Confidence Interval help

2

1 The contents of jars of honey may be assumed to be normally distributed. The contents, in grams, of a random sample of 8 jars were as follows:

458, 450, 457, 456, 460, 459, 458, 456

a) Calculate a 95% confidence interval for the mean contents of all jars:

95% CI:

x̅ = 456.75 Sx: 3.059 n = 8 ν = 7

using calculator: Multiplier = 2.365

95% CI = 456.75 +- 2.365 x (3.059)/root(8)

= (454.192, 459.308).

b) On each jar it states `contents 454 grams.' Comment on this statement using the given sample and your results to part a):

Evidence from part a) shows that the mean is above 454g, but some jars will contain less.

c) Given that the mean contents of all jars is 454 grams, state the probability that a 95% confidence interval calculated from the contents of a random sample of jars will not contain 454 grams.:

I know the probability is 0.05, but my question is why? Surely the probability of a jar content not lying in the confidence interval calculated in part (a) is 0.05 right? So why is the probability that X doesn't equal 454 0.05??

• Hint: you can answer (c) without doing any calculations. – whuber Feb 17 '19 at 16:22
• I know the answer is 0.05, but I can't tell why. There's 95% confidence the contents of the jar in grams will be in that interval right? So why is it 5% of it not being 454 grams considering 454 isn't within the interval in (a) and it's only 1 value, it's not a interval. – Sam Connell Feb 17 '19 at 16:58
• If this question is part of course work please include the self-study tag. – André.B Feb 17 '19 at 22:18
• It's not, I just want to understand the theory of it. And why it's 0.05 – Sam Connell Feb 17 '19 at 22:26
• From a t-procedure in R, I get the 95% CI $(454.1927,\, 459.3073),$ so your computation is essentially correct. As for (c), a 95% CI is derived from the statement $P\left(-2.365 < \frac{\bar X - \mu}{S/\sqrt{n}} < 2.365\right) = 0.95.$ Any 'probability' statement you make needs to be based on that. A CI based on another sample would have a different sample mean $\bar X,$ a different sample SD $S,$ and possibly a different $n.$ Once $\bar X$ and $S$ are observed, some statisticians object to making 'probability' statements about $\mu$--hence the word 'confidence'. – BruceET Feb 17 '19 at 23:57