H0: μ ≥ 20 (claim)
Ha: μ < 20
10-6 Confidence Intervals and Hypothesis Testing
1. LUNCH A sample of 145 high school seniors was
asked how many times they go out for lunch per
week. The mean number of times was 2.4 with a
standard deviation of 0.7. Use a 90% confidence
level to calculate the maximum error of estimate.
SOLUTION: E
≈ 0.096
Maximum Error of
Estimate
z = 1.645, s = 0.7,
and n = 145
Simplify.
The maximum error of estimate is approximately
0.096.
Identify the null and alternative hypotheses for
each statement. Then identify the statement
that represents the claim.
3. Lori thinks it takes a fast-food restaurant less than 2
minutes to serve her meal after she orders it.
SOLUTION: less than 2 minutes: μ < 2,
not less than 2 minutes: μ ≥ 2
The claim is μ < 2, and it is the alternative hypothesis
because it does not include equality. The null
hypothesis is μ ≥ 2, which is the complement of μ <
2.
H0: μ ≥ 2
Ha: μ < 2 (claim)
5. Mrs. Hart’s review game takes at least 20 minutes to
complete.
CCSS REASONING Identify the hypotheses and claim, decide whether to reject the null
hypothesis, and make a conclusion about the
claim.
7. COMPACT DISCS A manufacturer of blank
compact discs claims that each disc can hold at least
84 minutes of music. Using a sample of 219 compact
discs, Cayla calculated a mean time of 84.1 minutes
per disc with a standard deviation of 1.9 minutes.
Test the hypothesis at 5% significance.
SOLUTION: State the claim and hypothesis.
H0: μ ≥ 84 (claim)
Ha: μ < 84
Determine the critical region.
The alternative hypothesis is μ < 84, so this is a lefttailed test. We are testing at 5% significance, so we
need to identify the z-value that corresponds with the
lower 5% of the distribution. Keystrokes: ` ú
InvNorm .05 e
The critical region is z < –1.645.
Calculate the z-statistic for the sample data.
z=
SOLUTION: at least 20 minutes: μ ≥ 20
less than 20 minutes: μ < 20
The claim is μ ≥ 20, and it is the null hypothesis because it does include equality. The alternative
hypothesis is μ < 20, which is the complement of μ ≥ 20.
H0: μ ≥ 20 (claim)
Ha: μ < 20
CCSS REASONING Identify the hypotheses and claim, decide whether to reject the null
hypothesis, and make a conclusion about the
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claim.
7. COMPACT DISCS A manufacturer of blank
compact discs claims that each disc can hold at least
Formula for z-statistic
=
= 84.1, µ = 84, s = 1.9,
and n = 219
≈ 0.78
Decide whether to reject the null hypothesis.
H0 is not rejected because the z-statistic for the
sample does not fall within the critical region.
Make a conclusion about the claim.
There is not enough evidence to reject the claim that
the compact discs can hold at least 84 minutes of
music.
9. MUSIC A sample of 76 albums had a mean runPage 1
time of 61.3 minutes with a standard deviation of 5.2
minutes. Use a 95% confidence level to calculate the
maximum error of estimate.
6.
Make a conclusion about the claim.
There is not enough evidence to reject the claim that
the compact discs can hold at least 84 minutes of
10-6music.
Confidence Intervals and Hypothesis Testing
9. MUSIC A sample of 76 albums had a mean run
time of 61.3 minutes with a standard deviation of 5.2
minutes. Use a 95% confidence level to calculate the
maximum error of estimate.
SOLUTION: The sample size n is 76 and the standard deviation s
is 5.2. The mean is not needed to calculate the
maximum error of estimate. The z-values which
correspond to 95% significance are ±1.96.
H0: μ ≥ 6 (claim)
Ha: μ < 6
13. A company advertisement states that it takes no
more than 2 hours to paint a 200-square-foot room.
