1. (25%) FlyHigh Airlines determined that the distance traveled per aircraft per year is normally
distributed, with a mean of 60 thousand miles and a standard deviation of 10 thousand miles.
a. What proportion of aircrafts can be expected to travel between 44 and 54 thousand miles in a
year?
Known:
σ = 10
µ = 60; a = 44 in thousand miles; b = 54 in thousand miles
First find Z by using
𝑎−µ
)
𝜎
by using formula P(a < Z < b) => P(
F(
54−60
)
10
44−60
)
10
- F(
−6
𝑏−µ
)
𝜎
< Z < P(
𝑏−µ
)
𝜎
=> F(
- F(
𝑎−µ
)
𝜎
−16
= F( 10 ) - F( 10 ) = F(-0.6) – F(-1.6)
Apply F(-0.6) and F(-1.6) in Z table
The result is F(-0.6) = 0.27425 and F(-1.6) = 0.0548, therefore 0.27425 - 0.0548 = 0.21945
Proportion of aircrafts can be expected to travel between 44 and 54 thousand miles in a year is
0.21945
b. What percentage of aircrafts can be expected to travel either less than 25 or more than 70
thousand miles in a year?
σ = 10
µ = 60; a = 25 in thousand miles; b = 70 in thousand miles
P(a < X < b) => P(X < a) + P(X > b) = P(X < 25) + P(X > 70) => P(𝑧 <
P(𝑧 <
25−60
)
10
+ P(𝑧 >
70−60
)
10
= P(𝑧 <
−35
)
10
10
𝑎−µ
)
𝜎
+ P(𝑧 >
𝑏−µ
)
𝜎
+ P(𝑧 > 10) = P(𝑧 < −3.5) + P(𝑧 > 1)
=
Apply P(-3.5) and P(1) in Z table
The result is P(-3.5) = 0.0002 and P(1) = 0.84134, since the percentage of aircrafts can be expected to
travel more than 70 thousand miles in a year (end tail), so P(1) > 0.84134 = 1 - 0.84134 = 0.15866.
Back to formula P(X < 25) + P(X > 70) = P(𝑧 < −3.5) + P(𝑧 > 1) = P(-3.5) + (1-P(1)) = 0.0002 + (1 0.84134) = 0.0002 + 0.15866 = 0.15889 or 15.889%
Percentage of aircrafts can be expected to travel either less than 25 or more than 70 thousand
miles in a year = 15.889%
c. How many miles will be traveled by at least 70 percent of the aircrafts?
Since “at least” is more than 70%, or 30% from the lower tail of curve
70%
30%
P(X > Z) = 0.3, apply to Z-table, the closest value of 0.3000 is -0.52 therefore Z value is -0.52
Next, find X by using
X = 60 + (-0.52)*10 = 60 + (-5.2) = 54.8
Miles will be traveled by at least 70 percent of the aircrafts is 54.8 in thousand miles.
d. What are your answers to (a) through (c) if the standard deviation is 12 thousand miles?
d.a) What proportion of aircrafts can be expected to travel between 44 and 54 thousand
miles in a year?
Known:
σ = 12
µ = 60; a = 44 in thousand miles; b = 54 in thousand miles
First find Z by using
by using formula P(a < Z < b) => P(
F(
54−60
)
12
44−60
)
12
- F(
−6
𝑎−µ
)
𝜎
𝑏−µ
)
𝜎
< Z < P(
𝑏−µ
)
𝜎
=> F(
- F(
𝑎−µ
)
𝜎
−16
= F( 12 ) - F( 12 ) = F(-0.5) – F(-1.33)
Apply F(-0.5) and F(-1.33) in Z table
The result is F(-0.5) = 0.30854 and F(-1.6) = 0.09176, therefore 0.30854 - 0.09176 = 0.21408
Proportion of aircrafts can be expected to travel between 44 and 54 thousand miles in a year is
0.21408
d.b) What percentage of aircrafts can be expected to travel either less than 25 or more
than 70 thousand miles in a year?
