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The fraction or ratio of a sample possessing a certain trait is called a:
Population.
Mean.
Confidence interval.
Proportion.
Proportion is the same as fraction is the same as ratio is the same as percentage.
References
Multiple Choice
Learning Objective: 08-03 Compute
and interpret a confidence interval
for a population mean when the
population standard deviation is
unknown.
An economist wants to estimate the proportion of Canadians who own their homes. A random sample of 800 people reveals 544 own their homes. Develop a 95%
confidence interval for the population proportion.
0.68 ± 0.053
0.68 ± 0.032
0.32 ± 0.032
0.68 ± 0.027
We have:
x = 544 people,
n
= 800 people, 95% confidence interval
For a 95% confidence interval z = 1.96.
First calculate the sample proportion: So, we estimate 68% of people own their homes. The 95% confidence interval is
.
References
Multiple Choice
Learning Objective: 08-03 Compute
and interpret a confidence interval
for a population mean when the
population standard deviation is
unknown.
✓
✓
3.
Award: 10 out of 10.00 points
4.
Award: 10 out of 10.00 points
5.
Award: 10 out of 10.00 points
When do we apply the finite population correction factor?
When our sample size is 5% or more of our finite population size.
When we know exactly what the population size is.
Anytime we take a sample.
Anytime we are calculating a sample size.
References
Multiple Choice
Learning Objective: 08-06 Adjust a
confidence interval for a finite
population.
We know the total population is 3,000. 30% of the 500 people sampled say that they prefer hot dogs to hamburgers. Prepare a 95% confidence interval for the proportion
that prefer hot dogs.
26.00% to 34.00% prefer hot dogs.
26.33% to 33.67% prefer hot dogs.
16.67%
Cannot be determined from the information provided.
References
Multiple Choice
Learning Objective: 08-04 Compute
and interpret a confidence interval
for a population proportion.
Learning Objective: 08-05 Calculate the required sample size to estimate
a population mean or population proportion.
The necessary sample size depends on:
The level of confidence desired
The allowable margin of error
The variability in the population of study
The level of confidence desired, the allowable margin of error, and the variability in the population of study.
The necessary sample size depends on all three of the factors listed.
References
Multiple Choice
Learning Objective: 08-06 Adjust a
confidence interval for a finite
population.
✓
✓
✓
6.
Award: 10 out of 10.00 points
7.
Award: 10 out of 10.00 points
A pilot study shows that 64% of people living in the downtown core are single. A market research company wants to verify this claim. The company requires a 95%
confidence interval. How many residents should be interviewed to keep the margin of error within 0.02 of the population proportion?
23
9604
2213
6147
We have:
p
= 0.64, E
= 0.02, 95% confidence interval
For a 95% confidence interval z = 1.96.
A random sample of 2213 people is required.
References
Multiple Choice
Learning Objective: 08-06 Adjust a
confidence interval for a finite
population.
A random sample of 300 drivers revealed that 96 of them had received a speeding ticket in the last 3 months. Construct a 95% confidence interval for the number of drivers
who receive speeding tickets over a three-month period.
0.32 ± 0.027
0.32 ± 0.001
0.32 ± 0.069
0.32 ± 0.053
We have:
x
= 96 drivers, n
= 300 drivers, 95% confidence interval
For a 95% confidence interval z = 1.96.
First calculate the sample proportion: So, we estimate 32% of drivers have received a speeding ticket in the past 3 months. The 95% confidence interval is
.
References
Multiple Choice
Learning Objective: 08-04 Compute
and interpret a confidence interval
for a population proportion.
✓
✓
8.
Award: 10 out of 10.00 points
9.
Award: 10 out of 10.00 points
10.
Award: 10 out of 10.00 points
American Express cardholders are reported to spend more per purchase than either VISA or MasterCard cardholders. The average purchase at a giftware store is $70. The
owner conducts a random survey of 120 people who use their American Express card and finds 84 of them make purchases in excess of the average $70. Conduct a 90%
confidence interval for the number of American Express cardholders that make purchases greater than $70.
0.70 ± 0.002
0.70 ± 0.003
0.70 ± 0.069
0.70 ± 0.042
We have:
x
= 84 cardholders, n
= 120 cardholders, 90% confidence interval
For a 90% confidence interval z = 1.645.
First calculate the sample proportion: So, we estimate 70% of American Express cardholders spend in excess of the average purchase. The 90% confidence interval is
.
References
Multiple Choice
Learning Objective: 08-04 Compute
and interpret a confidence interval
for a population proportion.
A finite population is:
a large population.
an exactly known population size.
5% of the total population.
a population from which samples were taken.
References
Multiple Choice
Learning Objective: 08-05 Calculate
the required sample size to estimate
a population mean or population
proportion.
Determining a sample size depends on all of these except:
confidence level.
population size.
maximum allowable error.
variation in the population.
Determining sample size does not depend on the population size. Population size has no bearing on how large or small a sample size will be. If there is no prior estimate of
the population proportion, we use p
= 0.5 to maximize the sample size.
References
Multiple Choice
Learning Objective: 08-06 Adjust a
confidence interval for a finite
population.
✓
✓
✓
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Related Questions
Hello! I need help with the following stats questions.
1. What is a confidence interval?
2. What is a confidence coefficient?
3. What do statisticians mean by the term accuracy?
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March, 2022. Based on a random sample of 30 students' marks and with a 95% confidence
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(a) After receiving this result, student B in the same class claimed that there was a 95%
chance that the true average marks of the midterm were between 50 and
80. How would you respond to this statement? Is it correct? Why or why not?
(b) Student C in the same class did not agree with student B and he claimed that there was a
95% chance that the true average marks of the next midterm were between 50 and 80.
How would you respond to this statement? Is it correct? Why or why not?
