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lesya [120]
3 years ago
6

You are told that a 95% confidence interval for a population proportion is (0.3775, 0.6225) What was the sample proportion that

lead to this confidence interval?. Also, what is the size of the sample used.
Mathematics
1 answer:
Maru [420]3 years ago
6 0
The mean of the confidence interval is (0.3775 + 0.6225) / 2 = 0.5. Therefore, the standard deviation of the proportion would have been sqrt[0.5*(1 - 0.5) / n], where n is the sample size. This expression simplifies to sqrt(0.25/n).

A 95% CI has a corresponding z = 1.96, so since the distance from 0.5 to 0.3775 (or 0.6225 to 0.5) is equal to 0.1225. Therefore, if we divide 0.1225 / 1.96 = 0.0625, we get the value of the SD, and this should be equal to sqrt(0.25/n).

0.0625 = sqrt(0.25/n)
n = 64
This means that the proportion was 0.5 and the sample size was 64.
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Answer:

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Step-by-step explanation:

The diagram showing the three pieces of paper of three digit in which two of the letters were covered can be seen below.

From the diagram; we have;

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Assuming the covered digits are _

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243 + 1_7 + _26 = 826

The objective is to determine the sum of the two unknown digits

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From ; 1_7 , we have the choice to pick from 0 - 9

Assuming the covered digit is 0; let's check if we are right

So; 243 + 1<u>0</u>7  +  _26 = 826

350 +   _26 = 826

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_26 = 476

So; the covered digit is in the position of 4 , but it doesn't tally together

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Let consider the covered digit to be 5 for example.

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The covered digit is in the position of 4

SO;

<u>4</u>26 = 426

Here , our assumption is right.

As such , the covered digits are 5 and 4

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Step-by-step explanation:

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Step-by-step explanation:

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