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In: Statistics and Probability

The research firm conducted a survey to determine the average mean amount of money smokers spend...

The research firm conducted a survey to determine the average mean amount of money smokers spend on cigarettes per week. A sample of 19 smokers reveals a mean and standard deviation of $20 and $5 respectively. find the intercept the 99% confidence interval for the mean.

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Expert Solution

Solution: n = 19 , x̄ = 20 , s = 5, df = n-1 = 18

C = 99% , using t distribution table , tc = 2.878

Margin of error = E = tc(s/√n) = 2.878(5/√19) = 3.3013

Lower Limit = x̄ - E = 20 - 3.3013 = 16.6987

Upper Limit = x̄ - E = 20 + 3.3013 = 23.3013

The 99% confidence interval for the mean

(16.6987 , 23.3013)

orchestra answered 2 years ago
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