Question

The proportion of defective computers built by Byte Computer Corporation is 0.15. In an attempt to lower the defective rate, the owner ordered some changes made in the assembly process. After the changes were put into effect, a random sample of 42 computers were tested revealing a total of 4 defective computers. Perform the appropriate test of hypothesis to determine whether the proportion of defective computer has been lowered. Use alpha=0.01
a. State the appropriate null and alternative hypothesis

i.H0=

ii.H1=

b. Appropriate test to test the hypothesis

c. Calculate the value of the test statistic. Test statistic =

d. P-value or t critical value=

e. What is your conclusion

f. Interpretation

Answer 1

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   $Z = \frac{p-P}{\sqrt{\frac{PQ}{n} } }$

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

                                                                $p = \frac{x}{n} = \frac{4}{42} = 0.095$

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   $Z = \frac{p-P}{\sqrt{\frac{PQ}{n} } }$

                       

c)      

                 $Z = \frac{0.095-0.15}{\sqrt{\frac{0.15(0.85)}{42} } }$

                 $z = \frac{-0.055}{\sqrt{0.00303} } = - 0.9991$      

                     

  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.