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A cell phone provider has the business objective of wanting to estimate the proportion of subscribers who would upgrade to a new cell phone with improved features if it were made available at a substantially reduced cost. Data are collected from a random sample of 500 subscribers. The results indicate that 120 of the subscribers would upgrade to a new cell phone at a reduced cost. Complete parts​ (a) and​ (b) below. a. Construct a 99​% confidence interval estimate for the population proportion of subscribers that would upgrade to a new cell phone at a reduced cost. Round to 4 decimal places.

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princessesther2011

Answer:

99% confidence interval estimate for the population proportion of subscribers that would upgrade to a new cell phone at a reduced cost is between a lower limit of 0.1908 and an upper limit of 0.2892

Explanation:

Confidence interval = p + or - zsqrt [p(1-p) ÷ n]

p is sample proportion = 120/500 = 0.24

n is the number of subscribers sampled = 500

Confidence level (C) = 99% = 0.99

Significance level = 1 - C = 1 - 0.99 = 0.01

To obtain z, divide the significance level by 2

0.01/2 = 0.005 = 0.5%

The critical value (z) at 0.5% significance level is 2.576

zsqrt[p(1-p) ÷ n] = 2.576sqrt[0.24(1-0.24) ÷ 500] = 2.576sqrt[0.1824 ÷ 500] = 2.576sqrt[3.648×10^-4] = 2.576×0.0191 = 0.0492

Lower limit = p - 0.0492 = 0.24 - 0.0492 = 0.1908

Upper limit = p + 0.0492 = 0.24 + 0.0492 = 0.2892

99% confidence interval is between 0.1908 and 0.2892

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