10:36 PM Fri Sep 20 < SDS 302 - Homework 2 3 2 0 All sampled values T T -20 0 20 40 60 80 100 Last sample mean 18.84 3 2 Γ T T T -20 0 20 40 60 80 100 2nd to last mean 15.89 3 2 0 Simulated Sampling distribution of the sample means -20 20 T -20 0 20 40 60 80 100 -20 20 T 0 20 40 60 80 100 3rd to last mean = 10 2 4th to last mean = 5.22 0 20 40 60 80 100 -20 0 20 40 60 80 100 b. What is the difference between the histograms when the population is Normal (question 1) and when the population is Negative Exponential (question 2)? 3. What will be the shape of the sampling distribution if we generate it with a sample size of 40? (2 pt) +8 6% 10:35 PM Fri Sep 20 - a. Frequency 1000 0 Frequency 1. Generate the sampling distribution for the sample mean by setting the population distribution to Normal, the statistic the mean, the mean Mu of 100, the standard deviation sigma of 15, the sample size to 5, and the samples to draw at a time to 1000. (1 pt) a. Copy and paste of the screenshot of the sampling distribution generated. All sampled values 60 230 Simulated Sampling distribution of the sample means 500 400 300 50 100 True Pop. mean = 100 Aggregate mean 100.09. Last sample mean = 91.2 2 150 2nd to last mean 102.05 L T 7 50 100 150 ° Γ 50 100 150 Frequency 100 09 ° 50 100 Samp. dist. mean -100.09 Samp. dist. SD-6.85 3rd to last mean = 92.11 ° ° Γ T . 50 100 150 150 4th to last. mean = 94.6 50 100 150 b. From what we learned, could you provide the expected shape, mean, and standard deviation of the sampling distribution? 2. Generate the sampling distribution for the sample mean by setting the population distribution to Negative Exponential (skewed), the sample size to 5, and the samples to draw at a time to 1000. (1 pt) Copy and paste of the screenshot of the sampling distribution generated. All sampled values 40 Frequency 8 Simulated Sampling distribution of the sample means -20 0 20 60 80 -20 0 20 40 60 80 100 Last sample mean 18.84 ° 20 0 20 40 60 80 100 2 2nd to last mean 15.89 3 2 3rd to last 4th to last mean 10 mean = 5.22 -20 0 20 40 60 80 100 -20 0 20 40 60 80 100 O 20 0 20 40 60 80 100 b. What is the difference between the histograms when the population is Normal (question 1) and when the population is Negative Exponential (question 2)? What will be ล 6%

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter10: Statistics
Section10.4: Distributions Of Data
Problem 22PFA
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Related questions
Question
10:36 PM Fri Sep 20
<
SDS 302 - Homework 2
3
2
0
All sampled values
T
T
-20
0
20
40
60
80
100
Last sample
mean 18.84
3
2
Γ T T T
-20 0 20 40 60 80 100
2nd to last
mean 15.89
3
2
0
Simulated Sampling distribution of the
sample means
-20
20
T
-20 0 20 40 60 80 100
-20
20
T
0
20
40
60
80
100
3rd to last
mean = 10
2
4th to last
mean = 5.22
0
20 40 60 80 100
-20 0 20
40 60 80 100
b.
What is the difference between the histograms when the population is Normal
(question 1) and when the population is Negative Exponential (question 2)?
3. What will be the shape of the sampling distribution if we generate it with a
sample size of 40? (2 pt)
+8
6%
Transcribed Image Text:10:36 PM Fri Sep 20 < SDS 302 - Homework 2 3 2 0 All sampled values T T -20 0 20 40 60 80 100 Last sample mean 18.84 3 2 Γ T T T -20 0 20 40 60 80 100 2nd to last mean 15.89 3 2 0 Simulated Sampling distribution of the sample means -20 20 T -20 0 20 40 60 80 100 -20 20 T 0 20 40 60 80 100 3rd to last mean = 10 2 4th to last mean = 5.22 0 20 40 60 80 100 -20 0 20 40 60 80 100 b. What is the difference between the histograms when the population is Normal (question 1) and when the population is Negative Exponential (question 2)? 3. What will be the shape of the sampling distribution if we generate it with a sample size of 40? (2 pt) +8 6%
10:35 PM Fri Sep 20
-
a.
