Normal Distribution Question Help

A car manufacturer introduces a new method of assembling a particular component. A sample of assembly times (minutes) taken after the new method had become established was:

27, 19, 68, 41, 17, 52, 35, 72, 38

a) Calculate a 99% Confidence Interval for the mean assembly time:

mean = 41

standard deviation = 19.723

n = 9

degrees of freedom = 8

using inverse t-distribution:

multiplier = 3.355

99% Confidence Interval = 41 +- 3.355 x (19.723/3)

= (18.943, 63.057)

b) State any assumptions it was necessary to make in order to calculate the confidence interval in a

Sample is random and population is normally distributed.

c) A larger random sample of 45 assembly times had a mean of 36.3 minutes with a standard deviation of 9.8 minutes:

n = 45, mean = 36.3, sd = 9.8. 99% CI:

work out using invNormal: area = 0.99, s.d. = 1, mean = 0.

multiplier = 2.326

99% CI = 36.3 +- 2.326 +- (9.8/(root 45))

99% CI = (32.902, 39.698)

answers for a) and b) are correct but answer for c is incorrect (32.5-40.1). What have I done wrong?

• You did not multiply but just added. Look at the formula for a CI interval Commented Jan 23, 2019 at 18:00
• Btw: You are aware that you used the standard error. So you get a CI interval for the 'true' mean. Commented Jan 23, 2019 at 18:01
• I meant 2.326 x (9.8/root 45) Commented Jan 23, 2019 at 18:04
• What is the question? And what is the answer of the book? Commented Jan 23, 2019 at 18:08
• I've wrote out the whole question and the answer in the book is stated at the end. Commented Jan 23, 2019 at 18:17