Question

A manufacturer finds that in a random sample of 100 of its CD players, 96% have no defects. The manufacturer wishes to make a claim about the percentage of nondefective CD players and is prepared to exaggerate. What is the highest rate of nondefective CD players that the manufacturer could claim under the following condition? His claim would not be rejected at the 0.05 significance level if this sample data were used. Assume that a left-tailed hypothesis test would be performed.

Answer 1

At p₀ = 0.982 the manufacturer's claim would not be rejected.

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

`\hat p` = 0.96

The test statistic is:

`Z=\frac{\hat p-p_(0)}{\sqrt{(p_(0)(1-p_(0)))/(n)}}=\frac{0.96-p_(0)}{\sqrt{(p_(0)(1-p_(0)))/(100)}}`

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:

`Z=\frac{\hat p-p_(0)}{\sqrt{(p_(0)(1-p_(0)))/(n)}}=\frac{0.96-0.97}{\sqrt{(0.97(1-p\0.97))/(100)}}=-0.5862`

At p₀ = 0.98 the value of Z is:

`Z=\frac{\hat p-p_(0)}{\sqrt{(p_(0)(1-p_(0)))/(n)}}=\frac{0.96-0.98}{\sqrt{(0.98(1-0.98))/(100)}}=-1.4286`

At p₀ = 0.981 the value of Z is:

`Z=\frac{\hat p-p_(0)}{\sqrt{(p_(0)(1-p_(0)))/(n)}}=\frac{0.96-0.981}{\sqrt{(0.981(1-0.981))/(100)}}=-1.5382`

At p₀ = 0.982 the value of Z is:

`Z=\frac{\hat p-p_(0)}{\sqrt{(p_(0)(1-p_(0)))/(n)}}=\frac{0.96-0.982}{\sqrt{(0.982(1-0.982))/(100)}}=-1.65474`

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.