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Alex
3 years ago
11

A manufacturer finds that in a random sample of 100 of its CD players, 96% have no defects. The manufacturer wishes to make a cl

aim 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.
Mathematics
1 answer:
Marta_Voda [28]3 years ago
8 0

Answer:

At <em>p</em>₀ = 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, <em>p₀</em>.

The hypothesis to test this claim is defined as:

<em>H₀</em>: The proportion of defective is <em>p₀</em>, i.e. <em>p</em> = <em>p₀</em>.

<em>Hₐ</em>: The proportion of defective is less than <em>p₀</em>, i.e. <em>p</em> < <em>p₀</em>.

The significance level of the test is, <em>α</em> = 0.05.

The information provided is:

<em>n</em> = 100

\hat p = 0.96

The test statistic is:

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}}}

Decision rule:

If the test statistic value is less than the critical value of <em>z</em>, i.e. <em>Z</em>₀.₀₅ = -1.645 (because of the left tail), then the null hypothesis will be rejected. and vice versa.

So, to reject <em>H</em>₀<em> Z </em>≤ <em>Z</em>₀.₀₅.

Use hit and trial method.

At <em>p</em>₀ = 0.97 the value of <em>Z</em> is:

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

At <em>p</em>₀ = 0.98 the value of <em>Z</em> is:

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

At <em>p</em>₀ = 0.981 the value of <em>Z</em> is:

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

At <em>p</em>₀ = 0.982 the value of <em>Z</em> is:

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

At <em>p</em>₀ = 0.982 the value of <em>Z</em> is -1.65474.

<em>Z</em> = -1.65474 > <em>Z</em>₀.₀₅ = -1.645

Thus, at <em>p</em>₀ = 0.982 the manufacturer's claim would not be rejected.

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