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In Summer 2016, a consulting firm surveyed 100 UMD students for their selections of presidential candidates. Of the 100 stude
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Answer #1

Question 1

Confidence interval for Population Proportion is given as below:

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Where, P is the sample proportion, Z is critical value, and n is sample size.

We are given

x = 21

n = 100

P = x/n = 21/100 = 0.21

Confidence level = 95%

Critical Z value = 1.96

(by using z-table)

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Confidence Interval = 0.21± 1.96* sqrt(0.21*(1 – 0.21)/100)

Confidence Interval = 0.21 ± 1.96* 0.0407

Confidence Interval = 0.21± 0.0798

Lower limit = 0.21 - 0.0798 = 0.130

Upper limit = 0.21 + 0.0798 = 0.290

Answer: 13.0% to 29.0%

Question 2

Confidence interval for Population Proportion is given as below:

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Where, P is the sample proportion, Z is critical value, and n is sample size.

We are given

x = 16

n = 100

P = x/n = 16/100 = 0.16

Confidence level = 95%

Critical Z value = 1.96

(by using z-table)

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Confidence Interval = 0.16 ± 1.96* sqrt(0.16*(1 – 0.16)/100)

Confidence Interval = 0.16 ± 1.96* 0.0367

Confidence Interval = 0.16 ± 0.0719

Lower limit = 0.16 - 0.0719 = 0.088

Upper limit = 0.16 + 0.0719 = 0.232

Answer: 8.8% to 23.2%

Question 3

Answer: Yes

Because 95% confidence interval for Hilary exceeds the 95% confidence interval for Donald Trump.

Question 4

Based on previous discussion, Hilary Clinton will win.

So, we have estimate for p = 0.21, q = 1 – 0.21 = 0.79

Margin of error = E = 2% = 0.02

Confidence level = 95%

Critical Z value = 1.96

(by using z-table)

Sample size formula is given as below:

n = p*q*(Z/E)^2

n = 0.21*0.79*(1.96/0.02)^2

n = 1593.304

Required sample size = n = 1594

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