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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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Bulk wind shear is calculated by finding the vector difference between the winds at two different heights. Using the supercell w
ivanzaharov [21]

Answer:

See explaination

Explanation:

2. 0-1 km shear value: taking winds at 1000mb and 850 mb

15 kts south easterly and 50 kts southerly

Vector difference 135/15 and 180/50 will be 170/61 or southerly 61 kts

3. 0-6 km shear value: taking winds at 1000 mb and 500 mb

15 kts south easterly and 40 kts westerly

Vector difference 135/15 and 270/40 will be 281/51 kts

please see attachment

5 0
3 years ago
A steel rod, which is free to move, has a length of 200 mm and a diameter of 20 mm at a temperature of 15oC. If the rod is heate
kherson [118]

Explanation:

thermal expansion ∝L = (δL/δT)÷L ----(1)

δL = L∝L + δT ----(2)

we have δL = 12.5x10⁻⁶

length l = 200mm

δT = 115°c - 15°c = 100°c

putting these values into equation 1, we have

δL = 200*12.5X10⁻⁶x100

= 0.25 MM

L₂ = L + δ L

= 200 + 0.25

L₂ = 200.25mm

12.5X10⁻⁶ *115-15 * 20

= 0.025

20 +0.025

D₂ = 20.025

as this rod undergoes free expansion at 115°c, the stress on this rod would be = 0

3 0
3 years ago
A 13.7g sample of a compound exerts a pressure of 2.01atm in a 0.750L flask at 399K. What is the molar mass of the compound?a. 3
Katarina [22]

Answer: Option D) 298 g/mol  is the correct answer

Explanation:

Given that;

Mass of sample m = 13.7 g

pressure P = 2.01 atm

Volume V = 0.750 L

Temperature T = 399 K

Now taking a look at the ideal gas equation

PV = nRT

we solve for n

n = PV/RT

now we substitute

n = (2.01 atm x 0.750 L) / (0.0821 L-atm/mol-K x 399 K )

= 1.5075 / 32.7579

= 0.04601 mol

we know that

molar mass of the compound = mass / moles

so

Molar Mass = 13.7 g / 0.04601 mol

= 297.7 g/mol  ≈ 298 g/mol

Therefore Option D) 298 g/mol  is the correct answer

4 0
2 years ago
A specimen of commercially pure copper has a strength of 240 MPa. Estimate its average grain diameter using the Hall-Petch equat
romanna [79]

Answer:

3.115× 10^{-3} meter

Explanation:

hall-petch constant for copper is given by

      S_0=25 MPa

      k=0.12 for copper

now according to hall-petch equation

S_Y=S_0 +\frac{K}{\sqrt{D}}

240=25+\frac{0.12}{\sqrt{D}}

D=3.115× 10^{-3} meter

so the grain diameter using the hall-petch equation=3.115×  10^{-3} meter

5 0
2 years ago
What 2 forces move the secondary piston ahead?
jekas [21]

Answer:

The primary piston activates one of the two subsystems. The hydraulic pressure created, and the force of the primary piston spring, moves the secondary piston forward.

5 0
3 years ago
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