≠ 100, with a sample size of
n=5 and an acceptance region of 98.5 ≤ x̄ ≤ 101.5, the type I error probability α can be
calculated using the Z-score and Z-table.
To calculate Z-score, we use the formula: Z = (x̄ - μ) / (σ / √n) where x̄ is the sample
mean, μ is the population mean, σ is the population standard deviation, and n is the
sample size.
For the null hypothesis, μ = 100 and Z = (x̄ - 100) / (2 / √5)
The Z-score corresponding to the lower limit of the acceptance region, 98.5, is calculated
as follows: Z = (98.5 - 100) / (2 / √5) = -1.32
The Z-score corresponding to the upper limit of the acceptance region, 101.5, is
calculated as follows: Z = (101.5 - 100) / (2 / √5) = 1.32
Using a Z-table, we can find the area under the standard normal distribution curve that
corresponds to each Z-score. The area to the left of Z = -1.32 is 0.0921 and the area to
the left of Z = 1.32 is 0.9053. Therefore, the type I error probability α is equal to 1 (0.9053 - 0.0921) = 0.0026.
(b) To find the value of β for the case where the true mean heat evolved is 103, we can
use the Z-score formula again. Z = (x̄ - 103) / (2 / √5) = (x̄ - 103) / 0.8944
The value of x̄ that corresponds to β = 0.05 is calculated using the Z-table. For a Z-score
of 1.65, the area to the left is 0.95. Therefore, using the Z-score formula, we can calculate
x̄ as follows: 103 + (1.65 * 0.8944) = 104.48
This means that if the true mean is 103, there is a 5% chance of obtaining a sample mean
less than or equal to 104.48, resulting in a type II error.
(c) To find the value of β for the case where the true mean heat evolved is 105, we can
use the Z-score formula again. Z = (x̄ - 105) / (2 / √5) = (x̄ - 105) / 0.8944
The value of x̄ that corresponds to β = 0.05 is calculated using the Z-table. For a Z-score
of 2.58, the area to the left is 0.99. Therefore, using the Z-score formula, we can
calculate x̄ as follows: 105 + (2.58 * 0.8944) = 107.23
This means that if the true mean is 105, there is a 1% chance of obtaining a sample mean
less than or equal to 107.23, resulting in a type II error.
The value of β for the case where the true mean is 105 is smaller than the one found in
part (b) above because the mean is farther away from the null hypothesis and the sample
size is smaller. The smaller sample size makes it more difficult to detect a difference,
resulting in a larger
9-8 (a) If the critical region is xbar = 185, then alpha is the probability of obtaining a
sample mean of 185 or higher, given that the true mean foam height is 175.
Using the standard normal table, we find that the z-score for xbar = 185 is (185 - 175) /
(20/sqrt(10)) = 1.5.
The probability of observing a z-score of 1.5 or greater is 0.0668, so the type I error
probability alpha is 0.0668.
(b) If the true mean foam height is 195 millimeters, then the type II error probability beta
is the probability of not rejecting the null hypothesis H0: mu = 175 millimeters, given that
the true mean is 195.
The z-score for xbar = 195 is (195 - 175) / (20/sqrt(10)) = 3.
The probability of observing a z-score less than 3 is 0.0013, so the type II error
probability beta is 1 - 0.0013 = 0.9987.
9-15 For the hypothesis test in Exercise 9-8, H0: mu = 175 millimeters, H1: mu > 175
millimeters, with n = 10 and a critical value of xbar = 185 millimeters, the probability of
accepting H0 (Type II error) can be calculated for various values of the true mean, mu.
The operating characteristic (OC) curve can be constructed by plotting the probability of
accepting H0 (1 - alpha) on the y-axis and the true mean, mu, on the x-axis for the given
values of mu.
Steps to calculate probability of accepting H0 for each value of mu:
1. Find the standard deviation of the sample mean, sigma/sqrt(n), where sigma = 20
millimeters (standard deviation of foam height) and n = 10.
2. For each value of mu, find the corresponding z-score using the formula z = (xbar mu)/(sigma/sqrt(n)).
3. Using the z-score, find the probability of accepting H0 using a standard normal table.
For each value of mu, repeat these steps and plot the resulting probabilities on the y-axis
versus the corresponding mu value on the x-axis to construct the OC curve.
9-9 (a) The conclusion would be that the sample data provide sufficient evidence to
reject the null hypothesis H0: mu= 175 millimeters, in favor of the alternative hypothesis
H1: mu >175 millimeters, since xbar = 190 millimeters falls in the critical region defined as
xbar >=185 millimeters.
(b) The probability of observing a sample average as large as 190 millimeters (or larger),
if the true mean foam height was really 175 millimeters, can be found using the standard
normal cumulative distribution function. The z-score for 190 millimeters is (190175)/(20/sqrt(10)) = 2.37. Therefore, the probability is given by the cumulative probability
for z >= 2.37, which is 0.00889. This means that it is very UNUSUAL that we would
observe a sample average as large as 190 millimeters when the true mean was really 175
millimeters.
เช็คตัวเลขดีๆนะ มันประมาณหยาบไป
9-10. Repeat Exercise 9-8 assuming that the sample size is n =16 and the boundary of
the critical region is the same.
A consumer products company is formulating a new shampoo and is interested in foam
height (in millimeters). Foam height is approximately normally distributed and has a
standard deviation of 20 millimeters. The company wishes to test H0: mu= 175
millimeters versus H1: mu >175 millimeters, using the results of n = 16 samples.
(a) Find the type I error probability alpha if the critical region is xbar =185.
We need to find the critical value for x̄ at the alpha=0.05 level of significance, where x̄ is
the sample mean. Since the critical region is xbar =185, the critical value is 185. To find
the type I error probability, we need to find the probability of observing a sample mean of
x̄ =185 or higher under the assumption that the true mean is mu= 175 millimeters (H0 is
true).
Using the z-score formula, we can find the z-score corresponding to the critical value
185:
z = (185 - 175) / (20/√16) = (185 - 175) / (20/4) = 2.5
Using a standard normal table, the probability of observing a z-score of 2.5 or higher is
0.0062, so the type I error probability alpha is 0.0062.
(b) What is the probability of type II error if the true mean is 195 millimeters?
We need to find the probability of failing to reject H0 (i.e., accepting H0 as true) when the
true mean is 195 millimeters. To do this, we find the z-score corresponding to the value of
x̄ that is just below the critical value, and then find the corresponding probability using a
standard normal table.
z = (185 - 195) / (20/√16) = (185 - 195) / (20/4) = -2.5
The probability of observing a z-score of -2.5 or lower is 0.0062, so the probability of
type II error (beta) if the true mean is 195 millimeters is 0.0062.
9-11. Consider Exercise 9-8, and suppose that the sample size is increased to n =16. a)
Where would the boundary of the critical region be placed if the type I error probability
were to remain equal to the value that it took on when n =10?
The type I error probability alpha remains the same as before, so we can use the formula
for the critical value for a one-tailed test with alpha = 0.05, a standard normal
distribution, and a sample size of n =16:
z = z(1-alpha) = z(0.95) = 1.833
This means that xbar must be greater than
xbar_critical = 175 + 1.833 * (20/sqrt(16)) = 185.75
to reject the null hypothesis.
b) Using n =16 and the new critical region found in part (a), find the type II error
probability beta if the true mean foam height is 195 millimeters.
The formula for the type II error probability beta is:
beta = P(accept H0 | H1 is true) = P(xbar < xbar_critical | mu = 195)
This can be calculated using a standard normal distribution table:
beta = P(Z < (185.75 - 195)/(20/sqrt(16))) = P(Z < -2.86) = 0.002
(c) Compare the value of beta obtained in part (b) with the value from Exercise 9-8 (b).
What conclusions can you draw?
The value of beta obtained in part (b) is smaller than the value from Exercise 9-8 (b),
meaning that the increased sample size has led to a smaller type II error probability. This
is expected, as increasing the sample size generally results in a more powerful test and a
lower probability of making a type II error.
9-12 a) To find alpha, we need to find the probability of observing a sample mean in the
acceptance region when the true mean is not equal to 5 Volts. Since the standard
deviation is 0.25 Volts and the sample size is 8, the standard error of the sample mean is
0.25/sqrt(8) = 0.10 Volts.
Using a z-test, we can standardize the difference between the sample mean and the true
mean and find the probability that the standardized difference falls within the critical
region. The critical region is defined by 4.85 <= xbar <= 5.15, so the critical z-values are
-1.8 and 1.8 respectively.
Using a standard normal table, the area to the left of -1.8 is 0.0366 and the area to the
right of 1.8 is 0.0366. The area between -1.8 and 1.8 is 1 - 2 * 0.0366 = 0.9268. So, alpha
is 1 - 0.9268 = 0.0732, the probability of observing a sample mean in the acceptance
region when the true mean is not equal to 5 Volts.
(b) To find the power of the test, we need to find the probability of observing a sample
mean outside of the acceptance region when the true mean is 5.1 Volts. Using the same
calculation method as in (a), the standardized difference between the sample mean and
the true mean is (5.1 - 5) / 0.10 = 1.
Using a standard normal table, the area to the left of 1 is 0.8413. So, the power of the test
is 1 - 0.8413 = 0.1587, the probability of observing a sample mean outside of the
acceptance region when the true mean is 5.1 Volts.
