1. a survey of 30 random shoppers found a 95% confidence interval that shoppers
ID: 3259492 • Letter: 1
Question
1. a survey of 30 random shoppers found a 95% confidence interval that shoppers will spend between $24.50 and $28.30. A second survey surveyed 25 random shoppers. What is the standard deviation of the sampling distribution? Round to 3 decimal digits.
2. a survey of 582 citizens found that 391 of them favor a new bill introduced by the city. We want to find a 95% confidence interval for the true proportion of the population who favor the bill. What is the lower limit of the interval? (round to 3 decimal digit).
Explanation / Answer
1)
margin of error = (28.3-24.5)/2 = 1.9
standard deviation/sqrt(n) * z-score = 1.9
standard deviation = 1.9*sqrt(30)/1.96
= 5.30956
standard deviation of sample = 5.30956/sqrt(25) = 1.062
2)
p = 391/582
standard error = sqrt(391/582 * 191/582 * 1/582) = 0.019463
lower limit = mean - z * standard error
= 391/582 - 1.96 * 0.019463
= 0.634
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