Question

Suppose for a random sample of 49 CMU students, the mean amount of time spent eating or drinking per day is 1.58 hours with a S.D. of 0.62 hours!

• What is the Margin of Error at a C.L. of 90%?
• What is the right boundary of the C.I. at a C.L. of 90%?

Answer 1

Solution :

Given that,

= 1.58

s = 0.62

n =49

Degrees of freedom = df = n - 1 = 49 - 1 = 48

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t _/2,df = t_0.05,48 = 1.677

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

= 1.677 * (0.62 / 49)

= 0.148

Margin of error = 0.15

The 90% confidence interval estimate of the population mean is,

- E < < + E

1.58 - 0.15 < < 1.58 + 0.15

1.43 < < 1.73

Right boundary = 1.73