Answer:
a)<em> Null hypothesis : H₀</em>: the proportion of defective item of computer has been lowered. That is P < 0.15
<u><em>Alternative hypothesis: H₁:</em></u> The proportion of defective item of computer
has been higher. That is P> 0.15 (Right tailed test)
b) Test statistic ![Z = \frac{p-P}{\sqrt{\frac{PQ}{n} } }](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7Bp-P%7D%7B%5Csqrt%7B%5Cfrac%7BPQ%7D%7Bn%7D%20%7D%20%7D)
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 t<em>he proportion of defective item of computer has been lowered. </em>
Step-by-step explanation:
<u>Step(i)</u>:-
<em>Given the sample size 'n' = 42</em>
Given random sample of 42 computers were tested revealing a total of 4 defective computers.
The defective computers 'x' = 4
<em>The sample proportion of defective computers </em>
![p = \frac{x}{n} = \frac{4}{42} = 0.095](https://tex.z-dn.net/?f=p%20%3D%20%5Cfrac%7Bx%7D%7Bn%7D%20%3D%20%5Cfrac%7B4%7D%7B42%7D%20%3D%200.095)
<em>Given The Population proportion 'P' = 0.15</em>
<em>The level of significance ∝=0.01</em>
<u>Step(ii)</u>:-
a)<em> Null hypothesis : H₀</em>: the proportion of defective item of computer has been lowered. That is P < 0.15
<u><em>Alternative hypothesis: H₁:</em></u> The proportion of defective item of computer
has been higher. That is P> 0.15 (Right tailed test)
b)
Test statistic ![Z = \frac{p-P}{\sqrt{\frac{PQ}{n} } }](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7Bp-P%7D%7B%5Csqrt%7B%5Cfrac%7BPQ%7D%7Bn%7D%20%7D%20%7D)
c)
![Z = \frac{0.095-0.15}{\sqrt{\frac{0.15(0.85)}{42} } }](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B0.095-0.15%7D%7B%5Csqrt%7B%5Cfrac%7B0.15%280.85%29%7D%7B42%7D%20%7D%20%7D)
Calculate the value of the test statistic Z = - 0.9991
|Z| = |- 0.9991| = 0.991
<u>Step(iii)</u>:-
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.
<u><em>Conclusion</em></u>:-
Hence the null hypothesis is accepted at 0.01 level of significance.
f)
<em> The proportion of defective item of computer has been lowered.</em>