1. Calculate a 97% confidence interval for the proportion of customers who are "highly satisfied" with a product, based on a survey of 62 randomly selected customers of whom forty-one were "highly satisfied": write the appropriate decimal number in each blank, rounded to the nearest 0.01 if necessary.

 

_______ +/-______  *sqrt((______ )*(______ )/(___________))

 

 

2. Calculate a 90% confidence interval for the proportion of students who rate a class "very useful", given a random sample of 35 students, 28 of whom rated the class "very useful": write the apropriate decimal number in each blank, rounded to the nearest 0.01 if necessary.

_______ +/-______  *sqrt((______ )*(______ )/(___________))

 

3. A survey of 1,215 US Adults found that 20.08% of respondents worked from home at least once a week. Compute a 95% confidence interval for this proportion, rounding your answers to the nearest 0.0001.

 

_______ +/-______ 

 

 

4. A confidence interval is different than a point estimate because it...

a. is only useful when calculating proportion problems

b. involves mathematical calculations

c. includes a margin of error, giving a range of reasonable answers

d. does not consider uncertainty

AccountingQueen

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Status NEW Posted 16 Aug 2017 06:08 PM My Price 5.00

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