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Morgarella [4.7K]
3 years ago
11

A paint company produces glow in the dark paint with an advertised glow time of 15 min. A painter is interested in finding out i

f the product behaves worse than advertised. She sets up her hypothesis statements as H0 : µ ≤ 15 and Ha : µ > 15, then calculates a test statistic of z = −2.30. What would be the conclusions of her hypothesis test at significance levels of α = 0.05, α = 0.01, and α = 0.001?
Engineering
1 answer:
andrew11 [14]3 years ago
3 0

Answer:

p_v = P(z

Now we can decide based on the significance level \alpha. If p_v we reject the null hypothesis and in other case we FAIL to reject the null hypothesis.

\alpha=0.05 we see that p_v< \alpha so then we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly less than 15

\alpha=0.01 we see that p_v> \alpha so then we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is NOT significantly less than 15

\alpha=0.001 we see that p_v> \alpha so then we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is NOT significantly less than 15

Explanation:

For this case they conduct the following system of hypothesis for the ture mean of interest:

Null hypothesis: \mu \leq 15

Alternative hypothesis: \mu >15

The statistic for this hypothesis is:

z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And on this case the value is given z = -2.30

For this case in order to take a decision based on the significance level we need to calculate the p value first.

Since we have a lower tailed test the p value would be:

p_v = P(z

Now we can decide based on the significance level \alpha. If p_v we reject the null hypothesis and in other case we FAIL to reject the null hypothesis.

\alpha=0.05 we see that p_v< \alpha so then we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly less than 15

\alpha=0.01 we see that p_v> \alpha so then we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is NOT significantly less than 15

\alpha=0.001 we see that p_v> \alpha so then we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is NOT significantly less than 15

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Liquid flows at steady state at a rate of 2 lb/s through a pump, which operates to raise the elevation of the liquid 100 ft from
Greeley [361]

Answer:

D) 1.04 Btu/s from the liquid to the surroundings.

Explanation:

Given that:

flow rate (m) = 2 lb/s

liquid specific enthalpy at the inlet (h_{1}=40.09 Btu/lb)

liquid specific enthalpy at the exit (h_{2}=40.94 Btu/lb)

initial elevation (z_1=0ft)

final elevation (z_2=100ft)

acceleration due to gravity (g) = 32.174 ft/s²

W_{cv} = 3 Btu/s

The energy balance equation is given as:

Q_{cv}-W{cv}+m[(h_1-h_2)+(\frac{V_1^2-V_2^2}{2})+g(z_1-z_2)]=0

Since  kinetic energy effects are negligible, the equation becomes:

Q_{cv}-W{cv}+m[(h_1-h_2)+g(z_1-z_2)]=0

Substituting values:

Q_{cv}-(-3)+2[(40.09-40.94)+\frac{32.174(0-100)}{778*32.174} ]=0\\Q_{cv}+3+2[-0.85-0.1285 ]=0\\Q_{cv}+3+2(-0.9785)=0\\Q_{cv}+3-1.957=0\\Q_{cv}+1.04=0\\Q_{cv}=-1.04\\

The heat transfer rate is 1.04 Btu/s from the liquid to the surroundings.

8 0
3 years ago
Two mass streams of the same ideal gas are mixed in a steady-flow chamber while receiving energy by heat transfer from the surro
loris [4]

Answer:

(a)The final temperature of mixture is T₃ =m₁T₁/m₃+ m₂T₂/m₃ + Qin/m₃Cp

(b) The final volume is V₃ =V₁ + V₂ + RQin/P₃Cp

(c) The volume flow rate at exit is V₃ =V₁ + V₂

Explanation:

Solution

Now

The system comprises of two inlets and on exit.

Mass flow rate enthalpy of fluid from inlet -1 be m₁ and h₁

Mass flow rate enthalpy of fluid from inlet -2 be m₂ and h₂

Mass flow rate enthalpy of fluid from  exit be m₃ and h₃

Mixing chambers do not include any kind of work (w = 0)

So, both  the kinetic and potential energies of the fluid streams are usually negligible (ke =0, pe =0)

(a) Applying the mass balance of mixing chamber, min = mout

Applying the energy balance of mixing chamber,

Ein = Eout

min hin =mout hout

miCpT₁ + m₂CpT₂ +Qin =m₃CpT₃

T₃ = miCpT₁/m₃CpT₃ + m₂CpT₂/m₃CpT₃ + Qin/m₃CpT₃ +

T₃ =m₁T₁/m₃+ m₂T/m₃ + Qin/m₃Cp

The final temperature of mixture is T₃ =m₁T₁/m₃+ m₂T₂/m₃ + Qin/m₃Cp

(b) From the ideal gas equation,

v =RT/PT

v₃ = RT₃/P₃

The volume flow rate at the exit, V₃ =m₃v₃

V₃ = m₃ RT₃/P₃

Substituting the value of T₃, we have

V₃=m₃ R/P₃ (=m₁T₁/m₃+ m₂T₂/m₃ + Qin/m₃Cp)

V₃ =  R/P₃ (m₁T₁+ m₂T₂ + Qin/Cp)

Now

The mixing process occurs at constant pressure P₃=P₂=P₁.

Hence V₃ becomes:

V₃=m₁RT₁/P₁ +m₂RT₂/P₂ + RQin/P₃Cp

V₃ =V₁ + V₂ + RQin/P₃Cp

Therefore, the final volume is V₃ =V₁ + V₂ + RQin/P₃Cp

(c) Now for an adiabatic mixing, Qin =0

Hence V₃ becomes:

V₃ =V₁ + V₂ + r * 0/P₃Cp

V₃ =V₁ + V₂ + 0

V₃ =V₁ + V₂

Therefore the volume flow rate at exit is V₃ =V₁ + V₂

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which of the following tools is used for measuring small diameter holes which a telescoping gauge cannot fit into? A. telescopin
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Answer:

C. Dial indicator

Explanation:

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Design a PI controller to improve the steady-state error. The system should operate with a damping ratio of 0.8. Compute the ove
blondinia [14]

Answer:

The MATLAB Code for this PI Controller will be:

Kp = 350;

Ki = 300;

Kd = 50;

C = pid(Kp,Ki,Kd)

T = feedback(C*P,1);

t = 0:0.01:2;

step(T,t)

Explanation:

When you are designing a PID controller for a given system, follow the steps shown below to obtain a desired response.

Obtain an open-loop response and determine what needs to be improved

Add a proportional control to improve the rise time

Add a derivative control to reduce the overshoot

Add an integral control to reduce the steady-state error

Adjust each of the gains $K_p$, $K_i$, and $K_d$ until you obtain a desired overall response.

The further explanation is attached in the Word File.

Download docx
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