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Is the value 6.5% a parameter or a statistic?
i.
ii.
iii.
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Supposed a study using a random sample of postgraduate students in Sydney found that the average amount of time spent using a mobile phone per day is 5.5 hours.
Indicate whether the quantity described above is a population parameter or a sample statistic, and write down the notation.
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Machines. A sample of 260 members showed that 40 of them use cardio machines. Calculate a
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Focusing on government spending as a percentage of GDP in the U.S., we observed that between 1820 and 1929 the: minimum value was 3.13%; mean value was 6.83%; and maximum value was 29.03%. In contrast, between 1930 and 2021 , the a) None of the above answers are correct. B) maximum value was 10.94%.c) minimum value was 2.12%. D) mean value was 30.56%.
Note:-
Do not provide handwritten solution. Maintain accuracy and quality in your answer. Take care of plagiarism.
Answer completely.
You will get up vote for sure.
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(Ch8) True or False? With a given sample, when we lower the confidence level, it will reduce the width of a confidence interval of the population proportion.
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18
We can make a confidence interval more precise (narrower) by,
a
increasing the sample size.
b
reducing the confidence level.
c
increasing the confidence level.
d
both (a) and (b) are correct.
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Population Sample
35
30
24
32
32
28
35
31
34
20
24
24
26
29
32
35
32
37
35
Using sample data determine the Sample Mean?
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X1=80 X2=180 A=400 B=200
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Ca = 40 + 0.80 (Y-T)
Ig = 30
Xn= 10
T=20
G=20
DI-Y-T
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The birthweight (in kg) of 55 babies are tabulated in the frequency distribution below:
Birthweight
(kg)
Class Midpoint
Frequency
M
(1– 1.5|
(1.5-2
1.25
6.
1.75
10
(2- 2.5)
2.25
1
(2.5-3)
2.75
15
10
(3-3.5]
3.25
3.
(3.5- 4)
3.75
55
Total
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Question 1
The distance between the sample mean and the true population is the sampling error,
O True
False
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10. 7 students were asked how many pencils they had. The responses were 2, 5, 8, 3, 1, 6,
and 4.
a. Calculate the sample mean
8. Find the sample standard deviation
C Find the Q1, Q3, and the interquartile range.
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If the t-statistic for a variable is -1.74, is the variable statistically
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O No
O Yes
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K
The first statistics exam had a mean of 71 and a standard deviation of 15 points; the second had a mean of 82 and a standard
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168 points on the two exams, but Anna claims that her total is better. Explain.
The total of Anna's z-scores is, which is greater than Megantotal of
(Round to one decimal place as needed.)
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see image below. i=120, ii=35, iii=135, iv=6
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Manufacturers of tires report that car tires should be able to last an average of 50,000 miles. A new tire company produces a
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Assuming the new type of tread does not improve the mileage of the tire, 200 sample means were simulated and displayed
on the dotplot.
Simulated Tire Mileage
+++ +++H
47,000
48,000
49,000
50,000
51,000
52,000
53,000
Mean mileage
Using the dotplot and the sample mean mileage, is there convincing evidence that the new type of tread improves the
mileage of the tire?
Yes, because a mean mileage of 51,500 or more occurred only 14 out of 200 times, the mean mileage is statistically
significant. There is convincing evidence the new type of tire tread improves mileage of the tire.
Yes, because a mean mileage of 51,500 or less occurred 186 out of 200 times, the mean mileage is statistically
significant. There is convincing evidence the new type of…
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Question:
In a certain factory there are two independent
processes manufacturing the same item. The
average weight in a sample of 250 items
produced from one process is found to be 120
ozs. with a standard deviation of 12 ozs. while
the corresponding figures in a sample of 400
items from the other process are 124 and 14.
Obtain the standard error of difference between
the two sample means. Is this difference
significant ? Also find the 99% confidence limits
for the difference in the average weights of
items produced by the two processes
respectively.
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What are the Confidence Intervals for the Population Mean?
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Question 3
A car company wants to know the monthly sales made in ($000), based on the brand of vehicle. The data collected was entered on a MINITAB spreadsheet for analysis. Exhibit II below was subsequently generated.
Exhibit 2
Model
N
Mean
Median
Tri.Mean
Std Dev
S.E. Mean
Hyundai
23
109
135
107.64
*
1.34
Toyota
27
165
124
143.65
9.5
**
You are required to test at the 5% level of significance the hypothesis that the average monthly earnings on Hyundai Vehicles is equal to $110,000 versus the alternative that it is different from $110,000.
Complete the following:
i. Give the null and alternative hypothesis of this test.
ii. Determine the critical value(s) of this test.
iii. Compute the value of the test statistic.
iv. State the decision rule.
v. Give your decision based on the available sample evidence.
vi. Hence, state your conclusion.
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QUESTION 8
Having an extreme score in a sample (outliers) will NOT effect which of the following?
standard deviation
mean
mode
variance
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2. The SAT scores for 12 randomly selected senior high school students are:
1221
1342
1298
987
786 1439
671 1299 1020 1380
889 1157
a.) sample mean =
b.) s²=
c.) s =
d.) Are the SAT scores normally distributed (Explain Why)? YES or NO
e.) Construct and interpret a 95% confidence interval for the mean SAT score.
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Date
No. of Clients
Oct. 23
13
Oct. 24
8
Oct. 25
10
Oct. 26
15
Oct. 27
10
Oct. 28
22
Find the mean, median, and mode. (If an answer does not exist, enter DNE.)
mean
clients
median
clients
mode
clients
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Question 4 of 10, step
Calculate the margin of error of a confidence interval for the difference between two population means using the given information. Round your answer to six decimal
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e, 14. n, 88 ₂ 10.72, ny71.c 0.95
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