
MATLAB: An Introduction with Applications
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Author: Amos Gilat
Publisher: John Wiley & Sons Inc
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Given the hypothesis testing solution ,give an example for each of the Type I and Type II errors.

Transcribed Image Text:By using t statistic as hypothesis test, o of the population data is not known and oʻx=o°y
Variable X
Variable Y
n=50, x=1197.32, s,=11510.67
m=50, y=1297.18, s, = 10689.86
Let ug = mean brain weights of female of age <18, µ, = mean brain weights of male age <18
Ho: Hx = Hy
H1: Hx < Hy (LTT)
49(11510. 67) + 49(10689. 86)
50 + 50 – 2
Pooled Variance, sp
= 11100.27
(1197. 32 – 1297. 18)–0
t=
-4.7391
V11100. 27( + )
p-value approach, a= 0.05
P(T<-4.7391) = 3.63594x10-6= 0.00000363594
p-value < 0.05, Hence reject null hypothesis Ho.
At a= 0.05, We have sufficient evidence to conclude that mean of brain weights of brain weights
of female of age <18 is lesser than mean brain weights of male of age <18

Transcribed Image Text:Condition checking 1:
n=50 (n 230) m=50 (n 230)
The sample size, n for the brain weights of female and male of age less than 18 years old
(variable X and variable Y) is equals to 50. Sample size greater than or equals to 30 is usually
considered as a large sample.
According to the Central Limit Theorem, the sampling distribution of X and Y for a relatively
large sample size is approximately normal, irrespective of the shape of the population
distribution.
To choose a distribution, first standard deviation, o of data is checked whether is known or not
known, then whether oʻy= o, or of;+ o, is checked.
Assume that variance for Variable X and variance of Variable Y are equal
Họ: ox= o°y
Based on the value obtained in descriptive statistic:
ox=11510.67, o, = 10689.86
11510. 67
F=
10689. 86
= 1.0768
d= 49 df-49
0.05 Fo.025,49,49 = 1.6073
Since F < 1.6073, we cannot reject null hypothesis Ho
Hence, with a= 0.05, we have sufficient statistical evidence to conclude that of=o.
Since the standard deviation, o of the population data is not known and o,=o°y, t test statistic is
used.
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