9-13 (a) With a sample size of n = 16, the standard deviation of the sample mean
decreases, resulting in a narrower acceptance region. To maintain the same alpha (type I
error rate), the boundaries of the acceptance region need to be adjusted accordingly.
To find the new boundaries, we can use the Z-distribution table to determine the Z-score
corresponding to alpha/2, and then use that Z-score to calculate the new critical values.
Let's assume that alpha remains the same as in Exercise 9-12, then:
Z = (5.15 - 5)/(0.25/sqrt(16)) = 2.56
Z = (4.85 - 5)/(0.25/sqrt(16)) = -2.56
The new boundaries would be 4.95 <= xbar <= 5.05, as calculated using Z-scores and
the standard deviation of the sample mean.
(b) To find the power of the test for detecting a true mean output voltage of 5.1 Volts, we
need to find the probability of not rejecting the null hypothesis when the true mean is 5.1
Volts.
Let's calculate the Z-score for a true mean of 5.1 Volts:
Z = (xbar - 5.1)/(0.25/sqrt(16)) = 1.28
Using the Z-distribution table, we find that the area under the curve to the right of Z =
1.28 is 0.8972. This is the probability of not rejecting the null hypothesis when the true
mean is 5.1 Volts.
The power of the test (1 - beta) is 1 - 0.8972 = 0.1028, meaning that there is a 10.28%
chance of detecting a true mean output voltage of 5.1 Volts if it exists.
9-14 To find the acceptance region, we need to know the critical value for a two-tailed
test with alpha = 0.05 and n = 8. Assuming a normal distribution for the sample mean, we
can use the standard normal distribution table to find the critical value.
The critical value for a two-tailed test with alpha = 0.05 and n = 8 is approximately 2.306.
This means that the acceptance region is located 2.306 standard deviations from the
mean on either side.
The standard deviation for the sample mean is equal to the standard deviation of the
population divided by the square root of the sample size, or (0.25/√8) = 0.125.
The acceptance region is then located between (5 - 2.306 0.125) = 4.92 and (5 + 2.306
0.125) = 5.08. Therefore, the acceptance region is 4.92 <= xbar <= 5.08.
9-15 For the hypothesis test in Exercise 9-8, H0: mu = 175 millimeters, H1: mu > 175
millimeters, with n = 10 and a critical value of xbar = 185 millimeters, the probability of
accepting H0 (Type II error) can be calculated for various values of the true mean, mu.
The operating characteristic (OC) curve can be constructed by plotting the probability of
accepting H0 (1 - alpha) on the y-axis and the true mean, mu, on the x-axis for the given
values of mu.
Steps to calculate probability of accepting H0 for each value of mu:
1. Find the standard deviation of the sample mean, sigma/sqrt(n), where sigma = 20
millimeters (standard deviation of foam height) and n = 10.
2. For each value of mu, find the corresponding z-score using the formula z = (xbar mu)/(sigma/sqrt(n)).
3. Using the z-score, find the probability of accepting H0 using a standard normal table.
For each value of mu, repeat these steps and plot the resulting probabilities on the y-axis
versus the corresponding mu value on the x-axis to construct the OC curve.
9-16 The power function of a hypothesis test is the probability of rejecting the null
hypothesis (H0) when the alternative hypothesis (H1) is true. In other words, it is the
probability of correctly detecting a difference when there is one.
The power function of the test in Exercise 9-8 can be calculated as follows:
For each value of the true mean (mu) from 178 to 199, calculate the z-score using the
formula: z = (xbar - mu) / (standard deviation / square root of sample size)
where xbar is the sample mean, mu is the true mean, standard deviation is 20 millimeters,
and sample size is 10.
Next, use the z-score to find the corresponding p-value using a standard normal
distribution table. The p-value represents the probability of observing a sample mean as
extreme or more extreme than xbar, assuming H0 is true.
Finally, the power function can be calculated as 1 - p-value. This represents the
probability of rejecting H0 when the true mean is equal to the value of mu.
The resulting values of the power function can be plotted against the values of the true
mean (mu) to create a power function plot. The plot will show how the power of the test
changes as the true mean deviates from the null hypothesis.
9-17 (a) The null hypothesis H0: p=0.6 and the alternative hypothesis H1: p>0.6, where p
is the proportion of voters who favor the use of oxygenated fuels. If exactly 60% of the
voters favor the use of these fuels, then the mean number of voters who favor the use of
these fuels is 500_0.6 = 300, and the standard deviation is sqrt(500_0.6*0.4) = 12.04.
