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Suppose you asked 100 commuters how much they spend each year and obtained a mean of $167 spent on transportation and a standard deviation of $40. Using the 2 SE rule of thumb, calculate a 95% confidence interval for the mean and select the values that come closest to those that would fill the spaces in the following interpretation: we can be 95% confident that the mean amount of money spent on transportation lies between _________ and _________.A. $149 and $185 B. $163 and $171 C. $163 and $170 D. $155 and $212

User LoVo
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Answer:

B. $163 and $171

Explanation:

from the question, we were given the following:

mean= $167

standard deviation, =$40

sample size, n = 100

significance level, α= i- confidence level= 1- 0.95=0.05

from the z table, we get;

critical value,
Z_(\alpha/2 ) = Z_(0.025) = 1.96

error margin = critical value ×
(standard \ deviation)/(√(sample\ size) )

= 1.96×
(40)/(√(100) ) = 7.84

thus lower limit = mean - error margin = $167 - $7.84 =159.16

the upper limit = mean + error margin = $167 + $7.84 = $174.84

the closest is B. $163 and $171

User Joe Shamuraq
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