Question

In a survey of 1118 U.S. adults conducted in 2012 by the Financial Industry Regulatory Authority, 810 said they always pay their credit cards in full each month. Construct a 99.8% confidence interval for the proportion of U.S. adults who pay their credit cards in full each month. Round the answers to three decimal places. A 99% confidence ineral for the proportion of U.S. adults who pay their credit cards in full each month is ____________< p < ________________

Answer 1

Answer:

99.8% confidence:

[0.6833, 0.7656]

99% confidence:

0.6902 < p < 0.7588

Step-by-step explanation:

Lets call p the probability that a U.S adult pay its credit catrd in full each month. Lets call Y the random variable that counts the total of persons that paid their credit card in full from a random sample of 1118 adults. Y is a random variable with distribution Y ≈ Bi(1118,p) . The expected value of Y is μ = 1118*p and its variance is σ² = 1118*p(1-p). The proportion of adults that paid their credit card is a random variable, lets call it X, obtained from Y by dividing by 1118. The expected value is p and the variance is p(1-p)/1118. The Central Limit Theorem states that X can be approximated by a random variable with Normal Distribution, with mean μ = p, and standard deviation σ = √(p(1-p)/1118).

The standarization of X, W, is a random variable with distribution (approximately) N(0,1) obtained from X by substracting μ and dividing by σ. Thus

$W = \frac{X - \mu}{\sigma} = \frac{X - p}{\sqrt{\frac{p(1-p)}{1118}}}$

If we want a 99.8% confidence interval, then we can find a value Z such that P(-Z < W < Z) = 0.998. If we do so, then P(W < Z) = 0.999, therefore, Ф(Z) = 0.999, were Ф is the cummuative distribution function of the standard normal distribution. The values of Ф can be found on the attached file. We can find that Ф(3.08) = 0.999, thus, Z = 3.08.

We have that P(-3.08 < W < 3.08) = 0.998, in other words

$P(-3.08 < \frac{X-p}{\sqrt{\frac{p(1-p)}{1118}}} < 3.08) = 0.998$

$P(-3.08 * \sqrt{\frac{p(1-p)}{1118}} < X-p < 3.08 * \sqrt{\frac{p(1-p)}{1118}}) = 0.998$

$P(-X -3.08 * \sqrt{\frac{p(1-p)}{1118}} < -p < -X + 3.08 * \sqrt{\frac{p(1-p)}{1118}}) = 0.998$

Taking out the - sign from the -p and reversing the inequalities, we finally obtain

$P(X -3.08 * \sqrt{\frac{p(1-p)}{1118}} < p < X + 3.08 * \sqrt{\frac{p(1-p)}{1118}}) = 0.998$

As a conclusion, replacing p by its approximation X, a 99.8% confidence interval for p is

$[X -3.08 * \sqrt{\frac{X(1-X)}{1118}}\, ,  \, X + 3.08 * \sqrt{\frac{X(1-X)}{1118}}]$

replacing X with the proportion of the sample, 810/1118 = 0.7245, we have that our confidence interval is

$[0.7245 -3.08 * \sqrt{\frac{0.7245(1-0.7245)}{1118}}\, ,  \, 0.7245 + 3.08 * \sqrt{\frac{0.7245(1-0.7245)}{1118}}]$

By solving the fraction and multypling by 3.08, we have

[0.7245 - 0.0411 < p < 0.7245 + 0.0411]

FInally, the 99.8 % confidence interval for p is

[0.6833, 0.7656]

If p is 0.99 (99% confidence), then we would want Z such that Ф(Z) = 0.995, by looking at the table we have that Z is 2.57, therefore the 99% interval for p is

$[X -2.57 * \sqrt{\frac{X(1-X)}{1118}}\, ,  \, X + 2.57 * \sqrt{\frac{X(1-X)}{1118}}]$

and, by replacing X by 0.7245 we have that  the 99% confidence interval is

[0.6902, 0.7588]

thus, 0.6902 < p < 0.7588

I hope i could help you!