Suppose a survey revealed that 20% of 481 respondents said they had in the past sold...

Suppose a survey revealed that 20% of 481 respondents said they had in the past sold unwanted gifts over the Internet. (a) Use the information to construct a 90% confidence interval for the population proportion who sold unwanted gifts over the Internet, rounding your margin of error to the nearest hundredth. (Round your answers to two decimal places.) ( , ) (b) Use the information to construct a 98% confidence interval for the population proportion who sold unwanted gifts over the Internet, rounding your margin of error to the nearest hundredth. (Round your answers to two decimal places.) ( , ) (c) Which interval has a higher probability of containing the true population proportion? the 90% confidence interval the 98% confidence interval (d) Which interval is narrower? the 90% confidence interval the 98% confidence interval

Expert Solution

Answer:

Given,

p = 0.2 => q = 0.8

n= 481

a) The 90% CI is = 0.2 +/- 1.645 * Sqrt(0.2*.80/481)

=.0.2 +/- 0.03

= (0.17, 0.23)

b) The 98% CI is = 0.2 +/- 2.33 * Sqrt(0.20*0.80/481)

= 0.2 +/- 0.04

= (0.16, 0.24)

c) The 98% has a higher probability of containing the true population proportion. The confidence amount (98%) is what we should look at

d) The 90% confidence interval is narrower

orchestra answered 1 year ago

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