σ = 12
µ = 60; a = 25 in thousand miles; b = 70 in thousand miles
P(a < X < b) => P(X < a) + P(X > b) = P(X < 25) + P(X > 70) => P(𝑧 <
P(𝑧 <
25−60
)
12
+ P(𝑧 >
70−60
)
12
= P(𝑧 <
−35
)
12
𝑎−µ
)
𝜎
+ P(𝑧 >
𝑏−µ
)
𝜎
=
10
+ P(𝑧 > 12) = P(𝑧 < −2.92) + P(𝑧 > 0.83)
Apply P(-2.92) and P(0.83) in Z table
The result is P(-2.92) = 0.0018 and P(0.83) = 0.79673, since the percentage of aircrafts can be
expected to travel more than 70 thousand miles in a year (end tail), so P(1) > 0.79673 = 1 - 0.79673 =
0.20327.
Back to formula P(X < 25) + P(X > 70) = P(𝑧 < −2.92) + P(𝑧 > 0.83) = P(-2.92) + (1-P(0.83)) =
0.0002 + (1 - 0.79673) = 0.0018 + 0.20327 = 0.20507 or 20.507%
Percentage of aircrafts can be expected to travel either less than 25 or more than 70 thousand
miles in a year = 20.507%
d.c) How many miles will be traveled by at least 70 percent of the aircrafts?
Since “at least” is more than 70%, or 30% from the lower tail of curve
70%
30%
P(X > Z) = 0.3, apply to Z-table, the closest value of 0.3000 is -0.52 therefore Z value is -0.52
Next, find X by using
X = 60 + (-0.52)*12 = 60 + (-6.24) = 53.76
Miles will be traveled by at least 70 percent of the aircrafts is 53.76 in thousand miles.
2. (25%) You plan to conduct a marketing experiment in which students will be asked to taste one
of two different brands of soft drink. Their task is to correctly identify the brand tasted. You select
a random sample of 190 students and assume that the students have no ability to distinguish
between the two brands (which means that the two brands are equally likely to be picked.)
a. What is the probability that between 50% and 60% of the identifications will be correct in the
sample?
Known:
n = 190
population proportion (π) = 0.5
Find standard error of proportion
0.5∗(1−0.5)
190
√
0.5∗(0.5)
190
=√
0.25
=
190
=√
√0.00132 = 0.0363
So σp = 0.0363
First find Z by using
by using formula P(a < Z < b) => P(
0.6−0.5
0.5−0.5
0.1
𝑎−µ
)
𝜎
𝑏−µ
)
𝜎
< Z < P(
𝑏−µ
)
𝜎
=> P(
𝑎−µ
)
𝜎
- P(
0
P( 0.0363 ) - P( 0.0363 ) = P(0.0363) - P(0.0363) = P(2.755) – P(0)
Apply P(2.755) and P(0) in Z table
The result is P(2.755) = 0.99711 and P(0) = 0.5, therefore 0.99711 - 0.5 = 0.49711
The probability that between 50% and 60% of the identifications will be correct in the sample is
0.49711
b. What are the symmetrical limits of the population percentage within which the sample
percentage is contained with a probability of 90%?
The probability is 90% that the sample percentage is contained between a and b:
P(a < 𝑝̂ < b) = 0.90
P( -z < Z < z) = 0.90
P(Z < -z) – P(Z > z) = 1 -0.90
2P(Z < -z) = 0.10
P(Z < -z) = 0.05
z = ± 1.645
By using formula
Subtitution:
X=π
µ = p = 0.5
σ = σp = 0.0363
z=
π−𝑝
𝜎𝑝
= z* σp = π − 𝑝 => π = z* σp + 𝑝
When π = a, and z = -1.645
= -1.645*0.0363 + 0.5 = -0.0597 + 0.5 = 0.4403 = 44.03% ≈ 44%
When π = b, and z = 1.645
= 1.645*0.0363 + 0.5 = 0.0597 + 0.5 = 0.5597 = 55.97% ≈ 56%
Therefore, the lower limit is 44% and the upper limit is 56%. The sample percentage will be
contained within 6% symmetrically around the population percentage 50%.
c. What is the probability that the sample percentage of correct identifications is greater than
65%?
The probability that the sample percentage of correct identifications is greater than 65%:
P(π > 0.65) = (z >
0.65−0.5
)=
0.0363
0.15
(z > 00363) = (z > 4.13), apply to z table
z = 0.0001
The probability that the sample percentage of correct identifications is greater than 65% is 0.0001
d. Which is more likely to occur: more than 58% correct identifications in the sample of 190 or
more than 54% correct identifications in a sample of 1,000? Explain.