SOLUTION: no more than 2 hours: μ ≤ 2
more than 2 hours: μ > 2
The claim is μ ≤ 2, and it is the null hypothesis because it does include equality. The alternative
hypothesis is μ > 2, which is the complement of μ ≤ 2.
H0: μ ≤ 2 (claim)
Ha: μ > 2
The maximum error of estimate is about 1.17.
Identify the null and alternative hypotheses for
each statement. Then identify the statement
that represents the claim.
11. Julian sends at least six text messages to his best
friend every day.
SOLUTION: at least 6 texts: μ ≥ 6
less than 6 texts: μ < 6
The claim is μ ≥ 6, and it is the null hypothesis because it does include equality. The alternative
hypothesis is μ < 6, which is the complement of μ ≥ 6.
H0: μ ≥ 6 (claim)
Ha: μ < 6
Identify the hypotheses and claim, decide
whether to reject the null hypothesis, and make
a conclusion about the claim.
15. PIZZA A pizza chain promises a delivery time of
less than 30 minutes. Using a sample of 38 deliveries,
Chelsea calculated a mean delivery time of 29.6
minutes with a standard deviation of 3.9 minutes.
Test the hypothesis at 1% significance.
SOLUTION: State the claim and hypothesis.
H0: μ ≥ 30
Ha: μ < 30 (claim)
Determine the critical region.
The alternative hypothesis is μ < 30, so this is a lefttailed test. We are testing at 1% significance, so we
need to identify the z-value that corresponds with the
lower 1% of the distribution. Keystrokes: ` ú
InvNorm .01 e
13. A company advertisement states that it takes no
more than 2 hours to paint a 200-square-foot room.
SOLUTION: no more than 2 hours: μ ≤ 2
more than 2 hours: μ > 2
The claim is μ ≤ 2, and it is the null hypothesis because it does include equality. The alternative
hypothesis is μ > 2, which is the complement of μ ≤ 2.
H0: μ ≤ 2 (claim)
Ha: μManual
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eSolutions
Identify the hypotheses and claim, decide
The critical region is z < –2.326.
Calculate the z-statistic for the sample data.
z=
=
Formula for z-statisticPage 2
= 29.6, µ = 30, s = 3.9,
2.
H0: μ ≤ 2 (claim)
10-6 Confidence Intervals and Hypothesis Testing
Ha: μ > 2
Identify the hypotheses and claim, decide
whether to reject the null hypothesis, and make
a conclusion about the claim.
15. PIZZA A pizza chain promises a delivery time of
less than 30 minutes. Using a sample of 38 deliveries,
Chelsea calculated a mean delivery time of 29.6
minutes with a standard deviation of 3.9 minutes.
Test the hypothesis at 1% significance.
SOLUTION: State the claim and hypothesis.
Make a conclusion about the claim.
There is not enough evidence to support the claim of
a delivery time of less than 30 minutes.
17. DECISION MAKING The number of peaches in
40 random cans is shown below. Should the
manufacturer place a label on the can promising
exactly 12 peaches in every can? Explain your
reasoning.
13, 14, 13, 14, 12, 12, 12, 11, 15, 12, 13, 13, 14, 13, 14,
12, 15, 11, 11, 14,
13, 14, 14, 13, 12, 12, 12, 12, 13, 13, 11, 14, 14, 13, 14,
13, 13, 14, 12, 12
SOLUTION: Find the mean and standard deviation of the sample
data.
Enter the data as L1 in a graphing calculator and
then select STAT CALC 1 to obtain the statistics.
H0: μ ≥ 30
Ha: μ < 30 (claim)
Determine the critical region.
The alternative hypothesis is μ < 30, so this is a lefttailed test. We are testing at 1% significance, so we
need to identify the z-value that corresponds with the
lower 1% of the distribution. Keystrokes: ` ú
InvNorm .01 e
The sample mean and standard deviation are 12.9
and 1.08, respectively.
State the claim and hypothesis.
