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 deviation 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).

Answers

Answer 1

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

A Frequenter Of A Pub Had Observed That The New Barman Poured In Average 0.47 Liters Of Beer Into The

Related Questions

Please help me it’s for science I only have a few minutes

Answers

Answer:

Rocks

Explanation:

I am not sure tho bc they are made out of coal and I think coal is a kind of rock

Answer:

I'm taking a guess for ya I shall say Metal or and Minerals

Explanation:

Good luck

A continuous and aligned fiber-reinforced composite is to be produced consisting of 30 vol% aramid fibers and 70 vol% of a polycarbonate matrix; mechanical characteristics of these two materials are as follows:
Modulus of Elasticity [GPa] Tensile Strength [MPa] Aramid fiber 131 3600 Polycarbonate 2.4 65
Also, the stress on the polycarbonate matrix when the aramid fibers fail is 45 MPa. For this composite, compute the following:
(a) the longitudinal tensile strength, and
(b) the longitudinal modulus of elasticity

Answers

Answer:

1. 1111.5MPa

2. 56.1GPa

Explanation:

1. Longitudinal tensile stress can be obtained by obtaining the strength and volume of the fiber reinforcement. The derived formula is given by;

σcl = σm (1 - Vf) + σfVf

Substituting the figures, we will have;

45(1 - 0.30) + 3600(0.30)

45(0.70) + 1080

31.5 + 1080

= 1111.5MPa

2. Longitudinal modulus of elasticity or Young's modulus is the ability of an object to resist deformation. The derived formula is given by;

Ecl = EmVm + EfVf

Substituting the formula gives;

= 2.4 (1 - 0.30) + 131 (0.30)

= 2.4(0.70) + 39.3

= 16.8 + 39.3

= 56.1GPa

Using the appropriate relation, the longitudinal tensile stress and the longitudinal modulus are 1111.50 and 56.10 respectively.

Longitudinal tensile stress can be obtained using the relation :

σcl = σm (1 - Vf) + σfVf

Substituting the values into the relation:

45(1 - 0.30) + 3600(0.30)

45 × 0.70 + 1080

31.5 + 1080

= 1111.50 MPa

2.)

Longitudinal modulus of elasticity is obtained using the relation :

Ecl = EmVm + EfVf

Substituting the values thus :

2.4 (1 - 0.30) + 131 (0.30)

= 2.4 × 0.70 + 39.3

= 16.8 + 39.3

= 56.10 GPa

Hence, the longitudinal tensile stress and the longitudinal modulus are 1111.50 and 56.10 respectively.

Learn more : https://brainly.com/question/22664384

What causes the charging system warning lamp to go out when the engine starts up?
Select one:
a. It turns off when ground is supplied to the lamp.

b. It turns off because voltage is applied to both sides of the lamp.
c. It turns off automatically after about 5 seconds.
d. It turns off because voltage is applied to one side of the bulb and ground to the other side.

Answers

B is the answer I believe so

Most methods of transportation rely on some sort of infrastructure to drive, steer, navigate, or direct at some point or another in a journey. Which category of transportation system is least reliant on infrastructure?(1 point)

Answers

Answer:

Most methods of transportation rely on some sort of infrastructure to drive, steer, navigate, or direct at some point or another in a journey. Which category of transportation system is least reliant on infrastructure?(1 point). road

Explanation:

An interior beam supports the floor of a classroom in a school building. The beam spans 26 ft. and the tributary width is 16 ft. Dead load is 20 psf. Find:

a. Basic floor live load Lo in psf
b. Reduced floor live load L in psf
c. Uniformly distributed total load to the beam in lb/ft.
d. Compare the loading in part c with the alternate concentrated load requried by the Code. Which loading is more critical for bending, shear, and deflection.?

Answers

Answer:

a. [tex]L_o[/tex]  = 40 psf

b. L ≈ 30.80 psf

c. The uniformly distributed total load for the beam = 812.8 ft./lb

d. The alternate concentrated load is more critical to bending , shear and deflection

Explanation:

The given parameters of the beam the beam are;

The span of the beam = 26 ft.

The width of the tributary, b = 16 ft.

The dead load, D = 20 psf.

a. The basic floor live load is given as follows;

The uniform floor live load, = 40 psf

The floor area, A = The span × The width = 26 ft. × 16 ft. = 416 ft.²

Therefore, the uniform live load, [tex]L_o[/tex]  = 40 psf

b. The reduced floor live load, L in psf. is given as follows;

[tex]L = L_o \times \left ( 0.25 + \dfrac{15}{\sqrt{k_{LL} \cdot A_T} } \right)[/tex]

For the school, [tex]K_{LL}[/tex] = 2

Therefore, we have;

[tex]L = 40 \times \left ( 0.25 + \dfrac{15}{\sqrt{2 \times 416} } \right) = 30.80126 \ psf[/tex]

The reduced floor live load, L ≈ 30.80 psf

c. The uniformly distributed total load for the beam, [tex]W_d[/tex] = b × [tex]W_{D + L}[/tex] =

∴  [tex]W_d[/tex] =  = 16 × (20 + 30.80) ≈ 812.8 ft./lb

The uniformly distributed total load for the beam, [tex]W_d[/tex] = 812.8 ft./lb

d. For the uniformly distributed load, we have;

[tex]V_{max}[/tex] = 812.8 × 26/2 = 10566.4 lbs

[tex]M_{max}[/tex] =  812.8 × 26²/8 = 68,681.6 ft-lbs

[tex]v_{max}[/tex] = 5×812.8×26⁴/348/EI = 4,836,329.333/EI

For the alternate concentrated load, we have;

[tex]P_L[/tex] = 1000 lb

[tex]W_{D}[/tex] = 20 × 16 = 320 lb/ft.

[tex]V_{max}[/tex] = 1,000 + 320 × 26/2 = 5,160 lbs

[tex]M_{max}[/tex] =  1,000 × 26/4 + 320 × 26²/8 = 33,540 ft-lbs

[tex]v_{max}[/tex] = 1,000 × 26³/(48·EI) + 5×320×26⁴/348/EI = 2,467,205.74713/EI

Therefore, the loading more critical to bending , shear and deflection, is the alternate concentrated load

if you are running and you fall and everyone↓↓↓↓↓↓↓ passes you how can you still be in first place??
ik the answer but lets see if you know it twooo

Answers

They either stopped or you already won..? I am not sure I am not smart enough for this
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