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Read the paragraph and answer the question below: Results from two CNN/USA Today/Gallup polls, one conducted in March 2003 and one in November 2003, were recently presented online. Both polls involved samples of 1001 adults, aged 18 years and older. In the March sample, 45% of those sampled claimed to be fans of professional baseball whereas 51% of those polled in November claimed to be fans. Construct a 99% confidence interval for the proportion of adults who professed to be baseball fans in November 2003 (after the World Series). Interpret this interval.

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Answer:

99% confidence interval for the proportion of adults who professed to be baseball fans in November 2003 is (46.9%, 55.1%)

This interval means that the lower limit of the proportion of adults who professed to be baseball fans is 46.9% and the upper limit of the proportion is 55.1%

Step-by-step explanation:

Confidence interval = P' + or - t×sqrt[P'(1-P') ÷ n]

P' is sample proportion = 51% = 0.51

n = 1001

confidence level = 99%

t-value corresponding to 99% confidence interval and infinity degree of freedom is 2.576

t × sqrt[P'(1-P') ÷ n] = 2.576 × sqrt[0.51(1-0.51) ÷ 1001] = 2.576 × 0.0158 = 0.041

Lower limit = P' - 0.041 = 0.51 - 0.041 = 0.469 = 46.9%

Upper limit = P' + 0.041 = 0.51 + 0.041 = 0.551 = 55.1%

99% Confidence interval is (46.9%, 55.1%)

The interval means that the proportion of adults who professed to be baseball fans in November 2003 is between 46.9% and 55.1%

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