A survey of 170 commuters shows that on average people travel 17.4 km with standard deviation...

A survey of 170 commuters shows that on average people travel 17.4 km with standard deviation 8.4 km. Find 99% confidence interval for the population mean commuting distance. State your conclusion.

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Expert Solution

Solution :

Given that,

Point estimate = sample mean = = 17.4
Population standard deviation = = 8.4
Sample size = n =170

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576 ( Using z table )

Margin of error = E = Z/2* ( / $\sqrt$ n)

= 2.576* (8.4 / $\sqrt$ 170)

= 1.66

At 99% confidence interval is,

- E < < + E

17.4-1.66 < < 17.4+1.66

15.74< < 19.06

(15.74 , 19.06 )

lower limit 15.74

lower limit 19.06

sample mean lies between these confidence interval

orchestra answered 1 year ago

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