H0: μ = 12 (claim)
The critical region is z < –2.326.
Calculate the z-statistic for the sample data.
z=
Formula for z-statistic
=
= 29.6, µ = 30, s = 3.9,
and n = 38
≈ –0.63
Ha: μ ≠ 12
Determine the critical region.
The alternative hypothesis is μ ≠ 12, so this is a twotailed test. Test at 10% significance. Identify the zvalue that corresponds with the upper and lower 5%
of the distribution. Keystrokes: ` ú InvNorm .05 e
and ` ú InvNorm .95 e.
Decide whether to reject the null hypothesis.
H0 is not rejected because the z-statistic for the
sample does not fall within the critical region.
Make a conclusion about the claim.
There is not enough evidence to support the claim of
a delivery time of less than 30 minutes.
MAKING
The
17. DECISION
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number of peaches in
40 random cans is shown below. Should the
manufacturer place a label on the can promising
exactly 12 peaches in every can? Explain your
The critical region is –1.645 > z or z > 1.645.
Calculate the z-statistic for the sample data.
Page 3
z=
Formula for z-statistic
≈ 0.70
= 50
Simplify.
10-6 Confidence Intervals and Hypothesis Testing
The critical region is –1.645 > z or z > 1.645.
Use the maximum error of estimate to find the
confidence interval CI.
Calculate the z-statistic for the sample data.
CI =
z=
Formula for z-statistic
=
= 12.9, µ = 12, s =
1.08, and n = 40
≈ 5.27
Decide whether to reject the null hypothesis.
H0 is rejected because the z-statistic for the sample
±E
= 25 ± 0.70
The 90% confidence interval is 24.30 ≤ µ ≤ 25.70.
n = 100:
Find the maximum error of estimate.
E
≈ 0.49
Use the maximum error of estimate to find the
confidence interval CI.
CI =
a. GRAPHICAL Graph the 90% confidence
interval for n = 50, 100, and 250 on a number line.
b. ANALYTICAL How does the sample size affect
the confidence interval?
c. GRAPHICAL Graph the 90%, 95%, and 99%
confidence intervals for n = 150.
d. ANALYTICAL How does the confidence level
affect the confidence interval?
e. ANALYTICAL How does decreasing the size of
the confidence interval affect the accuracy of the
confidence interval?
Maximum Error of
Estimate
z = 1.645, s = 3, and
n = 100
Simplify.
19. MULTIPLE REPRESENTATIONS In this
problem, you will explore how the confidence interval
is affected by the sample size and the confidence
level. Consider a sample of data where
and s
=3.
= 25 and E = 0.70
falls within the critical region. It would also be
rejected at 5% and 1% significance.
Make a conclusion about the claim.
There is enough evidence to reject the claim that
each can contains exactly 12 peaches. Therefore, the
company should not put the claim on the label.
Confidence Interval for
Population Mean
±E
= 25 ± 0.49
Confidence Interval for
Population Mean
= 25 and E = 0.49
The 90% confidence interval is 24.51 ≤ µ ≤ 25.49.
n = 250:
Find the maximum error of estimate.
E
≈ 0.31
Maximum Error of
Estimate
z = 1.645, s = 3, and
n = 250
Simplify.
Use the maximum error of estimate to find the
confidence interval CI.
SOLUTION: a. n = 50:
Find the maximum error of estimate.
CI =
±E
= 25 ± 0.31
Confidence Interval for
Population Mean
= 25 and E = 0.31
E
≈ 0.70
Maximum Error of
Estimate
z = 1.645, s = 3, and n
= 50
Simplify.
The 90% confidence interval is 24.69 ≤ µ ≤ 25.31.
Use the maximum error of estimate to find the
confidence interval CI.
Confidence Interval for
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CI =
± E by Cognero
Population Mean
= 25 ± 0.70
= 25 and E = 0.70
b. Sample answer: With everything else held Page 4
constant, increasing the same size will decrease the
size of the confidence interval.
