Suppose 1,600 of 2,000 registered voters sampled said they planned to vote for the Republican candidate for president. Using the 0.95 degree of confidence, what is the interval estimate for the population proportion (to the nearest 10th of a percent)? A. 78.2% to 81.8% B. 76.5% to 83.5% C. 77.7% to 82.3% D. 69.2% to 86.4%

option A

Step-by-step explanation:

Sample proportion (p') = 1600/2000 = 0.8

Sample size (n) = 2000

q' = 1 - p' = 1 - 0.8 = 0.2

To construct a 95% confidence level for population proportion, we use the formulae below.

p = p' + Zα/2 × s/√n For the upper limit

p = p' - Zα/2 × s/√n For the lower limit

Where s is the sample standard deviation and is given below as

s = √ p' (1-p')

s = √ 0.8 × 0.2

s = √ 0.16

s = 0.4

Even though we are making use of our sample standard deviation, the critical value use is that of a z test and that's because the sample size is greater than 30 ( n = 2000.)

Hence the value for critical value Zα/2 is 1.96.

By substituting the parameters, we have that

For lower limit

p = 0.8 - 1.96× 0.4/√2000

p = 0.8 - 1.96 ( 0.0089)

p = 0.8 - 0.018

p = 0.782 = 78.2%

For upper limit

p = 0.8 + 1.96× 0.4/√2000

p = 0.8 + 1.96 ( 0.0089)

p = 0.8 + 0.018

p = 0.818= 81.8%

Hence the 95% confidence interval for proportion is between 78.2% and 81.8%

answer: no, but if it does it's a obvious pattern.

step-by-step explanation: