MATH302 week 5 knowledge check
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American Military University *
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Subject
Mechanical Engineering
Date
Feb 20, 2024
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18
Uploaded by ColonelGrouse24
Week 5 Knowledge Check Homework Practice Quest…
1
2
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Written Jan 6, 2024 10:15 PM - Jan 6, 2024 11:29 PM
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Attempt 1 of 4
Attempt Score
20 / 20 - 100 %
Overall Grade (Highest Attempt)
20 / 20 - 100 %
Question 1
1 / 1 point
There is no prior information about the proportion of Americans who
support gun control in 2018. If we want to estimate 92% confidence
interval for the true proportion of Americans who support gun control in
2018 with a 0.2 margin of error, how many randomly selected Americans
___
20___
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Question 2
1 / 1 point
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231___
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must be surveyed? Answer: (Round up your answer to nearest whole
number) Z-Critical Value =NORM.S.INV(.96) = 1.750686
n = n = p
∗
q
∗
Z
2
ME
2
.5
∗
.5
∗
1.750686
2
.2
2
The population standard deviation for the height of college basketball
players is 3.5 inches. If we want to estimate 97% confidence interval for
the population mean height of these players with a 0.5 margin of error, how
many randomly selected players must be surveyed? (Round up your answer
to nearest whole number) Answer: Z-Critical Value = NORM.S.INV(.985) = 2.17009
n =
Question 3
1 / 1 point
___
239___
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n = SD
2
∗
Z
2
ME
2
3.5
2
∗
2.17009
2
.5
2
The population standard deviation for the height of college basketball
players is 3 inches. If we want to estimate a 99% confidence interval for the
population mean height of these players with a 0.5 margin of error, how
many randomly selected players must be surveyed? (Round up your answer
to nearest whole number) Answer: Z-Critical Value = NORM.SINV(.995) = 2.575 n = n = SD
2
∗
Z
2
ME
2
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Question 4
1 / 1 point
___
11___
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Question 5
1 / 1 point
3
2
∗
2.575
2
.5
2
There is no prior information about the proportion of Americans who
support Medicare-for-all in 2019. If we want to estimate 95% confidence
interval for the true proportion of Americans who support Medicare-for-all
in 2019 with a 0.3 margin of error, how many randomly selected Americans
must be surveyed? Answer: (Round up your answer to nearest whole
number) Z-Critical Value = NORM.S.INV(.975) = 1.96
n = n = p
∗
q
∗
Z
2
ME
2
.5
∗
.5
∗
1.96
2
.3
2
___
178___
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Question 6
1 / 1 point
___
8___
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The population standard deviation for the height of college basketball
players is 3.4 inches. If we want to estimate 95% confidence interval for
the population mean height of these players with a 0.5 margin of error, how
many randomly selected players must be surveyed? (Round up your answer
to nearest whole number) Answer: Z-Critical Value = NORM.S.INV(.975) = 1.96
n = n = SD
2
∗
Z
2
ME
2
3.4
2
∗
1.96
2
.5
2
There is no prior information about the proportion of Americans who
support gun control in 2018. If we want to estimate 95% confidence
interval for the true proportion of Americans who support gun control in
2018 with a 0.36 margin of error, how many randomly selected Americans
must be surveyed? Answer: (Round up your answer to nearest whole
number)
Question 7
1 / 1 point
___
140___
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Z-Critical Value =NORM.S.INV(.975) = 1.96
n = n = p
∗
q
∗
Z
2
ME
2
.5
∗
.5
∗
1.96
2
.36
2
A researcher would like to estimate the proportion of all children that have
been diagnosed with Autism Spectrum Disorder (ASD) in their county. They
are using 95% confidence level and the CDC national estimate that 1 in 68
≈
0.0147 children are diagnosed with ASD. What sample size should the
researcher use to get a margin of error to be within 2%? Round up to the
nearest integer. Answer Z-Critical Value = NORM.S.INV(.975) = 1.96
n =
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Question 8
1 / 1 point
A recent study of 750 Internet users in Europe found that 35% of Internet
users were women. What is the 95% confidence interval estimate for the
true proportion of women in Europe who use the Internet?
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.0147
∗
.9853
∗
1.96
2
.02
2
0.321 to 0.379
0.305 to 0.395
0.316 to 0.384
0.309 to 0.391
Z-Critical Value = NORM.S.INV(.975) = 1.96
LL = 0.35 - 1.96*
UL = 0.35 +1.96*
√
.35
∗
.65
750
√
.35
∗
.65
750
Question 9
1 / 1 point
___
0.274___
(50
%)
___
0.526___
(50 %)
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Question 10
1 / 1 point
A random sample found that forty percent of 100 Americans were satisfied
with the gun control laws in 2017. Compute a 99% confidence interval for
the true proportion of Americans who were satisfied with the gun control
laws in 2017. Fill in the blanks appropriately.