Case 1:
n = 190
population proportion (π) = 0.5
Find standard error of proportion
0.5∗(1−0.5)
190
√
0.5∗(0.5)
190
=√
0.25
= √ 190 = √0.00132 = 0.0363
So σp = 0.0363
The probability that the sample percentage of correct identifications is greater than 58%:
P(π > 0.58) = (z >
Z = 0.0139
0.58−0.5
)=
0.0363
0.08
(z > 00363) = (z > 2.2), apply to z table
Case 2:
n = 1000
population proportion (π) = 0.5
Find standard error of proportion
0.5∗(1−0.5)
1000
√
0.5∗(0.5)
1000
=√
0.25
= √1000 = √0.00025 = 0.0158
So σp = 0.0158
The probability that the sample percentage of correct identifications is greater than 54%:
P(π > 0.54) = (z >
0.54−0.5
)=
0.0158
0.04
(z > 0.0158) = (z > 2.53), apply to z table
Z = 0.0057
Since Z for more than 58% correct identifications in the sample of 190 bigger than (>) Z for more
than 54% correct identifications in a sample of 1,000, therefore which is more likely to occur - more
than 58% correct identifications in the sample of 190
3. (25%) A study reports that the cost of repairing a hybrid vehicle is falling even while typical
repairs on conventional vehicles are getting more expensive. One of the most common hybrid
repairs, replacing the hybrid inverter assembly, had a mean repair cost of $1,357 in 2014. Suppose
a sample of 100 hybrid inverter assembly repairs completed in the last month was selected. The
sample mean repair cost was $1,280 with the sample standard deviation of $300.
a. Using the critical value approach, at the 0.05 level of significance, is there evidence that the
population mean repair cost is less than $1,357?
First, define the null hypothesis and alternative hypothesis
H0: µ ≥ $1,357
HA: µ < $1,357
Data provided:
N = 100
𝑋̅ = $1,280
S = $300
µ = $1,357
α= 0.05
Determine degree of freedom (df) by using formula N – 1 => 100 – 1 = 99.
This uses one tailed test alternative hypothesis (lower tailed test) since HA contain “<” sign
α = 0.05 = 5%, since one tailed test, we use 5% = 0.05
By using t table, with df = 99 and α = 0.05, the critical value = 1.6604
Since this is lower tailed test, so the result of critical value become negative, therefore the result is
-1.6604
Area of
rejection
Area of nonrejection
-1.6604
Therefore test criteria is H0 rejected if Tstat< -1.6604
We need to use T test since σ unknown by formula
=>
1280−1357
300
√100
=
−77
300
10
=
−77
30
= -2.567
Comparing test criteria to critical value, Tstat = -2.567 and critical value = -1.6604 so -2.567 < -1.6604,
it seems Tstat result fall in rejection region.
Decision is Reject H0; Interpretation: there is significant evidence that the population mean repair
cost is less than $1,357.
b. Perform the same test as (a) using the p-value approach.
First, define the null hypothesis and alternative hypothesis
H0: µ ≥ $1,357
HA: µ < $1,357
ρ-value ≤ α; H0 is rejected
ρ-value > α H0 is not rejected
Data provided:
N = 100
𝑋̅ = $1,280
S = $300
µ = $1,357
α= 0.05
We need to use T test since σ unknown by formula
=>
1280−1357
300
√100
=
−77
300
10
=
−77
30
= -2.567
Tstat = -2.567
Find probability of Tstat by using t-score table for -2.567 ≈ -2.57
probability of Tstat by using z-score table for -2.567 ≈ -2.57 is 0.0051
So ρ-value is 0.0051
Decision:
ρ-value = 0.0051 and α= 0.05
0.0051 < 0.05 therefore ρ-value < α; Decision is Reject H0; Interpretation: there is significant
evidence that the population mean repair cost is less than $1,357.