Using the normal approximation to the binomial distribution, we can calculate the
probability of observing more than 400 voters who favor the use of these fuels as:
P(X > 400 | p=0.6) = 1 - P(X <= 400 | p=0.6)
Where X is the number of voters who favor the use of these fuels, and P(X <= 400 |
p=0.6) can be calculated using the cumulative distribution function (CDF) of the normal
distribution:
P(X <= 400 | p=0.6) = CDF(Z), where Z = (400 - 300) / 12.04 = 1.66
Therefore,
P(X > 400 | p=0.6) = 1 - CDF(1.66)
Using a standard normal table, the CDF(1.66) can be found, which gives us the type I
error probability alpha:
alpha = 1 - CDF(1.66)
(b) The type II error probability beta is the probability of accepting the null hypothesis H0
when the true proportion of voters who favor the use of oxygenated fuels is 0.75. The
mean number of voters who favor the use of these fuels under H1 is 500_0.75 = 375, and
the standard deviation is sqrt(500_0.75*0.25) = 12.5.
Using the normal approximation to the binomial, we can calculate the probability of
observing less than or equal to 400 voters who favor the use of these fuels as:
beta = P(X <= 400 | p=0.75) = CDF((400 - 375) / 12.5)
The value of beta can be found using a standard normal table.
9-18 a) The type I error probability is the probability of rejecting the null hypothesis when
it is true, i.e., the probability of concluding that p > 0.3 when p = 0.3. To calculate this, we
use the cumulative distribution function (CDF) of the binomial distribution with
parameters n = 10 and p = 0.3. The type I error probability is given by:
P(X <= 1 | p = 0.3) = 1 - P(X > 1 | p = 0.3) = 1 - P(X >= 2 | p = 0.3) = 1 - CDF(2, 10, 0.3) = 1
- 0.2955
(b) The type II error probability is the probability of accepting the null hypothesis when it
is false, i.e., the probability of not rejecting the null hypothesis when p = 0.2. This is given
by:
P(X <= 1 | p = 0.2) = CDF(1, 10, 0.2) = 0.125
(c) The power of the test is 1 - beta, where beta is the type II error probability. Hence, the
power of the test is:
power = 1 - beta = 1 - 0.125 = 0.875
9-19 a) The type I error probability for this procedure can be found by computing the
probability of obtaining a sample that falls outside the acceptance region (4 to 8
graduates) when the true proportion is p = 0.4.
Since the sample size is small (n = 15), we will use the normal approximation to the
binomial distribution. Let p̂ be the sample proportion of graduates. Then, the standard
deviation of p̂ is sqrt(p(1-p)/n), where p = 0.4.
The critical region is defined by p̂ < 4/15 or p̂ > 8/15. To find the type I error probability,
we need to find the area under the standard normal distribution curve to the left of z =
(4/15 - 0.4)/(sqrt(0.4(1-0.4)/15)) and to the right of z = (8/15 - 0.4)/(sqrt(0.4(1-0.4)/15)).
Using a standard normal table, we find that the area to the left of z = -1.47 is 0.0759, and
the area to the right of z = 1.47 is 0.0759. So, the type I error probability is alpha = 2 *
0.0759 = 0.1518.
(b) To find the type II error probability, we need to find the probability of obtaining a
sample that falls within the acceptance region (4 to 8 graduates) when the true
proportion is p = 0.2.
The standard deviation of p̂ is still sqrt(p(1-p)/n), where p = 0.2. We need to find the area
under the standard normal distribution curve between z = (4/15 - 0.2)/(sqrt(0.2(10.2)/15)) and z = (8/15 - 0.2)/(sqrt(0.2(1-0.2)/15)).
Using a standard normal table, we find that the area between z = -0.69 and z = 1.08 is
0.8291. So, the type II error probability is beta = 1 - 0.8291 = 0.1709.
9-20 (a) To test the hypothesis, we can use a one-sample z-test with the null hypothesis
H0: μ = 100 and the alternative hypothesis H1: μ > 100.
The z-score for the sample mean of 98°F is calculated as follows:
z = (98 - 100) / (2 / sqrt(9)) = -1.7888543819998317
Since the z-score is negative, it falls in the left tail of the standard normal distribution. We
can use a standard normal table to find the area to the left of the z-score, which is the Pvalue.
The P-value is equal to 0.0404.
Since the P-value is less than the significance level α = 0.05, we reject the null hypothesis
and conclude that the water temperature is not acceptable.
(b) The P-value for the test is 0.0404.
(c) To find the probability of accepting the null hypothesis at α = 0.05 when the true
mean temperature is 104°F, we need to calculate the z-score for the sample mean in this
scenario.
z = (104 - 100) / (2 / sqrt(9)) = 1.7888543819998317
The area to the right of the z-score is 0.0404, which is the probability of accepting the
null hypothesis (H0: μ = 100) at α = 0.05 when the true mean temperature is 104°F.
revise 8-9 3rd editio for 9-21
To find the 95% two-sided confidence interval on the true mean yield, we will use the
formula for a t-confidence interval for the mean:
x̄ ± t* (s/√n)
where x̄ is the sample mean, s is the sample standard deviation, n is the sample size, and
t is the t-critical value for the desired level of confidence and degrees of freedom (df = n1).
