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bonufazy [111]
3 years ago
5

Ricky withdrew a total of $175 from his bank account over 5 days. He withdrew the same amount each day. By how much did the amou

nt in his bank account change each day
Mathematics
1 answer:
kifflom [539]3 years ago
8 0

Answer:

$35

Step-by-step explanation:

175/5 = 35

have a nice day! (maybe get this double checked in case I did it wrong :D)

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The desired percentage of sio2 in a certain type of aluminous cement is 5.5. to test whether the true average percentage is 5.5
LekaFEV [45]
Given that t<span>he desired percentage of sio2 in a certain type of aluminous cement is 5.5. to test whether the true average percentage is 5.5 for a particular production facility, 16 independently obtained samples are analyzed. suppose that the percentage of sio2 in a sample is normally distributed with σ = 0.32 and that \bar{x}=5.24.

</span>
<span>To investigate whether this indicate conclusively that the true average percentage differs from 5.5.



Part A:

From the question, it is claimed that </span><span>t<span>he desired average percentage of sio2 in a certain type of aluminous cement is 5.5</span></span> and we want to test whether the information from the random sample <span>indicate conclusively that the true average percentage differs from 5.5.

Therefore, the null hypothesis and the alternative hypothesis is given by:

H_0:\mu=5.5 \\  \\ H_a:\mu\neq5.5



Part B:

The test statistics is given by:

z= \frac{\bar{x}-\mu}{\sigma/\sqrt{n}}  \\  \\ =\frac{5.25-5.5}{0.32/\sqrt{16}} \\  \\ = \frac{-0.25}{0.32/4} = -\frac{0.25}{0.08}  \\  \\ =-3.125



Part C:

The p-value is given by

P(z\ \textless \ -3.125)=1-P(z



Part D:

Because the p-value is less than the significant level α, we reject the null hypothesis and conclude that "</span><span>There is sufficient evidence to conclude that the true average percentage differs from the desired percentage."



Part E:

</span>If the true average percentage is μ = 5.6 and a level α = 0.01 test based on n = 16 is used, what is the probability of detecting this departure from H0? (Round your answer to four decimal places.)

The probability of detecting the departure from H_0 is given by

1-\phi\left(z_{1-\frac{\alpha}{2}}+ \frac{\mu_0-\mu_1}{\sigma/\sqrt{n}} \right)+\phi\left(-z_{1-\frac{\alpha}{2}}+ \frac{\mu_0-\mu_1}{\sigma/\sqrt{n}} \right) \\  \\ =1-\phi\left(z_{1-\frac{0.01}{2}}+ \frac{5.5-5.6}{0.32/\sqrt{16}} \right)+\phi\left(-z_{1-\frac{0.01}{2}}+ \frac{5.5-5.6}{0.32/\sqrt{16}} \right) \\  \\ =1-\phi\left(z_{1-0.005}+ \frac{-0.1}{0.32/4} \right)+\phi\left(-z_{1-0.005}+ \frac{-0.1}{0.32/4} \right)

=1-\phi\left(z_{0.995}+ \frac{-0.1}{0.08} \right)+\phi\left(-z_{0.995}+ \frac{-0.1}{0.08} \right) \\  \\ =1-\phi(2.576-1.25)+\phi(-2.576-1.25) \\  \\ =1-\phi(1.326)+\phi(-3.826) \\  \\ =1-0.90758+0.00007 \\  \\ =0.0925



Part F:

What value of n is required to satisfy α = 0.01 and β(5.6) = 0.01? (Round your answer up to the next whole number.)

The value of n is required to satisfy α = 0.01 and β(5.6) = 0.01 is given by

n=\left[ \frac{\sigma(z_{0.005}+z_{0.01})}{\mu_0-\mu} \right]^2 \\  \\ = \left[\frac{0.32(-2.576-2.326)}{5.5-5.6} \right]^2 \\  \\ =\left[\frac{0.32(-4.902)}{-0.1} \right]^2=\left[\frac{-1.56864}{-0.1} \right]^2 \\  \\ =(15.6864)^2=247
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