A survey of an urban university (population of 25,450) showed that 870 of 1,100 students sampled supported a fee increase to fund improvements to the student recreation center. Using the 95% level of confidence, what is the confidence interval for the proportion of students supporting the fee increase? A. [0.714, 0.866] B. [0.759, 0.822] C. [0.767, 0.815] D. [0.771, 0.811]

95% confidence interval for the proportion of students supporting the fee increase is [0.767, 0.815]. Option C

Step-by-step explanation:

The confidence interval for a proportion is given as [p +/- margin of error (E)]

p is sample proportion = 870/1,100 = 0.791

n is sample size = 1,100

confidence level (C) = 95% = 0.95

significance level = 1 - C = 1 - 0.95 = 0.05 = 5%

critical value (z) at 5% significance level is 1.96.

E = z × sqrt[p(1-p) ÷ n] = 1.96 × sqrt[0.791(1-0.791) ÷ 1,100] = 1.96 × 0.0123 = 0.024

Lower limit of proportion = p - E = 0.791 - 0.024 = 0.767

Upper limit of proportion = p + E = 0.791 + 0.024 = 0.815

95% confidence interval for the proportion of students supporting the fee increase is between a lower limit of 0.767 and an upper limit of 0.815.

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step-by-step explanation:

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