Can you help me with the:1.mean difference, show how it was calculated2.confidence interval of the difference FemaleHeightX-mean (X-mean)?MaleHeightX-mean (X-mean)?1150-4.9524.501164-5.2627.6721550.050.002167.64-1.622.62162.567.6157.91165.10-4.1617.314147.30-7.6558.524169-0.260.07152.40-2.556.50165.10-4.1617.31154.94-0.010.00177.808.5472.93716510.05101.007154.94-14.32205.068150-4.9524.501583.059.308164-5.2627.6710149.86-5.0925.919.1700.740.5511150-4.9524.5010168-1.261.5912149-5.9535.40111700.740.5513150-4.9524.50121788.7476.3914149-5.9535.40131799.7494.8715155.450.50.25141777.7459.9116154.94-0.010.00Total:2369.58-0.06604.48171561.051.10Mean:169.26181627.0549.70191550.050.002016813.05170.30Variance:46.5021143-11.95142.80Standard deviation:6.8222150-4.9524.50DEGREE OF FREEDOM:23157.482.536.4024144.78-10.17103.43df = (41-1) + (14-1)df = 40 + 13251627.0549.70df =53261561.051.1027149-5.9535.40281605.0525.502916510.05101.003016813.05170.3031149-5.9535.4032160.025.0725.70331627.0549.7034145-9.9599.0035153-1.953.8036152.4-2.556.5037150-4.9524.503816510.05101.0039154.94-0.010.00401594.0516.4041153-1.953.80Total:6353.070.121675.31Mean:154.95Variance:41.88Standard deviation:6.47

solution Given Mean(X1)= 154.95 , s1 = 6.47, n1=1 =41 Mean(X2)= 169.26 , s2 = 6.82, n2= 14 As per…

Answered: confidence interval of the difference

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Question

Can you help me with the:

1.mean difference, show how it was calculated

2.confidence interval of the difference

Female
Height
X-mean (X-mean)?
Male
Height
X-mean (X-mean)?
1
150
-4.95
24.50
1
164
-5.26
27.67
2
155
0.05
0.00
2
167.64
-1.62
2.62
162.56
7.61
57.91
165.10
-4.16
17.31
4
147.30
-7.65
58.52
4
169
-0.26
0.07
152.40
-2.55
6.50
165.10
-4.16
17.31
154.94
-0.01
0.00
177.80
8.54
72.93
7
165
10.05
101.00
7
154.94
-14.32
205.06
8
150
-4.95
24.50
158
3.05
9.30
8
164
-5.26
27.67
10
149.86
-5.09
25.91
9.
170
0.74
0.55
11
150
-4.95
24.50
10
168
-1.26
1.59
12
149
-5.95
35.40
11
170
0.74
0.55
13
150
-4.95
24.50
12
178
8.74
76.39
14
149
-5.95
35.40
13
179
9.74
94.87
15
155.45
0.5
0.25
14
177
7.74
59.91
16
154.94
-0.01
0.00
Total:
2369.58
-0.06
604.48
17
156
1.05
1.10
Mean:
169.26
18
162
7.05
49.70
19
155
0.05
0.00
20
168
13.05
170.30
Variance:
46.50
21
143
-11.95
142.80
Standard deviation:
6.82
22
150
-4.95
24.50
DEGREE OF FREEDOM:
23
157.48
2.53
6.40
24
144.78
-10.17
103.43
df = (41-1) + (14-1)
df = 40 + 13
25
162
7.05
49.70
df =53
26
156
1.05
1.10
27
149
-5.95
35.40
28
160
5.05
25.50
29
165
10.05
101.00
30
168
13.05
170.30
31
149
-5.95
35.40
32
160.02
5.07
25.70
33
162
7.05
49.70
34
145
-9.95
99.00
35
153
-1.95
3.80
36
152.4
-2.55
6.50
37
150
-4.95
24.50
38
165
10.05
101.00
39
154.94
-0.01
0.00
40
159
4.05
16.40
41
153
-1.95
3.80
Total:
6353.07
0.12
1675.31
Mean:
154.95
Variance:
41.88
Standard deviation:
6.47

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

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