A 99% confidence interval for the true proportion of Americans who were
satisfied with the gun control laws in 2017 is ( ) (round to 3 decimal places)
Z-Critical Value = NORM.S.INV(.995) = 2.575
LL = 0.4 - 2.575*
UL = 0.4 -+2.575*
√
.40
∗
.60
100
√
.40
∗
.60
100
Suppose a marketing company wants to determine the current proportion
of customers who click on ads on their smartphones. It was estimated that
___
0.334___
(50 %)
___
0.506___
(50 %)
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Question 11
1 / 1 point
A random sample of 150 people was selected and 12% of them were left
handed. Find the 90% confidence interval for the proportion of left-handed
people.
the current proportion of customers who click on ads on their smartphones
is 0.42 based on a random sample of 100 customers.
Compute a 92% confidence interval for the true proportion of customers
who click on ads on their smartphones and fill in the blanks appropriately.
< p < (round to 3 decimal
places)
Z-Critical Value = NORM.S.INV(.96) = 1.750686
LL = 0.42 - 1.750686 *
UL = 0.42 + 1.750686 *
√
.42
∗
.58
100
√
.42
∗
.58
100
(–1.645, 1.645)
(0.0764, 0.1636)
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Question 12
1 / 1 point
A survey of 85 families showed that 36 owned at least one DVD player. Find the 99%
confidence interval estimate of the true proportion of families who own at least one DVD
player. Place your limits, rounded to 3 decimal places
, in the blanks. Place the lower limit
in the first blank___ and the upper limit in the second blank___
When entering your answer do not use any labels or symbols other than the decimal
point. Simply provide the numerical values. For example, 0.123 would be a legitimate
entry. Make sure you put a 0 before the decimal.
(.12, .88)
(0.0436, 0.1164)
(0.068, 0.172)
Z-Critical Value = NORM.S.INV(.95) = 1.645
LL = .12 - 1.645*
UL = .12 + 1.645*
√
.12
∗
.88
150
√
.12
∗
.88
150
___
Answer for blank # 1: 0.285
(50 %)
Answer for blank # 2: 0.562
(50 %)
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Question 13
1 / 1 point
If a sample has 25 observations and a 99% confidence estimate for
is needed, the appropriate value of the t-multiple required is___. Place your
answer, rounded to 3 decimal places, in the blank. For example, 3.456
would be an appropriate entry.
Z-Critical Value =NORM.S.INV(.995) = 2.575
LL = .4235 - 2.575*
UL = .4235 + 2.575*
√
.4235
∗
.5765
85
√
.4235
∗
.5765
85
μ
___
Answer: 2.797
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Question 14
1 / 1 point
A sample of 40 country CD recordings of Willie Nelson has been examined.
The average playing time of these recordings is 51.3 minutes, and the
standard deviation is s = 5.8 minutes.
Using an appropriate t-multiplier, construct a 95% confidence interval for
the mean playing time of all Willie Nelson recordings.
Place your LOWER limit, in minutes, rounded to 2 decimal places, in the
first blank. For example, 56.78 would be a legitimate entry.___.
Place your UPPER limit, in minutes, rounded to 2 decimal places, in the
second blank. For example, 67.89 would be a legitimate entry.___
___
Answer for blank # 1: 49.45
(50 %)
Answer for blank # 2: 53.15
(50 %)
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In Excel,
=T.INV.2T(0.01,24)
T-Critical Value =T.INV.2T(.05,39) = 2.022691
LL = 51.3 - 2.022691*
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Question 15
1 / 1 point
The personnel department of a large corporation wants to estimate the
family dental expenses of its employees to determine the feasibility of
providing a dental insurance plan. A random sample of 12 employees
reveals the following family dental expenses (in dollars).
See Attached Excel for Data.
dental expense data.xlsx
Construct a 90% confidence interval estimate for the mean of family dental
expenses for all employees of this corporation.
Place your LOWER limit, in dollars rounded to 1 decimal place, in the first
blank. Do not
use a dollar sign, a comma, or any other stray mark. For
example, 123.4 would be a legitimate entry.___
Place your UPPER limit, in dollars rounded to 1 decimal place, in the second
blank. Do not
use a dollar sign, a comma, or any other stray mark. For
example, 567.8 would be a legitimate entry.___
___
UL = 51.3 + 2.022691*
5.8
√
40
5.8
√
40
Answer for blank # 1: 231.5
(50 %)
Answer for blank # 2: 384.9
(50 %)
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Question 16
1 / 1 point
The percent defective for parts produced by a manufacturing process is
targeted at 4%. The process is monitored daily by taking samples of sizes n
= 160 units. Suppose that today's sample contains 14 defectives.
How many units would have to be sampled to be 95% confident that you
can estimate the fraction of defective parts within 2% (using the
information from today's sample--that is using the result that T-Critical Value = T.INV.2T(.10,11) = 1.795885
LL = 308.1667 - 1.795885 *
UL = 308.1667 +1.795885 *
147.928
√
12
147.928
√
12
)?
Place your answer, as a whole number, in the blank. For example 567 would
be a legitimate entry.