For a 95% confidence interval, with df = 5-1 = 4, the t-critical value from a t-distribution
table is 2.776.
We can calculate the sample mean and standard deviation as follows: x̄ = (91.6 + 88.75 +
90.8 + 89.95 + 91.3) / 5 = 90.724 s = sqrt(((91.6 - x̄)^2 + (88.75 - x̄)^2 + (90.8 - x̄)^2 +
(89.95 - x̄)^2 + (91.3 - x̄)^2) / (5-1)) = 0.7944
Finally, substituting the values in the formula, we have:
x̄ ± t (s/√n) = 90.724 ± 2.776 (0.7944 / sqrt(5)) = 90.724 ± 2.776 * 0.4115 = (89.05, 92.4)
So the 95% two-sided confidence interval for the true mean yield is (89.05, 92.4). This
means we can be 95% confident that the true mean yield of the chemical process lies
between 89.05% and 92.4%.
9-21 (a) To test the hypothesis that the mean yield is not 90%, we will perform a onesample t-test. The null hypothesis is H0: μ = 90% and the alternative hypothesis is H1: μ
≠ 90%. The test statistic t is calculated as follows:
t = (x̄ - μ) / (s / √n) = (91.1 - 90) / (3 / √5) = 2.67
The degrees of freedom for this test are n - 1 = 4. Using a t-distribution table with 4
degrees of freedom and a two-tailed test with alpha = 0.05, the critical values are ±2.776.
Since the calculated t-value of 2.67 is not within the critical region, we fail to reject the
null hypothesis.
(b) The P-value is the probability of obtaining a test statistic as extreme or more extreme
than the one we calculated under the null hypothesis. The P-value can be calculated
using a t-distribution table or using software. With a calculated t-value of 2.67 and 4
degrees of freedom, the P-value is approximately 0.06.
(c) To detect a true mean yield of 85% with probability 0.95, we need to find the sample
size required to achieve a desired power of 0.95. The formula for power is given by (1 - β),
where β is the type II error probability. To solve for n, we need to find the t-value
corresponding to a power of 0.95 with 4 degrees of freedom (since the sample size is
unknown) and a two-tailed test with alpha = 0.05. Then, we can use the formula for tvalue given above to solve for n.
(d) The type II error probability β is the probability of failing to reject the null hypothesis
when it is false. To find β for a true mean yield of 92%, we need to find the t-value that
corresponds to a type II error probability of β with 4 degrees of freedom and a two-tailed
test with alpha = 0.05. Then, we can use the formula for t-value given above to see if the
calculated t-value is within the critical region. If it is not within the critical region, we
would fail to reject the null hypothesis, and the type II error probability would be β.
(e) The decision made in part (c) is not directly comparable to the 95% CI on mean yield
that was constructed in Exercise 8-7. The 95% CI gives an interval estimate of the
population mean yield with a certain degree of confidence, while the hypothesis test in
part (c) gives a decision about the population mean yield based on a single sample.
However, if the 95% CI does not contain the value of 90% (the null hypothesis), this
would suggest evidence against the null hypothesis, and we would reject the null
hypothesis in a hypothesis test with alpha = 0.05.
9-22 a) To test the hypothesis H0: μ = 3 vs. H1: μ ≠ 3 with a significance level of α =
0.05, we need to find the critical value of the test statistic. Since we have the population
standard deviation (σ = 0.9) and the sample size is large enough (n = 15), we can use the
z-test statistic:
z = (x̄ - μ) / (σ/√n) = (2.78 - 3) / (0.9/√15) = -2.13
Using a standard normal table, we can find the p-value for this test statistic: p-value = 2
P(Z < -2.13) = 2 0.0156 = 0.0312
Since the p-value is greater than the significance level (α = 0.05), we fail to reject the null
hypothesis. We do not have enough evidence to suggest that the mean wear is different
from 3.
(b) The power of the test is defined as the probability of rejecting the null hypothesis
when it is false. The power of the test when μ = 3.25 can be calculated as:
z = (x̄ - μ) / (σ/√n) = (2.78 - 3.25) / (0.9/√15) = -3.33
p-value = 2 P(Z < -3.33) = 2 0.0005 = 0.001
Since the p-value is smaller than the significance level (α = 0.05), we reject the null
hypothesis. The power of the test is the probability of making this correct decision and is
equal to 1 - β, where β is the type II error probability. Therefore, the power of the test
when μ = 3.25 is 1 - β = 1 - 0.0005 = 0.999.
