Respuesta :
Answer:
a) Null hypothesis : H₀: the proportion of defective item of computer has been lowered. That is P < 0.15
Alternative hypothesis: H₁: The proportion of defective item of computer
has been higher. That is P> 0.15 (Right tailed test)
b) Test statistic [tex]Z = \frac{p-P}{\sqrt{\frac{PQ}{n} } }[/tex]
c) Calculate the value of the test statistic = 0.991
d) The critical value at 0.01 level of significance = Z₀.₀₁ = 2.57
e) Null hypothesis accepted at 0.01 level of significance
f) we accepted null hypothesis.
Hence the proportion of defective item of computer has been lowered.
Step-by-step explanation:
Step(i):-
Given the sample size 'n' = 42
Given random sample of 42 computers were tested revealing a total of 4 defective computers.
The defective computers 'x' = 4
The sample proportion of defective computers
[tex]p = \frac{x}{n} = \frac{4}{42} = 0.095[/tex]
Given The Population proportion 'P' = 0.15
The level of significance ∝=0.01
Step(ii):-
a) Null hypothesis : H₀: the proportion of defective item of computer has been lowered. That is P < 0.15
Alternative hypothesis: H₁: The proportion of defective item of computer
has been higher. That is P> 0.15 (Right tailed test)
b)
Test statistic [tex]Z = \frac{p-P}{\sqrt{\frac{PQ}{n} } }[/tex]
c)
[tex]Z = \frac{0.095-0.15}{\sqrt{\frac{0.15(0.85)}{42} } }[/tex]
[tex]z = \frac{-0.055}{\sqrt{0.00303} } = - 0.9991[/tex]
Calculate the value of the test statistic Z = - 0.9991
|Z| = |- 0.9991| = 0.991
Step(iii):-
d)
The critical value at 0.01 level of significance = Z₀.₀₁ = 2.57
e) Calculate the value of the test statistic Z = 0.991 < 2.57 at 0.01 level of significance.
Conclusion:-
Hence the null hypothesis is accepted at 0.01 level of significance.
f)
The proportion of defective item of computer has been lowered.
