Question

You measure 29 turtles' weights, and find they have an average weight of 56 grams. Assume the population standard deviation is 5 grams. Based on this, what is the margin of error associated with a 90% confidence interval for the true population mean turtle weight. The sample size is sufficiently large that the normal distribution value can be used in deriving the margin of error.

Give your answer as a decimal, to two places

± ______ grams

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 56

Population standard deviation =    = 5
Sample size = n =29

At 90% confidence level the z is

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

Margin of error = E = Z/2    * ( /n)

= 1.645 * (5 /  29 )

=1.53

± 1.53__ grams