Question

You measure 22 watermelons' weights, and find they have a mean weight of 36 ounces. Assume the population standard deviation is 12.6 ounces. Based on this, what is the maximal margin of error associated with a 99% confidence interval for the true population mean watermelon weight. Give your answer as a decimal, to two places ± ± ounces

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 36

Population standard deviation =    = 12.6

Sample size = n =22

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576   ( Using z table )

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

= 2.576* ( 12.6/ 22)

= 6.92

At 99% confidence interval is,

±  E

36± 6.92

answer 36± 6.92

note (29.08 ,42.92)