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Whitepunk [10]
2 years ago
14

A university council wants to determine the need for a new staff room. The council conducted a survey of the staff by randomly s

peaking to 860 staff members out of 2,844 staff members. The council found that 70% of those surveyed were in favor of building a new staff room. Assuming a 90% confidence level, which statement holds true?
As the sample size is appropriately large, the margin of error is 0.036.


As the sample size is appropriately large, the margin of error is 0.026.


As the sample size is too small, the margin of error is 0.026.


As the sample size is too small, the margin of error cannot be trusted.

To find a margin of error for a sample proportion, these assumptions must be met, where n is the sample size and is the sample proportion.




First rewrite the proportion of positive responses as a decimal number.




Now substitute the sample proportion and the sample size, 860, into the formulas for the assumptions to determine whether they are met.




Both values are greater than 10, so the assumptions are met, and the margin of error can be calculated.


Find the margin of error using this formula, where z is the critical value.




For a 90% confidence level, . Substitute the values of z, n, and into the formula for margin of error and evaluate.
Mathematics
1 answer:
iogann1982 [59]2 years ago
7 0

Answer:

As the sample size is appropriately large, the margin of error is 0.026

Step-by-step explanation:

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Answer:

P=1/42.

Step-by-step explanation:

We know that the student council has 10 members where 5 of the members are Seniors. They need to choose a President, Vice President, Secretary and Treasurer. We calculate the probability that the President is a Senior:

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A research group needs to determine a 90% confidence interval for the mean repair cost for all car insurance small claims. From
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Answer:

a) z = 1.645

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Step-by-step explanation:

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

\alpha = \frac{1-0.9}{2} = 0.05

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 1-0.05 = 0.95, so z = 1.645, which means that the answer of question a is z = 1.645.

Now, find  the margin of error 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.

(b) If the group wants their estimate to have a maximum error of $12, how many small claims should they sample?

They should sample at least n small claims, in which n is found when

M = 12, \sigma = 124.88. So

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

12 = 1.645*\frac{124.88}{\sqrt{n}}

12\sqrt{n} = 205.43

\sqrt{n} = \frac{205.43}{12}

\sqrt{n} = 17.12

\sqrt{n}^{2} = (17.12)^{2}

n = 293

The should sample at least 293 small claims.

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