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In a survey of 616 males ages 18-64, 392 say they have

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In a survey of 616 males ages...
In a survey of
616
males ages 18-64,
392
say they have gone to the dentist in the past year.
Construct 90% and 95% confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals.
Submitted: 3 years ago.Category: Calculus and Above
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6/18/2015
Tutor: Ashok Kumar, Electrical Engineer replied 3 years ago
Ashok Kumar
Ashok Kumar, Electrical Engineer
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Customer reply replied 3 years ago
2 hrs
Tutor: Ashok Kumar, Electrical Engineer replied 3 years ago
Here p = 392/616 = 0.6363
standard deviation = sqrt[p(1-p)/n]
= sqrt(0.6363*0.3636/616)
= 0.01938
Confidence interval: p±z*standard deviation
For 90% confidence interval, z = 1.645
Hence, 90% confidence interval: (0.6363±1.645*0.01938)
: (0.6044, 0.6682)
For 95% confidence interval, z = 1.96
Hence, 95% confidence interval: (0.6363±1.96*0.01938)
: (0.5983, 0.6743)
90% confidence interval means that 90 times out of 100, the probability will be between 0.6044 and 0.6682. Similarly 95% confidence means.
95% confidence interval is wider.
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Tutor: Ashok Kumar, Electrical Engineer replied 3 years ago
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