Respuesta :
Answer:
At p₀ = 0.982 the manufacturer's claim would not be rejected.
Step-by-step explanation:
The manufacturer wishes to make a claim about the percentage of non-defective CD players and is prepared to exaggerate.
Let the claim made by him be denoted as, p₀.
The hypothesis to test this claim is defined as:
H₀: The proportion of defective is p₀, i.e. p = p₀.
Hₐ: The proportion of defective is less than p₀, i.e. p < p₀.
The significance level of the test is, α = 0.05.
The information provided is:
n = 100
[tex]\hat p[/tex] = 0.96
The test statistic is:
[tex]Z=\frac{\hat p-p_{0}}{\sqrt{\frac{p_{0}(1-p_{0})}{n}}}=\frac{0.96-p_{0}}{\sqrt{\frac{p_{0}(1-p_{0})}{100}}}[/tex]
Decision rule:
If the test statistic value is less than the critical value of z, i.e. Z₀.₀₅ = -1.645 (because of the left tail), then the null hypothesis will be rejected. and vice versa.
So, to reject H₀ Z ≤ Z₀.₀₅.
Use hit and trial method.
At p₀ = 0.97 the value of Z is:
[tex]Z=\frac{\hat p-p_{0}}{\sqrt{\frac{p_{0}(1-p_{0})}{n}}}=\frac{0.96-0.97}{\sqrt{\frac{0.97(1-p\0.97)}{100}}}=-0.5862[/tex]
At p₀ = 0.98 the value of Z is:
[tex]Z=\frac{\hat p-p_{0}}{\sqrt{\frac{p_{0}(1-p_{0})}{n}}}=\frac{0.96-0.98}{\sqrt{\frac{0.98(1-0.98)}{100}}}=-1.4286[/tex]
At p₀ = 0.981 the value of Z is:
[tex]Z=\frac{\hat p-p_{0}}{\sqrt{\frac{p_{0}(1-p_{0})}{n}}}=\frac{0.96-0.981}{\sqrt{\frac{0.981(1-0.981)}{100}}}=-1.5382[/tex]
At p₀ = 0.982 the value of Z is:
[tex]Z=\frac{\hat p-p_{0}}{\sqrt{\frac{p_{0}(1-p_{0})}{n}}}=\frac{0.96-0.982}{\sqrt{\frac{0.982(1-0.982)}{100}}}=-1.65474[/tex]
At p₀ = 0.982 the value of Z is -1.65474.
Z = -1.65474 > Z₀.₀₅ = -1.645
Thus, at p₀ = 0.982 the manufacturer's claim would not be rejected.