(c) To determine the required sample size to detect a true mean of 3.75 with a power of
0.9, we can use the formula for the sample size of a one-sided z-test:
n = (z1 - z2)^2 * σ^2 / (μ1 - μ0)^2
where z1 and z2 are the critical values for the desired power and significance level, μ1 is
the true mean, and μ0 is the null hypothesis mean. Solving for n, we have:
n = (2.33 - 1.64)^2 * 0.9^2 / (3.75 - 3)^2 = 83.3
Therefore, a sample size of n = 84 is required to detect a true mean of 3.75 with a power
of 0.9 at the α = 0.05 significance level.
9-23 a) To test H0: μ = 155 versus H0: μ ≠ 155 using α = 0.01, we can use a one-sided ztest. The test statistic is calculated as:
z = (x̄ - μ) / (σ / √n) = (154.2 - 155) / (1.5 / √10) = -0.67
Since the test is one-tailed, the critical value of z at α = 0.01 is 2.33. Because the
calculated z value of -0.67 is not greater than the critical value, we fail to reject the null
hypothesis.
b) The P-value for this test is the probability of observing a test statistic as extreme or
more extreme than the one calculated, given that the null hypothesis is true. In this case,
the P-value is calculated as the area under the standard normal curve to the left of -0.67.
Using a standard normal table or calculator, the P-value can be calculated to be
approximately 0.25.
c) The β-error is the probability of failing to reject the null hypothesis when the
alternative hypothesis is true. In this case, β is the probability of failing to reject H0: μ =
155 when μ = 150. The β-error is calculated using a z-test, with the test statistic:
z = (x̄ - μ) / (σ / √n) = (154.2 - 150) / (1.5 / √10) = 1.6
Using a standard normal table or calculator, β can be calculated to be approximately 0.06.
d) To find the value of n required to get β < 0.1 when μ = 150, we can use the equation for
the test statistic and solve for n:
n = (z * (σ / √n) / (μ - x̄))^2
Substituting the values for z = 1.96 (the critical value at α = 0.01 for a two-tailed test), σ =
1.5, μ = 150, and x̄ = 154.2, we get:
n = (1.96 * (1.5 / √n))^2 / (150 - 154.2)^2
Solving for n, we get n = approximately 37.
9-24 a) To test the claim that battery life exceeds 40 hours, we can perform a one-tailed
test with the null hypothesis H0: μ = 40 and alternative hypothesis H1: μ > 40, where μ is
the true mean life of the battery.
Using α = 0.05, the critical value for a one-tailed test with 9 degrees of freedom (df = n 1) can be found using a t-distribution table. The critical value is 1.833.
We calculate the t-statistic as follows:
t = (x̄ - μ) / (σ / √n) = (40.5 - 40) / (1.25 / √10) = 4
Since 4 > 1.833, we reject the null hypothesis H0 and conclude that there is evidence to
support the claim that battery life exceeds 40 hours.
(b) The P-value is the probability of observing a test statistic as extreme as the one we
obtained, or more extreme, assuming the null hypothesis is true. The P-value can be
calculated as:
P-value = P(t > 4, df = 9) = 1 - P(t < 4, df = 9)
Using a t-distribution table, the P-value is approximately 0.0002, which is much less than
α = 0.05.
(c) The β-error is the probability of failing to reject the null hypothesis when the
alternative hypothesis is true. If the true mean life is 42 hours, then the β-error is
calculated as follows:
β = P(t < 1.833, df = 9 | μ = 42) = P(t > -1.833, df = 9 | μ = 42)
Using a t-distribution table, the β-error is approximately 0.08.
(d) To ensure that β does not exceed 0.10 when the true mean life is 44 hours, we need
to calculate the minimum sample size required.
Let's call the minimum sample size n. We can use the equation for t-statistic to solve for
n:
t = (x̄ - μ) / (σ / √n) = 1.833
Rearranging, we get:
n = (tσ / √(x̄ - μ))^2
Substituting the known values:
n = (1.833 * 1.25 / √(44 - 40.5))^2 = (2.291)^2 = 5.288 ≈ 6
Therefore, a sample size of n = 6 would be required to ensure that β does not exceed
0.10.
(e) To answer the question in part (a) by calculating an appropriate confidence bound on
life, we can use a t-confidence interval. A 95% t-confidence interval for the mean life of
the battery is calculated as follows:
CI = x̄ ± t (σ / √n) = 40.5 ± 1.833 (1.25 / √10) = (37.77, 43.23)
Since the confidence interval does not contain 40, we conclude that there is evidence to
support the claim that battery life exceeds 40 hours.
9-25 a) Testing the hypothesis:
H0: μ = 3500 psi (the mean tensile strength is equal to 3500 psi) Ha: μ ≠ 3500 psi (the
mean tensile strength is not equal to 3500 psi)
Significance level: α = 0.01
We can use a two-sample t-test to test this hypothesis:
t = (x̄ - μ) / (s / √n)
Where x̄ is the sample mean (3250 psi), μ is the population mean (3500 psi), s is the
sample standard deviation (not known), and n is the sample size (12).
