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# Confidence interval watch

1. Hi, can someone check for my answer? Because my answers do not agree with the answers given. So I'm not sure whether what I did is right or not. Thanks.

1) The numbers 24.4, 18.9, 12.8, 20.5, 19.1, 15.2, 21.7, 14.6 from a random sample of values of a normally distributed r.v. Find a 98% C.I. for the mean mu of X.

So the sample mean is 147.2/8 = 18.4
s² = [2816.56 - (147.2)²/8] / 7 = 15.44

So the 98% C.I.
= [18.4 ± 2.326sqrt(15.44/8) ]
= [15.17, 21.63]

But the answer given is [14.23, 22.57]

2) The mature weight of a certain breed of dog has a normal distribution. A sample of these dogs was taken and the following results found:
Sample size = 10
Sample mean = 3.84
Sample standard deviation = 0.4673
Construct a 95% C.I. for the mean mature weight of this type of dogs.

So 95% C.I.
= [3.84 ± 1.96*0.4673/sqrt(10) ]
= [3.55, 4.13]

But the answer given is [3.506, 4.174]
2. You don't know the true variance so you should be looking at the t distribution tables.
3. (Original post by SsEe)
You don't know the true variance so you should be looking at the t distribution tables.
But I thought if the distribution is normal then we don't use the t-distribution? I'm confused.
4. If the distribution is normal then you use the normal tables if you know the true variance and the t tables if you only know the sample variance.
5. (Original post by SsEe)
If the distribution is normal then you use the normal tables if you know the true variance and the t tables if you only know the sample variance.
Oh I see. thank you.

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