The article “Students Increasingly Turn to Credit Cards” (San Luis Obispo Tribune, July 21, 2006) reported that 37% of college freshmen and

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.

in progress 0
Athena 1 month 2021-10-18T13:19:50+00:00 1 Answer 0 views 0

Answers ( )

    0
    2021-10-18T13:21:31+00:00

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

    Z_{1-\alpha /2}= Z_{0.95}= 1.648

    0.37±1.648*\sqrt{\frac{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{\frac{p'_2(1-p'_2)}{n_2} }

    0.48±1.648* \sqrt{\frac{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{\frac{p'_1(1-p'_1)}{n_1} } = \sqrt{\frac{0.37*0.63}{1000} } = 0.01527 = 0.015

    Seniors: \sqrt{\frac{p'_2(1-p'_2)}{n_2} } = \sqrt{\frac{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.

Leave an answer

45:7+7-4:2-5:5*4+35:2 =? ( )