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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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A plate (A-C) is connected to steelflat bars by pinsat A and B. Member A-E consists of two 6mm by 25mm parallel flat bars. At C,
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Answer:

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Explanation:

Given:

- The missing figure is in the attachment.

- The dimensions of member AC = ( 6 x 25 ) mm x 2

- The diameter of the pin d = 19 mm

- Load at point A is P = 2 kN

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-  Find the axial stress in AE and the shear stress in pin C.

Solution:

- The stress in member AE can be calculated using component of force P along the member AE  as follows:

                                    stress_ac = P*cos(Q) / A_ae

Where, Angle Q: A_E_B   and A_ac: cross sectional area of member AE.

                                    cos(Q) = 4 / 5   ..... From figure ( trigonometry )

                                    A_ae = 0.006*0.025*2 = 3*10^-4 m^2

Hence,

                                    stress_ae = 2*(4/5) / 3*10^-4

                                    stress_ae = 5.333 MPa

- The force at pin C can be evaluated by taking moments about C equal zero:

                                   (M)_c = P*6 - F_eb*3

                                      0 = P*6 - F_eb*3

                                      F_eb = 0.5*P

- Sum of horizontal forces for member AC is zero:

                                      P - F_eb - F_c = 0

                                      F_c = 0.5*P

- The shear stress of double shear bolt is given by an expression:

                                     shear stress = shear force / 2*A_pin

Where, The area of the pin C is:

                                     A_pin = pi*d^2 / 4

                                     A_pin = pi*0.019^2 / 4 = 2.8353*10^-4 m^2

Hence,

                                     shear stress = 0.5*P / 2*A_pin

                                     shear stress = 0.5*2 / 2*2.8353*10^-4

                                    shear stress = 1.763 MPa

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