Question

The article "Students Increasingly Turn to Credit Cards" (San Luis Obispo Tribune, July 21, 2006) reported that 37% of college freshmen and 48% of college seniors carry a credit card balance from month to month. Suppose that the reported percentages were based on random samples of 1000 college freshmen and 1000 college seniors. a. Construct a 90% confidence interval for the proportion of college freshmen who carry a credit card balance from month to month. b. Construct a 90% confidence interval for the proportion of college seniors who carry a credit card balance from month to month.c. Explain why the two 90% confidence intervals from Parts (a) and (b) are not the same width.

Answer 1

Explanation:

Hello!

There are two variables of interest:

X₁: number of college freshmen that carry a credit card balance.

n₁= 1000

p'₁= 0.37

X₂: number of college seniors that carry a credit card balance.

n₂= 1000

p'₂= 0.48

a. You need to construct a 90% CI for the proportion of freshmen who carry a credit card balance.

The formula for the interval is:

p'₁±
`Z_(1-\alpha /2)*\sqrt{(p'_1(1-p'_1))/(n_1) }`

`Z_(1-\alpha /2)= Z_(0.95)= 1.648`

0.37±1.648*
`\sqrt{(0.37*0.63)/(1000) }`

0.37±1.648*0.015

[0.35;0.39]

With a confidence level of 90%, you'd expect that the interval [0.35;0.39] contains the proportion of college freshmen students that carry a credit card balance.

b. In this item, you have to estimate the proportion of senior students that carry a credit card balance. Since we work with the standard normal approximation and the same confidence level, the Z value is the same: 1.648

The formula for this interval is

p'₂±
`Z_(1-\alpha /2)*\sqrt{(p'_2(1-p'_2))/(n_2) }`

0.48±1.648*
`\sqrt{(0.48*0.52)/(1000) }`

0.48±1.648*0.016

[0.45;0.51]

With a confidence level of 90%, you'd expect that the interval [0.45;0.51] contains the proportion of college seniors that carry a credit card balance.

c. The difference between the width two 90% confidence intervals is given by the standard deviation of each sample.

Freshmen:
`\sqrt{(p'_1(1-p'_1))/(n_1) } = \sqrt{(0.37*0.63)/(1000) } = 0.01527 = 0.015`

Seniors:
`\sqrt{(p'_2(1-p'_2))/(n_2) } = \sqrt{(0.48*0.52)/(1000) }= 0.01579 = 0.016`

The interval corresponding to the senior students has a greater standard deviation than the interval corresponding to the freshmen students, that is why the amplitude of its interval is greater.