4. (25%) A recent study found that 61 children who watched a commercial for potato chips
featuring a long-standing sports celebrity endorser ate a mean of 38 grams of potato chips as
compared to a mean of 26 grams for another group of 51 children who watched a commercial for
an alternative food snack. Suppose that the sample standard deviation for the children who
watched the sports celebrity–endorsed potato chips commercial was 21.5 grams and the sample
standard deviation for the children who watched the alternative food snack commercial was 12.9
grams.
a. Assuming that the population variances are equal and = 0.05, is there evidence that the mean
amount of potato chips eaten was significantly higher for the children who watched the sports
celebrity–endorsed potato chips commercial?
First, define the null hypothesis and alternative hypothesis
H0: µ1 = µ2 or µ1 - µ2 = 0
HA: µ1 ≠ µ2
Find degree of freedom by using formula:
(n1 - 1)+(n2 - 1) = (61 - 1)+(51 - 1) = 60 + 50 = 110
So df is equal to 110
α = 0.05 level of significance, formula used is α/2; therefore 0.025
Find the critical value by using formula in Excel = TINV(α;df) = TINV(0.05;110) equal to 1.982
So the critical value for two tailed test are 1.982 and -1.982
Area of
rejection
Area of
rejection
Area of nonrejection
1.982
-1.982
Decision rule is:
Reject H0 if TSTAT > 1.982 or if TSTAT < -1.982; otherwise, do not reject H0.
𝑋̅
S
n
childern who watched a commercial
for potato chips featuring a longstanding sports celebrity endorser
children who watched a
commercial for an
alternative food snack
38
21.5
61
26
12.9
51
Find Tstat by using formula as follow:
̅̅̅
𝑋1 = 38
𝑆1 = 21.5
𝑛1 = 61
̅̅̅
𝑋2 = 26
𝑆2 = 12.9
𝑛2 = 51
The difference µ1 - µ2 comes from the null hypothesis. In this type of test, we assume no
difference in the population means, so we can substitute 0 for µ1 - µ2
Find 𝑆𝑝2 first =
(61−1)∗21.52 +(51−1)∗12.92
(61−1)+(51−1)
=
60∗462.25+50∗166.41
60+50
=
27735+8320.5
110
=
36055.5
110
= 327.777
Next is to find Tstat:
(38−26)−0
12
1 1 =
√327.777(61+51) √327.777(0.0164+0.0196)
12
3.435
=
12
√327.777(0.0164+0.0196)
=
12
√327.777∗0.036
=
12
√11.8
= 3.493
The result of Tstat is 3.493
Comparing to test criteria is Tstat > 1.982, it seems Tstat result fall in rejection region (3.493 > 1.982)
The decision is H0 is rejected. Interpretation: there is evidence that the mean amount of potato chips
eaten was significantly higher for the children who watched the sports celebrity–endorsed potato
chips commercial.
b. Assuming that the population variances are equal, construct a 95% confidence interval estimate
of the difference between the mean amount of potato chips eaten by children who watched the
sports celebrity–endorsed potato chips commercial and children who watched the alternative food
snack commercial.
Known:
̅̅̅
𝑋1 = 38
𝑆1 = 21.5
𝑛1 = 61
̅̅̅
𝑋2 = 26
𝑆2 = 12.9
𝑛2 = 51
𝑆𝑝2 =
(61−1)∗21.52 +(51−1)∗12.92
(61−1)+(51−1)
=
60∗462.25+50∗166.41
60+50
Find degree of freedom by using formula:
(n1 - 1)+(n2 - 1) = (61 - 1)+(51 - 1) = 60 + 50 = 110
So df is equal to 110
α = 0.05 level of significance
=
27735+8320.5
110
=
36055.5
110
= 327.777
Find the critical value by using formula in Excel = TINV(α;df) = TINV(0.05;110) equal to 1.982
T α/2 = 1.982
In order to find Confident Interval or CI, use the formula below:
1
1
(38 – 26) ± 1.982√327.777(61 + 51) = 12 ± 1.982√327.777(0.0164 + 0.0196) = 12 ± √11.8 =
12 ± 3.435 => 8.565 ≤ µ1 - µ2 ≤ 15.435
Interpretation:
Children who watched the sports celebrity–endorsed potato chips commercial has mean amount of
between 8.565 and 15.435 grams more potato chips eaten than children who watched the
alternative food snack commercial.