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Leno4ka [110]
3 years ago
10

A null hypothesis is being tested. How does the confidence level affect the testing process?

Mathematics
1 answer:
never [62]3 years ago
4 0
The confidence level C and the significance level alpha are linked through the equation

alpha = 1-C

So for instance, if the confidence level is C = 95% = 0.95 then alpha is 

alpha = 1-C
alpha = 1-0.95
alpha = 0.05

meaning we have a 5% significance level. The larger C gets, the smaller alpha gets and vice versa. It turns out that 0 < C < 1 and also 0 < alpha < 1. 

The closer C gets to 1, the alpha value gets closer to 0. The smaller alpha gets, the harder it is to reject the null. Why is that? If we have a fixed p value, say p = 0.02 then we reject the null if alpha > pvalue. But we fail to reject the null when alpha < pvalue. For very small alpha values, we're going to fail to reject H0 no matter how small the pvalue is. The pvalue would have to be really small for H0 to be rejected.

In short, I'm saying that if the confidence level is high, then the chance of rejecting the null hypothesis is low (or rare)

This is why the answer is choice A
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A quality control engineer at a potato chip company tests the bag filling machine by weighing bags of potato chips. Not every ba
jekas [21]

Answer:

The test statistic is z = 1.865.

Step-by-step explanation:

The test statistic is:

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

In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation and n is the size of the sample.

H0: p = 0.11

This means that 0.11 is tested at the null hypothesis, and so:

\mu = 0.11

\sigma = \sqrt{0.11*0.89} = 0.3129

The engineer weighs 94 bags and finds that 16 of them are over-filled.

This means that:

n = 94, X = \frac{16}{94} = 0.1702

What is the test statistic?

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

z = \frac{0.1702 - 0.11}{\frac{0.3129}{\sqrt{94}}}

z = 1.865

The test statistic is z = 1.865.

6 0
3 years ago
A model for tumor growth is given by the Gompertz equation dV dt = a (In b-, In V) V where a &gt;0 and b &gt;0 are constants and
Xelga [282]

Answer:

V(t) = b^{(1-e^{-at})}

Step-by-step explanation:

We are given the following information in the question:

\displaystyle\frac{dV}{dt} = a (In b - In V) V

where a > 0 and b > 0.

\displaystyle\frac{dV}{dt} = a (In b-In V) V\\\\\displaystyle\frac{dV}{dt} = -aV(ln\frac{V}{b})\\\\\frac{dV}{V(ln\frac{V}{b})} = (-a)dt\\\\\text{Put } ln\frac{V}{b} = z\\\\\text{Integrating both sides}\\\\\int \frac{dV}{V(ln\frac{V}{b})} = \int (-a)dt\\\\\text{We get}\\\\\int \frac{dz}{z} = \int (-a)dt\\\\\\\text{where C is the constant of integration}

V(0) = 1~ mm^3\\V(t) = b.e^{e^{-at+C}}\\\text{Putting t =0, V(0) = 1}\\\\V(0) = 1 = b.e^{e^{C}}\\\\V(t) = b^{(1-e^{-at})}

where v(t) is the required tumor volume as a function of time that has an initial tumor volume of V(0) = 1 cubic mm.

5 0
3 years ago
Can someone help me please ill give brainy
Gnom [1K]
Y = 6

Since there is no slope the equation would look like this y=0x+6

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6 0
3 years ago
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The local gasoline distributor just built a new storage tank. For safety reasons they need to enclose the tank within a circular
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Answer:

62.8 feet

Step-by-step explanation:

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6+4 = 10 feet

Circumference = 2×pi×r

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= 62.83185307 feet

3 0
3 years ago
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