Answer:
Step-by-step 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±[tex]Z_{1-\alpha /2}*\sqrt{\frac{p'_1(1-p'_1)}{n_1} }[/tex]
[tex]Z_{1-\alpha /2}= Z_{0.95}= 1.648[/tex]
0.37±1.648*[tex]\sqrt{\frac{0.37*0.63}{1000} }[/tex]
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±[tex]Z_{1-\alpha /2}*\sqrt{\frac{p'_2(1-p'_2)}{n_2} }[/tex]
0.48±1.648* [tex]\sqrt{\frac{0.48*0.52}{1000} }[/tex]
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: [tex]\sqrt{\frac{p'_1(1-p'_1)}{n_1} } = \sqrt{\frac{0.37*0.63}{1000} } = 0.01527 = 0.015[/tex]
Seniors: [tex]\sqrt{\frac{p'_2(1-p'_2)}{n_2} } = \sqrt{\frac{0.48*0.52}{1000} }= 0.01579 = 0.016[/tex]
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.
