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BabaBlast [244]
3 years ago
9

When you construct a confidence interval for the difference between two proportions, what do you use as an unbiased estimate of

the difference between the two proportions?
Mathematics
1 answer:
Liono4ka [1.6K]3 years ago
3 0

Answer:

Thus difference in sample proportions would be the unbiased estimate of the difference between the two proportions

Step-by-step explanation:

Given that you construct a confidence interval for the difference between two proportions

Let the proportions be p1 and p2

Sample sizes be n1 and n2

For samples of larger size randomly drawn we know that

p1 follows Normal with mean = sample proportion = p1 and

p2 is N with mean = sample proportion p2

Assuming these two proportions are independent we have

p1-p2 will follow a normal distribution with mean = p1-p2

Thus difference in sample proportions would be the unbiased estimate of the difference between the two proportions

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Pioneer company has given an aptitude test to 71 potential job applicants. The test score had an average of 81 and a standard de
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Answer:

Follows are the solution to the given points:

Step-by-step explanation:

In point A:

\to df = n - 1 = 71-1= 70

In point B:

\to t = \frac{(x -X)}{s} = \frac{(93-81)}{8} = \frac{12}{8}= 1.5

In point C:

For df = 70, the top 5% critical t score

tcrit = 1.666914479

Thus,

\to 1.666914479 = \frac{(x - 81)}{8}\\\\\to 1.666914479 \times 8 = (x - 81)\\\\\to  13.335315832 = (x - 81)\\\\\to  13.335315832 +81 =x \\\\\to x= 94.335315832

In point D:

For df = 70, the top 5% critical t score

tcrit = -1.666914479

\to -1.666914479 = \frac{(x - 81)}{8}\\\\\to -1.666914479 \times 8= (x - 81)\\\\\to -13.335315832= (x - 81)\\\\\to -13.335315832+81 = x \\\\\to x = 67.664684168\\\\

In point E:

The lower cutoff is 0.10 in the center, which would be around 80 %. The critical point therefore is

tcrit = -1.293762898

\to -1.293762898 = \frac{(x-81)}{8}\\\\\to -1.293762898 \times 8= (x-81) \\\\\to -1.293762898 \times 8= (x-81) \\\\\to -10.350103184=x-81\\\\\to -10.350103184+81=x\\\\\to x=70.649896816

In point F:

The lower cutoff is 0.90 in the center, which would be around 80 %. The critical point therefore is

tcrit = 1.293762898

\to 1.293762898 = \frac{(x-81)}{8}\\\\\to 1.293762898 \times 8= x-81\\\\\to 10.350103184 =x-81\\\\\to 10.350103184 +81=x\\\\\to x = 91.350103184\\

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3 years ago
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