Since we don't know the population standard deviation σ, we need to estimate it using
the sample standard deviation:
s = sqrt[(sum(x_i - x̄)^2) / (n - 1)]
where x_i are the observations.
We can calculate the t-statistic:
t = (3250 - 3500) / (s / sqrt(12))
We don't have the exact value for s, so we'll use the standard deviation estimate 60 psi.
t = (3250 - 3500) / (60 / sqrt(12)) = -5.16
We can use a t-distribution table with 11 degrees of freedom (n - 1 = 12 - 1 = 11) to find
the p-value:
p-value = P(t < -5.16) = 0.000079
Since the p-value is less than the significance level (0.000079 < 0.01), we reject the null
hypothesis. There is evidence to suggest that the mean tensile strength is not equal to
3500 psi.
(b) Smallest level of significance at which you would be willing to reject the null
hypothesis:
To find the smallest level of significance at which you would be willing to reject the null
hypothesis, we can find the critical value at which the p-value becomes greater than the
significance level.
For example, at α = 0.05, the critical value would be:
t_critical = t_distribution_table(0.05, 11 degrees of freedom) = -1.833
Since the calculated t-statistic (-5.16) is less than the critical value (-1.833), we would still
reject the null hypothesis even at a significance level of 0.05.
(c) Confidence Interval:
To answer the question in part (a) with a two-sided confidence interval, we can calculate
a 95% confidence interval (α = 0.05) for the mean tensile strength:
CI = x̄ ± t_critical * (s / sqrt(n))
Using the same standard deviation estimate of 60 psi and the same sample mean x̄ =
3250 psi, we get:
CI = 3250 ± t_critical (60 / sqrt(12)) = 3250 ± 1.833 (60 / sqrt(12)) = (3157, 3343)
The 95% confidence interval for the mean tensile strength is (3157, 3343) psi. Since 3500
psi is not within the confidence interval, we have evidence to suggest that the mean
tensile strength is not equal to 3500 psi.
9-26 To solve this problem, we need to find the sample size required to detect a true
mean strength of 3500 psi with a power of 0.8. Let's use the following notation:
μ0: null hypothesis mean strength = 3500 psi
μA: alternative hypothesis mean strength ≠ 3500 psi
α: level of significance
β: type II error probability
n: sample size
We know that:
Power = 1 - β, where Power is the probability of correctly rejecting the null
hypothesis if the alternative is true.
Power = 0.8 (the desired value)
α = 0.01 (from Exercise 9-25)
σ = 60 psi (known standard deviation of tensile strength)
We want to find the minimum sample size (n) that will give us a power of 0.8. To do this,
we can use the formula for the sample size needed to achieve a certain level of power:
n = (z1 - z2)^2 * σ^2 / (μA - μ0)^2
where z1 and z2 are the Z-scores corresponding to α/2 and β, respectively. To find z1 and
z2, we can use a Z-table or use a statistical software package.
Let's use a Z-table:
For α/2 = 0.005, z1 = 2.58
For β = 0.2, z2 = 0.84
Substituting the values in the formula:
n = (2.58 - 0.84)^2 * 60^2 / (μA - 3500)^2
Since μA is unknown, we can choose a value that makes the calculation easier. For
example, let's choose μA = 3700 psi. Then:
n = (2.58 - 0.84)^2 * 60^2 / (3700 - 3500)^2 = 49.44
Rounding up to the nearest integer, the sample size required to detect a true mean
strength of 3500 psi with power 0.8 is n = 50.
9-27 a) The null hypothesis H0: μ = 100 and alternative hypothesis H1: μ ≠ 100 are tested
using a one-sample z-test. The test statistic is calculated as follows:
z = (x̄ - μ) / (σ/√n) = (102.2 - 100) / (4/√8) = 2.8
Since we have a two-tailed test, the critical value at α = 0.05 is ±1.96. Since z = 2.8 >
1.96, we reject the null hypothesis H0.
(b) The power of the test can be calculated as 1 - β, where β is the probability of failing to
reject the null hypothesis when it is false. If the true mean speed is μ = 95, the test
statistic becomes:
z = (102.2 - 95) / (4/√8) = 5.55
Since z > 1.96, the null hypothesis is rejected, and the power of the test is 1 - β = 1.
(c) To find the sample size required to detect a true mean speed as low as 95 with a
power of at least 0.85, we use the formula for the sample size n:
n = (z1-α/2 + z1-β)^2 * σ^2 / (μ1 - μ0)^2
where z1-α/2 = 1.96, z1-β = norm.ppf(1-0.15), μ1 = 102.2, and μ0 = 95.
n = (1.96 + norm.ppf(1-0.15))^2 * (4^2) / (102.2 - 95)^2 n = 44.7
The sample size required is approximately 45.
