Question

In a random sample of 18 people, the mean commute time to work was 33.1 minutes and the standard deviation was 7.2 minutes. Assume the population is normally distributed and use a t-distribution to construct a 98% confidence interval for the population mean mu . What is the margin of error of mu? Interpret the results.

The confidence interval for the population mean mu is:

The margin of error of mu is:

Answer #1

Solution :

degrees of freedom = n - 1 = 18 - 1 = 17

t/2,df
= t_{0.01,17} = 2.567

Margin of error = E = t_{/2,df}
* (s /n)

= 2.567 * ( 7.2 / 18)

Margin of error = E = 4.36

The 98% confidence interval estimate of the population mean is,

± E

= 33.1 ± 4.36

= ( 28.74, 37.46 )

We are 98% confidence that the true mean commute time to work was between 28.74 and 37.46 minutes.

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