E
10-6 Confidence Intervals and Hypothesis Testing
≈ 0.63
Maximum Error of
Estimate
z = 2.576, s = 3, and
n = 150
Simplify.
b. Sample answer: With everything else held
constant, increasing the same size will decrease the
size of the confidence interval.
Use the maximum error of estimate to find the
confidence interval CI.
CI =
c. 90%:
Find the maximum error of estimate.
Maximum Error of
Estimate
z = 1.645, s = 3, and
n = 150
Simplify.
E
≈ 0.40
±E
= 25 ± 0.63
Confidence Interval for
Population Mean
= 25 and E = 0.63
The 99% confidence interval is 24.37 ≤ µ ≤ 25.63.
Use the maximum error of estimate to find the
confidence interval CI.
CI =
±E
= 25 ± 0.40
Confidence Interval for
Population Mean
= 25 and E = 0.40
The 90% confidence interval is 24.60 ≤ µ ≤ 25.40.
95%:
Find the maximum error of estimate.
Maximum Error of
Estimate
z = 1.96, s = 3, and n
= 150
Simplify.
E
≈ 0.48
Use the maximum error of estimate to find the
confidence interval CI.
CI =
±E
= 25 ± 0.48
d. Sample answer: With everything else held
constant, increasing the confidence level will
decrease the size of the confidence interval.
e. Expanding the confidence interval reduces the
accuracy of the estimate. So decreasing the size of
the confidence interval increases the accuracy of the
estimate.
21. CHALLENGE A 95% confidence interval for the
mean weight of a 20-ounce box of cereal was 19.932
≤ μ ≤ 20.008 with a sample standard deviation of 0.128 ounces. Determine the sample size that led to
this interval.
SOLUTION: The range of values for the mean weight is 19.932 to
20.008. Half of this range is the maximum error of
estimate.
Confidence Interval for
Population Mean
= 25 and E = 0.48
The 95% confidence interval is 24.52 ≤ µ ≤ 25.48.
Use the formula for calculating the maximum error
of estimate to determine the same size. The z-values
which correspond to 95% significance are ±1.96.
99%:
Find the maximum error of estimate.
Maximum Error of
Estimate
z = 2.576, s = 3, and
n = 150
Simplify.
E
≈ 0.63
Use the maximum error of estimate to find the
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confidence
CI.Cognero
CI
Confidence Interval for
Page 5
e. Expanding the confidence interval reduces the
accuracy of the estimate. So decreasing the size of
confidenceIntervals
interval increases
the accuracy
of the
10-6the
Confidence
and Hypothesis
Testing
estimate.
21. CHALLENGE A 95% confidence interval for the
mean weight of a 20-ounce box of cereal was 19.932
≤ μ ≤ 20.008 with a sample standard deviation of 0.128 ounces. Determine the sample size that led to
this interval.
used in a decision-making process?
SOLUTION: Sample answer: You can use a statistical test to help
you to determine the strength of your decision.
25. GEOMETRY In the graph below, line l passes
through the origin. What is the value of
?
SOLUTION: The range of values for the mean weight is 19.932 to
20.008. Half of this range is the maximum error of
estimate.
Use the formula for calculating the maximum error
of estimate to determine the same size. The z-values
which correspond to 95% significance are ±1.96.
A –4
B
C
D4
SOLUTION: The coordinates of the point O is (0, 0).
Find the slope of the line using the points (0, 0) and
(1, 4).
The slope of the line is
.
Find the slope of the line using the points (0, 0) and
(a, b).
23. WRITING IN MATH How can a statistical test be
used in a decision-making process?
SOLUTION: Sample answer: You can use a statistical test to help
you to determine the strength of your decision.
25. GEOMETRY In the graph below, line l passes
through the origin. What is the value of
?
But
Therefore,
.
Option C is the correct answer.