___
___
Answer: 767
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Question 17
1 / 1 point
^
p
= 0.0875
Z-Critical Value = NORM.S.INV(.975) = 1.96
n = n = p
∗
q
∗
Z
2
ME
2
.0875
∗
.9125
∗
1.96
2
.02
2
A randomly selected sample of college basketball players has the following
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___
63.846___
(50 %)
___
65.841___
(50 %)
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Question 18
1 / 1 point
A random sample of stock prices per share (in dollars) is shown. Find the
90% confidence interval for the mean stock price. Assume the population
of stock prices is normally distributed.
See Attached Excel for Data
stock price data.xlsx
heights in inches.
See Attached Excel for Data.
height data.xlsx
Compute a 93% confidence interval for the population mean height of
college basketball players based on this sample and fill in the blanks
appropriately.
< μ
< (round to 3
decimal places)
T-Critical Value =T.INV.2T(.07,31) = 1.876701
LL
= 64.84375
−
1.876701
∗
3.006542
√
32
UL
= 64.84375 + 1.876701
∗
3.006542
√
32
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Question 19
1 / 1 point
A random sample of college football players had an average height of 66.35
inches. Based on this sample, (65.6, 67.1) found to be a 90% confidence
interval for the population mean height of college football players. Select
the correct answer to interpret this interval.
(17.884, 40.806)
(27.512, 31.178)
(13.582, 45.108)
(16.572, 42.118)
(–1.833, 1.833)
T-Critical Value = T.INV.2T(.10,9) = 1.833113
LL = 29.345 - 1.833113*
UL = 29.345 + 1.833113*
22.03462
√
10
22.03462
√
10
Question 20
1 / 1 point
A random sample of college football players had an average height of 64.55
inches. Based on this sample, (63.2, 65.9) found to be a 92% confidence
interval for the population mean height of college football players. Select
the correct answer to interpret this interval.
Done
We are 90% confident that the population mean height of college
football players is between 65.6 and 67.1 inches.
There is a 90% chance that the population mean height of college
football players is between 65.6 and 67.1 inches.
We are 90% confident that the population mean height of college
football palyers is 66.35 inches.
A 90% of college football players have height between 65.6 and 67.1
inches.
We are 92% confident that the population mean height of college
football players is between 63.2 and 65.9 inches.
We are 92% confident that the population mean height of college
football palyers is 64.55 inches.
A 92% of college football players have height between 63.2 and 65.9
inches.
There is a 92% chance that the population mean height of college
football players is between 63.2 and 65.9 inches.
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- I need parts 1, 2, and 3 answered pertaining to the print provided. NOTE: If you refuse to answers all 3 parts and insist on wasting my question, then just leave it for someone else to answer. I've never had an issue until recently one single tutor just refuses to even read the instructions of the question and just denies it for a false reasons or drags on 1 part into multiple parts for no reason.arrow_forwardall one question help asap please!arrow_forwardThe network diagram for a project is shown in the accompanying figure, with three timeestimates for each activity. Activity times are in weeks. Do the following:a. Compute the expected time for each activity and the expected duration for each path.b. Identify the critical path.c. Compute the variance of each activity and the variance and standard deviation of eachpath.arrow_forward
- Josh and Jake are both helping to build a brick wall which is 6 meters in height. They lay 250 bricks each, but Josh finishes this task in three (3) hours while Jake requires 4.5 hours to complete his part. select the BEST response below: Jake does more work than Josh O Josh does more work than Jake Both Josh and Jake do the same amo O of work and have the same amount of power Both Josh and Jake does the same O amount of work, however, Josh has m power than Jake.arrow_forwardASK Question Four The following data were recorded during the tensile test of a 14-mm diameter mild steel rod. The gage length was 50 mm. at fracture, the elongation between the gage marks was 4.0 mm and the minimum diameter was 11.3 mm a) Plot the conventional stress-strain curve for the mild steel b) determine the proportional limit c) Modulus of elasticity d) Yield stress at 0.2% c) Ultimate stress f) Breaking strength g) Percent elongation in 50 mm and reduction in area Load (N) 0 Elongation (mm) 0 6310 0.01 12600 0.02 18800 0.03 25100 0.04 31300 0.05 37900 0.06 40100 0.163 41600 0.433 46200 1.25 $2400 2.5 58500 4.5 68000 75 2arrow_forwardI need GF,CF,CD asap pleasearrow_forward
- TRUE OR FALSE Directions: Choose true is the statement is correct, otherwise choose false. 1. Literature is not related to a certain region’s culture, tradition, history and people. 2. All written, oral, and multimedia forms of work are considered literature. 3. Poetry commonly utilizes rhymes and figures of speech while prose uses formal sentence structures. 4. Drama is delivered on a theatrical stage with actors and actresses. 5. Fictional works have the same elements of non-fictional works. 6. Facts, information, and statistics are focused on non-fictional works. 7. Human history and advancements were possible without the role literature. 8. The standard of universality conveys that literature must be perpetual and timely. It is forever relevant and it appeals to one and all, anytime, and anywhere. 9. Intellectual beauty expresses that a literary piece must stimulate thought about life and human nature, while suggestiveness pertains to how it evokes emotion of the reader. 10. All…arrow_forwardI need soolutions for CD,CF,GFarrow_forwardAccess Pearson Mastering Engineering Back to my courses Course Home Course Home Scores ■Review Next >arrow_forward
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