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Charra [1.4K]
3 years ago
15

A company manufacturing computer chips finds that 8% of all chips manufactured are defective. In an effort to decrease the perce

ntage of defective chips, management decides to provide additional training to those employees hired within the last year. After training was implemented, a sample of 450 chips revealed only 27 defects. A hypothesis test is performed to determine if the additional training was effective in lowering the defect rate. Which of the following statement is true about this hypothesis test?
A. It is a two tailed test about a proportion.
B. It is a one tailed test about a mean.
C. It is a one tailed test about a proportion.
D. It is a two tailed test about a mean.
E. None of the above.
Mathematics
1 answer:
Lemur [1.5K]3 years ago
3 0

Answer:

A. It is a two tailed test about a proportion

Step-by-step explanation:

We are given;

Population proportion; p= 8% = 0.08

Sample size; n = 450

Defective ones in the sample = 27

Sample proportion; p^ = 27/450 = 0.06

Let's define the hypotheses.

Null Hypothesis; H0: p = 0.08

Alternative hypothesis;Ha: p ≠ 0.08

≠ is used for the alternative hypothesis because we can't really ascertain if the training was going to make things worse.

Thus, this is a two tailed hypothesis test about a proportion.

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Zanzabum

Answer:

\beta=45\degree\:\:or\:\:\beta=135\degree

Step-by-step explanation:

We want to solve \tan \beta \sec \beta=\sqrt{2}, where 0\le \beta \le360\degree.

We rewrite in terms of sine and cosine.

\frac{\sin \beta}{\cos \beta} \cdot \frac{1}{\cos \beta} =\sqrt{2}

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\frac{\sin \beta}{1-\sin^2\beta}=\sqrt{2}

\implies \sin \beta=\sqrt{2}(1-\sin^2\beta)

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This is a quadratic equation in \sin \beta.

By the quadratic formula, we have:

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\sin \beta=\frac{1}{\sqrt{2} } or \sin \beta=-\frac{2}{\sqrt{2} }

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When \sin \beta=\frac{\sqrt{2}}{2} , \beta=\sin ^{-1}(\frac{\sqrt{2} }{2} )

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A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 201.9-cm and a standard dev
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Answer:

There is a 0.08% probability that the average length of a randomly selected bundle of steel rods is greater than 204.1-cm.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

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In this problem, we have that:

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 201.9-cm and a standard deviation of 2.1-cm. This means that \mu = 201.9, \sigma = 2.1.

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This probability is 1 subtracted by the pvalue of Z when X = 204.1.

Z = \frac{X - \mu}{\sigma}

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Z = 3.14

Z = 3.14 has a pvalue of 0.9992. This means that there is a 1-0.9992 = 0.0008 = 0.08% probability that the average length of a randomly selected bundle of steel rods is greater than 204.1-cm.

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