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Alisiya [41]
3 years ago
14

A frequenter of a pub had observed that the new barman poured in average 0.47 liters of beer into the glass with a standard devi

ation equal to 0.09 liters instead of a half a liter with the same standard deviation. The frequenter had used a random sample of 47 glasses of beer in his experiment. Consider the one-sided hypothesis test for volume of beer in a glass: H0: u=0.5 against H1: u<0.5. Determine the P-value of this test.
Round your answer to four decimal places (e.g. 98.7654).
Engineering
1 answer:
aalyn [17]3 years ago
6 0

Answer:

P-value = 0.0011

Explanation:

Formula for the test statistic is;

z = (x¯ - μ)/(σ/√n)

We have;

Sample mean;x¯ = 0.47

Population mean; μ = 0.5

Standard deviation; σ = 0.09

Sample size; n = 47

Thus;

z = (0.46 - 0.5)/(0.09/√47)

z = -3.05

From z-distribution table attached, the p-value corresponding to z = -3.05 is;

P = 0.00114

To four decimal places gives;

P-value = 0.0011

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A long corridor has a single light bulb and two doors with light switch at each door.
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Answer:

The answer is below

Explanation:

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A                    B                       C (output)

0                    0                        0

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Chemical milling is used in an aircraft plant to create pockets in wing sections made of an aluminum alloy. The starting thickne
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Answer:

a) metal removal rate is 1915.37 mm³/min

b) the time required to etch to the specified depth is 500 min or 8.333 hrs

Explanation:

Given the data in the question;

starting thickness of one work part of interest = 20 mm

depth of series of rectangular-shaped pockets = 12 mm

dimension of pocket = 200 mm by 400 mm

radius of corners of each rectangle = 15 mm

penetration rate = 0.024 mm/minute

etch factor = 1.75

a)

To get the metal removal rate MRR;

The initial area will be smaller compare to the given dimensions of 200mm by 400mm and the metal removal rate would increase during the cut as area is increased. so'

A = 200 × 400 - ( 30 × 30 - ( π × 15² ) )

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A = 79807 mm²

Hence, metal removal rate MRR = penetration rate × A

MRR = 0.024 mm/minute × 79807 mm²

MRR = 1915.37 mm³/min

Therefore, metal removal rate is 1915.37 mm³/min

b) To get the time required to etch to the specified depth;

Time to machine ( etch ) =  depth of series of rectangular-shaped pockets / penetration rate

we substitute

Time to machine ( etch ) = 12 mm / 0.024 mm/minute

Time to machine ( etch ) = 500 min or 8.333 hrs

Therefore, the time required to etch to the specified depth is 500 min or 8.333 hrs

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