Therefore, confidence interval is from 25.69 to 30.31 with a width of 4.62. LET'S TRY THIS Identify the confidence interval estimate and compute its width based on the given data below. 1. Confidence level= 97% 0= 16.2 n= 42 X= 45 2. Confidence level= 90% s= 12.5 n= 13 X= 88 3. Confidence level= 98% 0= 8 n= 140 え= 70.2 4. Confidence level= 95% S= 4.1 n= 22 x= 23.5 width Problem solving The average hour spent in computer by 20 senior high school students during online class is 7 hours with a standard deviation of 2 hours. Construct a 98% confidence interval of the average hour spent by all senior high school students. Identify the width of the interval. Sample Size
Therefore, confidence interval is from 25.69 to 30.31 with a width of 4.62. LET'S TRY THIS Identify the confidence interval estimate and compute its width based on the given data below. 1. Confidence level= 97% 0= 16.2 n= 42 X= 45 2. Confidence level= 90% s= 12.5 n= 13 X= 88 3. Confidence level= 98% 0= 8 n= 140 え= 70.2 4. Confidence level= 95% S= 4.1 n= 22 x= 23.5 width Problem solving The average hour spent in computer by 20 senior high school students during online class is 7 hours with a standard deviation of 2 hours. Construct a 98% confidence interval of the average hour spent by all senior high school students. Identify the width of the interval. Sample Size
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
Related questions
Question
Estimation of parameters
![METERS
= 25.69 <u < 30.31
Therefore, confidence interval is from 25.69 to 30.31 with a
width of 4.62.
LET'S TRY THIS
10
40
Identify the confidence interval estimate and compute its width based
on the given data below.
1. Confidence level= 97%
0= 16.2
n= 42
x=45
2. Confidence level= 90%
s= 12.5
n= 13
え= 88
3. Confidence level= 98%
0= 8
n= 140
x= 70.2
4. Confidence level= 95%
S= 4.1
n= 22
i= 23.5
a width
Problem solving
The average hour spent in computer by 20 senior high school students
during online class is 7 hours with a standard deviation of 2 hours.
Construct a 98% confidence interval of the average hour spent by all
senior high school students. Identify the width of the interval.
Appropriate Sample Size
Lonath of the Interval](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb86e8680-303f-4455-bc99-84d9554bc6d8%2F4cee4638-f9a7-486e-a110-99600cc77fba%2Fqjllcgo_processed.jpeg&w=3840&q=75)
Transcribed Image Text:METERS
= 25.69 <u < 30.31
Therefore, confidence interval is from 25.69 to 30.31 with a
width of 4.62.
LET'S TRY THIS
10
40
Identify the confidence interval estimate and compute its width based
on the given data below.
1. Confidence level= 97%
0= 16.2
n= 42
x=45
2. Confidence level= 90%
s= 12.5
n= 13
え= 88
3. Confidence level= 98%
0= 8
n= 140
x= 70.2
4. Confidence level= 95%
S= 4.1
n= 22
i= 23.5
a width
Problem solving
The average hour spent in computer by 20 senior high school students
during online class is 7 hours with a standard deviation of 2 hours.
Construct a 98% confidence interval of the average hour spent by all
senior high school students. Identify the width of the interval.
Appropriate Sample Size
Lonath of the Interval
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