(d) To answer the question in part (a) with a one-sided confidence bound, we can
calculate a one-sided upper confidence bound on the mean speed. The confidence
bound can be calculated as follows:
x̄ + zα/2 σ/√n = 102.2 + 1.96 4/√8 = 106.82
The one-sided upper confidence bound on the mean speed is 106.82 meters per second,
which is higher than 100 meters per second. This suggests that there is evidence to
support the claim that the mean speed exceeds 100 meters per second.
9-28 a) To test the hypotheses H0: mu = 1.5 versus H1: mu ≠ 1.5 using alpha = 0.01, we
can use a Z-test for the mean. The test statistic Z is calculated as Z = (x̄ - μ) / (σ / √n),
where x̄ is the sample mean, μ is the population mean, σ is the population standard
deviation, and n is the sample size. In this case, Z = (1.4975 - 1.5) / (0.01 / √25) = -2.5.
Since Z < -2.576, which is the critical value for alpha = 0.01 with 24 degrees of freedom,
we reject the null hypothesis H0.
b) To compute the power of the test if the true mean diameter is 1.495 inches, we need
to find the probability of rejecting the null hypothesis H0 if the true mean is 1.495 inches.
The test statistic Z is Z = (1.4975 - 1.495) / (0.01 / √25) = 1. The power of the test is the
complement of the probability of a type II error, which is the probability of failing to reject
H0 when H0 is false. The power of the test can be calculated using a Z-distribution table
or a software package.
c) To find the sample size required to detect a true mean diameter as low as 1.495 inches
if we wanted the power of the test to be at least 0.9, we can use the formula for sample
size n = (Z1-α/2 + Z1-β)^2 σ^2 / (μ0 - μ1)^2, where Z1-α/2 and Z1-β are the critical values of the Zdistribution corresponding to alpha/2 and beta, respectively. Solving for n, we get n = (2.576 + 1.282)^2
0.01^2 / (1.5 - 1.495)^2 = (3.858)^2 * 0.0001 / 0.005^2 = approximately 65.
d) To answer the question in part (a) by constructing a two-sided confidence interval on
the mean diameter, we can use the formula for the confidence interval as x̄ ± Zα/2 (σ / √n),
where x̄ is the sample mean, Zα/2 is the critical value of the Z-distribution corresponding to alpha/2, and
n is the sample size. In this case, the 95% two-sided confidence interval on the mean diameter is 1.4975 ±
2.576 (0.01 / √25) = [1.4867, 1.5083]. Since the interval does not contain the value 1.5, we
reject the null hypothesis H0 and conclude that the true mean diameter is different from
1.5.
9-29 a) To test the hypotheses H0: mu = 1.5 versus H1: mu ≠ 1.5 using alpha = 0.01, we
can use a Z-test for the mean. The test statistic Z is calculated as Z = (x̄ - μ) / (σ / √n),
where x̄ is the sample mean, μ is the population mean, σ is the population standard
deviation, and n is the sample size. In this case, Z = (1.4975 - 1.5) / (0.01 / √25) = -2.5.
Since Z < -2.576, which is the critical value for alpha = 0.01 with 24 degrees of freedom,
we reject the null hypothesis H0.
b) To compute the power of the test if the true mean diameter is 1.495 inches, we need
to find the probability of rejecting the null hypothesis H0 if the true mean is 1.495 inches.
The test statistic Z is Z = (1.4975 - 1.495) / (0.01 / √25) = 1. The power of the test is the
complement of the probability of a type II error, which is the probability of failing to reject
H0 when H0 is false. The power of the test can be calculated using a Z-distribution table
or a software package.
c) To find the sample size required to detect a true mean diameter as low as 1.495 inches
if we wanted the power of the test to be at least 0.9, we can use the formula for sample
size n = (Z1-α/2 + Z1-β)^2 σ^2 / (μ0 - μ1)^2, where Z1-α/2 and Z1-β are the critical values of the Zdistribution corresponding to alpha/2 and beta, respectively. Solving for n, we get n = (2.576 + 1.282)^2
0.01^2 / (1.5 - 1.495)^2 = (3.858)^2 * 0.0001 / 0.005^2 = approximately 65.
d) To answer the question in part (a) by constructing a two-sided confidence interval on
the mean diameter, we can use the formula for the confidence interval as x̄ ± Zα/2 (σ / √n),
where x̄ is the sample mean, Zα/2 is the critical value of the Z-distribution corresponding to alpha/2, and
n is the sample size. In this case, the 95% two-sided confidence interval on the mean diameter is 1.4975 ±
2.576 (0.01 / √25) = [1.4867, 1.5083]. Since the interval does not contain the value 1.5, we
reject the null hypothesis H0 and conclude that the true mean diameter is different from
1.5.
9-30 CD