TMA 05: Hypothesis Testing and Distribution Analysis

Surname 1
Student’s Name
Professor’s Name
Course
Date
TMA 05
Question 1
(a)
(i) Null hypothesis (H0 ¿: μ = 0.26 (the mean breaking strength of the new type of
polyester fiber and existing type is the same).
Alternative hypothesis ( Ha ¿: μ ≠ 0.26 (the mean breaking strength of the new
type of polyester fiber differs from the mean breaking strength of the existing
type of polyester fiber.
(ii) The sample size = 30 (sufficiently large sample). Also, the normality
assumption is met since the Shapiro Wilk = 0.972 with p-value > 0.05.
(iii) Hypothesis and significance levels α=0.05
Statistic: z = √ n ( x−μ )
s
Where: x – sample mean, s – sample standard deviation
(iv) Z ~ N (0, 1),
f ( z ) = 1
√ 2 π e
1
2 z2
, zε R
Null distribution is the probability distribution of the test statistic when
the null hypothesis is true. The term null is used to take care of the fact that
true distribution is often unknown and thus the distribution used is a guess.
(v) The figure below shows the output

Surname 2
(vi) The mean strength of the new fiber is 0.3659 greater than the hypothesized
0.26. The Z = 2.16 with corresponding p-value = 0.031
(vii) The p-value = 0.031 is less than α=0.05 thus at 5% significance level reject
the null hypothesis and conclude that there is sufficient evidence to support the
claim; mean breaking strength of the new type of polyester fiber differs from
the mean breaking strength of the existing type of polyester fiber. However, at
1% significance level (p-value = 0.031 > 0.01) fail to reject the null hypothesis
and conclude that the mean breaking strength of the new type of polyester
fiber and existing type is the same (0.26 grams/denier).
(b) Given p0=0.4 , n=60 , failures( x )=39, Hypothesis;
H0 : p=0.4 , vs H0 : p >0.4
(i) ^p= x
n =39
60 =0.65
Statistic Z0 = √n ( ^p−p0 )
√ p0( 1− p0 )
Z0 = √ 60 ( 0.65−0.4 )
√ 0.4(1−0.4) =3.95285
(ii) Figure below shows the rejection region shaded