Question

For a random sample of 400 people, the mean cost for textbooks during the first semester of college was found to be $371.75, and the sample standard deviation was $38.44. Assuming that the population is normally distributed, find the margin of error of a 90% confidence interval for the population mean.

please with explanation and which table you have used.

Answer 1

solution :

Given that,

s =$38.44

n = 400

Degrees of freedom = df = n - 1 =400 - 1 = 399

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

t /2,df = t0.05,90 = 1.649    ( using student t table)

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

=1.649 * ( 38.44/ 400)

E= 3.1694

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