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kati45 [8]
3 years ago
15

Dairy cows at large commercial farms often receive

Mathematics
1 answer:
IrinaK [193]3 years ago
3 0

Answer:

a) The 99% confidence interval for the true mean milk production is (26.35 kg/d, 29.65 kg/d).

b) They would need to use a confidence level of 94.52%.

Step-by-step explanation:

a)

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.99}{2} = 0.005

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.005 = 0.995, so z = 2.575

Now, find M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample. So

M = 2.575*\frac{2.25}{\sqrt{12}} = 1.65

The lower end of the interval is the mean subtracted by M. So it is 28-1.65 = 26.35 kg/d

The upper end of the interval is the mean added to M. So it is 28 + 1.65 = 29.65 kg/d.

The 99% confidence interval for the true mean milk production is (26.35 kg/d, 29.65 kg/d).

(b) If the farms want the confidence interval to be no wider than 1.25 kg/d, what level of confidence would they need to use?

Now we want to find the value of z when M = 1.25. So:

1.25 = z*\frac{2.25}{\sqrt{12}}

0.6495z = 1.25

z = 1.92, that has a pvalue of 0.9726.

So the confidence level should be 1 - 2*(1-0.9726) = 0.9452

They would need to use a confidence level of 94.52%.

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Step-by-step explanation:

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The procedure consists in finding an appropriate function that depends on only one variable. Then, the first derivative of the function will be found, equated to 0 and find the maximum or minimum values.

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