1. a survey of 30 random shoppers found a 95% confidence interval that shoppers

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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