Question

1.) A random sample of 25 college graduates found 18 were proficient with Microsoft Access. Determine a 95% confidence interval for the percent college graduate proficient with Microsoft Access.

Select one:

A. [0.854, 0.934]

B. 0.910, 0.978]

C. 0.544, 0.896]

D. 0.904, 0.984]

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2.)

A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than examine the records for all students, the dean randomly selects 200 students out of total of 4100 university students and finds that 118 of them are receiving financial aid. Use a 90% confidence interval to estimate the true proportion of students who receive financial aid.

Select one:

A. 0.59 + 0.0572

B. 0.59 + 0.0553

C. 0.59 + 0.0682

D. 0.59 + 0.0659

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3.) A confidence interval was used to estimate the proportion of statistics students that are females. A random sample of 72 statistics students generated the following 90% confidence interval: (0.438, 0.642). Which of the following interpretations is correct?

Select one:

A. 90% of the sampled students are female.

B. We are 90% confidence that proportion of all statistics female students falls in the interval 0.438 to 0.642

C. 90% of all statistic students are female.

D. We are 90% confident that the sample proportion of statistics female students falls in interval 0.438 to 0.642.

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4.) The Friday after Thanksgiving is the biggest shopping day of the year. You are interested in the number of people who claim to have finished their Christmas shopping at the end of this weekend. On Monday, you take a random sample of people by standing at a toll booth at 7:00 a.m. and asking every third commuter if he or she has finished Christmas shopping. Of the 67 commuters you poll, 14 claim to have finished Christmas shopping. A 90% confidence interval of the people in the surrounding communities who have finished Christmas shopping is

Select one:

A. [0.1116, 0.3064]

B. [0.1273, 0.2907]

C. [0.0616, 0.2566]

D. None of the above is the correct answer.

Answer 1

1)

sample proportion, = 0.72
sample size, n = 25
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.72 * (1 - 0.72)/25) = 0.0898

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

Margin of Error, ME = zc * SE
ME = 1.96 * 0.0898
ME = 0.176

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.72 - 1.96 * 0.0898 , 0.72 + 1.96 * 0.0898)
CI = (0.544 , 0.896)

2)

sample proportion, = 0.59
sample size, n = 200
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.59 * (1 - 0.59)/200) = 0.0348

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, Zc = Z(α/2) = 1.64

Margin of Error, ME = zc * SE
ME = 1.64 * 0.0348
ME = 0.0571

A. 0.59 + 0.0572

3)

D. We are 90% confident that the sample proportion of statistics female students falls in interval 0.438 to 0.642.

4)

sample proportion, = 0.209
sample size, n = 67
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.209 * (1 - 0.209)/67) = 0.0497

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, Zc = Z(α/2) = 1.645

Margin of Error, ME = zc * SE
ME = 1.645 * 0.0497
ME = 0.0818

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.209 - 1.645 * 0.0497 , 0.209 + 1.645 * 0.0497)
CI = (0.1273 , 0.2907)