Question

A set of 29 students were sampled to see how many minutes they
traveled to get to school. The mean time spent was 17.69 minutes
with a standard deviation of 11.98 minutes. Assume that time spent
traveling is normally distributed. Find a 90% confidence interval
for the true mean travel time for all students at the school.

Answer #1

b )Given that,

= 17.69

s =11.98

n = 29

Degrees of freedom = df = n - 1 =29 - 1 = 28

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

t
/2,df = t0.05,28 = : **1.701** ( using
student t table)

Margin of error = E = t/2,df * (s /n)

= **1.701** * (11.98 /
29) = 3.78

The 90% confidence interval is,

- E < < + E

17.69 - 3.78 < < 17.69+3.78

13.91 < < 21.47

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