Frequency
1000
0
Frequency
1.
Generate the sampling distribution for the sample mean by setting the
population distribution to Normal, the statistic the mean, the mean Mu of 100, the
standard deviation sigma of 15, the sample size to 5, and the samples to draw at a
time to 1000. (1 pt)
a.
Copy and paste of the screenshot of the sampling distribution generated.
All sampled values
60
230
Simulated Sampling distribution of the
sample means
500
400
300
50
100
True Pop. mean = 100
Aggregate mean 100.09.
Last sample
mean = 91.2
2
150
2nd to last
mean 102.05
L
T
7
50
100
150
°
Γ
50
100
150
Frequency
100
09
°
50
100
Samp. dist. mean -100.09
Samp. dist. SD-6.85
3rd to last
mean = 92.11
°
°
Γ
T
.
50
100
150
150
4th to last.
mean = 94.6
50
100
150
b.
From what we learned, could you provide the expected shape, mean, and
standard deviation of the sampling distribution?
2.
Generate the sampling distribution for the sample mean by setting the
population distribution to Negative Exponential (skewed), the sample size to 5, and
the samples to draw at a time to 1000. (1 pt)
Copy and paste of the screenshot of the sampling distribution generated.
All sampled values
40
Frequency
8
Simulated Sampling distribution of the
sample means
-20
0
20
60
80
-20
0
20
40
60
80
100
Last sample
mean 18.84
°
20 0 20 40 60 80 100
2
2nd to last
mean
15.89
3
2
3rd to last
4th to last
mean 10
mean = 5.22
-20 0
20
40 60 80 100
-20 0 20 40 60 80 100
O
20 0 20 40 60 80 100
b.
What is the difference between the histograms when the population is Normal
(question 1) and when the population is Negative Exponential (question 2)?
What will be
ล 6%
Transcribed Image Text:10:35 PM Fri Sep 20 - a. Frequency 1000 0 Frequency 1. Generate the sampling distribution for the sample mean by setting the population distribution to Normal, the statistic the mean, the mean Mu of 100, the standard deviation sigma of 15, the sample size to 5, and the samples to draw at a time to 1000. (1 pt) a. Copy and paste of the screenshot of the sampling distribution generated. All sampled values 60 230 Simulated Sampling distribution of the sample means 500 400 300 50 100 True Pop. mean = 100 Aggregate mean 100.09. Last sample mean = 91.2 2 150 2nd to last mean 102.05 L T 7 50 100 150 ° Γ 50 100 150 Frequency 100 09 ° 50 100 Samp. dist. mean -100.09 Samp. dist. SD-6.85 3rd to last mean = 92.11 ° ° Γ T . 50 100 150 150 4th to last. mean = 94.6 50 100 150 b. From what we learned, could you provide the expected shape, mean, and standard deviation of the sampling distribution? 2. Generate the sampling distribution for the sample mean by setting the population distribution to Negative Exponential (skewed), the sample size to 5, and the samples to draw at a time to 1000. (1 pt) Copy and paste of the screenshot of the sampling distribution generated. All sampled values 40 Frequency 8 Simulated Sampling distribution of the sample means -20 0 20 60 80 -20 0 20 40 60 80 100 Last sample mean 18.84 ° 20 0 20 40 60 80 100 2 2nd to last mean 15.89 3 2 3rd to last 4th to last mean 10 mean = 5.22 -20 0 20 40 60 80 100 -20 0 20 40 60 80 100 O 20 0 20 40 60 80 100 b. What is the difference between the histograms when the population is Normal (question 1) and when the population is Negative Exponential (question 2)? What will be ล 6%
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