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3 years ago

Which figure shows that

Mathematics

2 answers:

FromTheMoon [43]3 years ago

7 0

It’s either Option 1 or 4

Makovka662 [10]3 years ago

3 0

The answer option 4 im pretty sire

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The term that describes the restoration of the insured person to the financial position that he or she was in before the loss occurred is called indemnity. This allows protection to the insurer in case of loss and damage and will protect against any legal quandry that may occur.

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3 years ago

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3 years ago

5x-19≤1 OR -4x+3 -9/4 D) No solution E) All values of x are solutions

sukhopar [10]

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3 years ago

What are accepted without proof in a logical system?

aleksklad [387]

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4 years ago

For each shipment of parts a manufacturer wants to accept only those shipments with at most 10% defective parts. A large shipmen

Stella [2.4K]

Answer:

(a) The p-value of the test statistic should be 0.33.

(b) No, there is not sufficient evidence to reject the entire shipment.

Step-by-step explanation:

We are given that for each shipment of parts a manufacturer wants to accept only those shipments with at most 10% defective parts.

A quality control manager randomly selects 50 of the parts from the shipment and finds that 6 parts are defective.

Let p = proportion of defective parts among the current shipment.

So, Null Hypothesis, $H_0$ : p $\leq$ 10% {means that the defective parts in the shipment is at most 10%}

Alternate Hypothesis, $H_A$ : p > 10% {means that the defective parts in the shipment is greater 10%}

The test statistics that would be used here One-sample z proportion statistics;

T.S. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of parts that are defective among the current shipment = $\frac{6}{50}$ = 12%

n = sample of parts from the shipment = 50

So, test statistics = $\frac{0.12-0.10}{\sqrt{\frac{0.12(1-0.12)}{50} } }$

= 0.44

The value of z test statistics is 0.44.

(a) Now, P-value of the test statistics is given by the following formula;

P-value = P(Z > 0.44) = 1 - P(Z $\leq$ 0.44)

= 1 - 0.67003 = 0.3299 ≈ 0.33

(b) Since, the P-value of test statistics is more than the level of significance as 0.33 > 0.05, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the defective parts in the shipment is at most 10%.

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3 years ago