Question

In a survey, 101 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $41 and standard deviation of $12. Construct a confidence interval in the form estimate ± margin of errorestimate ± margin of error at a 90% confidence level.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 41

Population standard deviation =    = 12

Sample size = n = 101

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z_{0.05 =} 1.645

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

= 1.645 * ( 12 /  101 )

= 1.96

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

  ± E

41 ± 1.96   

( 39.04, 42.96 )