27. The Service Club at Jake’s school was founded 8
years ago. The number of members of the club by
year is shown in the table. Which linear equation best
models the data?
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A –4
A y = 1.4x
Therefore,
.
10-6 Confidence Intervals and Hypothesis Testing
Option C is the correct answer.
27. The Service Club at Jake’s school was founded 8
years ago. The number of members of the club by
year is shown in the table. Which linear equation best
models the data?
Option B is the correct answer.
29. HEALTH The heights of students at Madison High
School are normally distributed with a mean of 66
inches and a standard deviation of 2 inches. Of the
1080 students in the school, how many would you
expect to be less than 62 inches tall?
SOLUTION: Given
.
The probability that a randomly selected value in the
distribution is less than μ – 2σ, that is, 66 – 2(2) or
62.
P(x < 62) = 2% + 0.5% = 2.5%
A y
B y
C y
D y
= 1.4x
= 1.4x + 10.4
= 1.6x
= 1.6x + 11.1
So, 27 students expected to be less than 62 inches
tall.
SOLUTION: Keystrokes:
Press STAT ENTER 0 ENTER 2 ENTER 4 ENTER 6 ENTER 8 ENTER ► 1 1 ENTER 1 3 ENTER 1 5 ENTER 1 9 ENTER 2 2 ENTER STAT ► 4 ENTER .
Find a n for each geometric sequence.
31. =
, r = 3, n = 8
SOLUTION: 33. = 16, r = 0.5, n = 8
SOLUTION: Option B is the correct answer.
Find a 1.
29. HEALTH The heights of students at Madison High
School are normally distributed with a mean of 66
inches and a standard deviation of 2 inches. Of the
1080 students in the school, how many would you
expect to be less than 62 inches tall?
SOLUTION: Given
.
The probability that a randomly selected value in the
distribution is less than μ – 2σ, that is, 66 – 2(2) or
62.
Find a 8.
P(x < 62) = 2% + 0.5% = 2.5%
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So, 27
students
expected
to be
tall.
less than 62 inches
Write each equation in standard form. State
whether the graph of the equation is a parabola,
Page 7
circle, ellipse, or hyperbola. Then graph the
equation.
2
10-6 Confidence Intervals and Hypothesis Testing
Write each equation in standard form. State
whether the graph of the equation is a parabola,
circle, ellipse, or hyperbola. Then graph the
equation.
Write an equation in slope-intercept form for
each graph.
2
35. x = 8y
SOLUTION: 37. SOLUTION: The line passes trough the origin. Therefore, b = 0.
Find the slope of the line using the points (0, 0) and
(2.5, 2).
The equation is in standard form of a parabola.
Therefore, it represents a parabola.
Graph the equation.
The equation of the line is
.
Write an equation in slope-intercept form for
each graph.
39. SOLUTION: For all values of x the value of y is –4.
Therefore, the equation of the line is y = –4.
Find each missing measure. Round to the
nearest tenth, if necessary.
37. SOLUTION: The line passes trough the origin. Therefore, b = 0.
Find the slope of the line using the points (0, 0) and
(2.5, 2).
41. SOLUTION: Use the Pythagorean Theorem.
2
The equation of the line is
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.
a +b =c
2
2
Pythagorean Theorem
2
2
2
a = 7 and b = 24
Simplify.
Take the square root of
Page 8
each side.
Evaluate the square
root.
7 + 24 = c
625 = c2
=c
25 = c
39. SOLUTION: 10-6For
Confidence
Intervals
andofHypothesis
Testing
all values of
x the value
y is –4.
Therefore, the equation of the line is y = –4.
Find each missing measure. Round to the
nearest tenth, if necessary.
41. SOLUTION: Use the Pythagorean Theorem.
a +b =c
2
2
2
Pythagorean Theorem
2
2
2
a = 7 and b = 24
Simplify.
Take the square root of
each side.
Evaluate the square
root.
7 + 24 = c
625 = c2
=c
25 = c
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