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Whitepunk [10]
3 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]3 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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(a) Average time to get to school
Average time (minutes) = Summation of the two means = mean time to walk to bus stop + mean time for the bust to get to school = 8+20 = 28 minutes

(b) Standard deviation of the whole trip to school
Standard deviation for the whole trip = Sqrt (Summation of variances)

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Standard deviation for the whole trip = Sqrt (2^2+4^2) = Sqrt (20) = 4.47 minutes

(c) Probability that it will take more than 30 minutes to get to school
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P(X>30) = 1-P(X=30) = 1-0.3264 = 0.6736 = 67.36%

With actual average time to walk to the bus stop being 10 minutes;

(d) Average time to get to school
Actual average time to get to school = 10+20 = 30 minutes

(e) Standard deviation to get to school
Actual standard deviation = Previous standard deviation = 4.47 minutes. This is due to the fact that there are no changes with individual standard deviations.

(f) Probability that it will take more than 30 minutes to get to school
Z(x=30) = (mean - 30)/Sd = (30-30)/4.47 = 0/4.47 = 0

From Z table, P(x=30) = 0.5
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