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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

Respuesta :

Answer:

B. $163 and $171

Step-by-step 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, [tex]Z_{\alpha/2 } = Z_{0.025} = 1.96[/tex]

error margin = critical value ×[tex]\frac{standard \ deviation}{\sqrt{sample\ size} }[/tex]

                     = 1.96×[tex]\frac{40}{\sqrt{100} }[/tex] = 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

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