Basic Stats/Probability Question.

TheSetMiner

A random sample of 16 people from Dublin had an average weight
of 70 kgs. The standard deviation of this sample was 8 kgs. Which
of the following is a 95% confidence interval for the average weight
of all people in Dublin if we assume their weights have a normal
distribution?

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So in this case is the S.D 8 or is it 8/sqrt(16) = 2?

the 95% C.I corresponds to 1.96 z value, right?

So then the answer should 70+- 2(1.96)

ie 70+- 3.92

For some reason, the software is telling me this answer is incorrect. Anyone any ideas? seemed pretty basic at first glance.

Ompala

What answer is the software giving you as a matter of interest?

TheSetMiner

Ompala wrote: »

What answer is the software giving you as a matter of interest?

Not giving an answer. Merely telling me I am incorrect within like 4% error tolerance.

Ompala

TheSetMiner wrote: »

Not giving an answer. Merely telling me I am incorrect within like 4% error tolerance.

Ah ok, the thing is though that since your sample size is smaller than 30 you wouldn't be using z tables, you would use t tables. Are you a leaving cert student?

TheSetMiner

Ompala wrote: »

Ah ok, the thing is though that since your sample size is smaller than 30 you wouldn't be using z tables, you would use t tables. Are you a leaving cert student?

No. I am a college student.
We have touched on t tables but I am not clear on when to use them vs when to use z tables.

Also, is the standard deviation here 8 or 2, can you confirm?

Ompala

TheSetMiner wrote: »

No. I am a college student.
We have touched on t tables but I am not clear on when to use them vs when to use z tables.

Also, is the standard deviation here 8 or 2, can you confirm?

You use t tables when the sample size is less than 30 as they allow for greater variation in your data. If sample size is greater than 30 use z tables.

The standard deviation of the sample is 8, however for our confidence interval we use the standard error of the standard deviation, s/sqrt(n) = 8/sqrt(16) = 8/4 = 2.

Your confidence interval should be 70 +- t(2), you have to find t using your t tables

TheSetMiner

Ah. Wow, that has clicked just now. The whole degrees of freedom t score chart makes sense to me finally. After a quick google I found that the t score at 95% confidence interval with 15 degrees of freedom is 2.131, which if multiplied by 2 gives 4.262 which the software has now accepted!

Thanks man!

Ompala

TheSetMiner wrote: »

Ah. Wow, that has clicked just now. The whole degrees of freedom t score chart makes sense to me finally. After a quick google I found that the t score at 95% confidence interval with 15 degrees of freedom is 2.131, which if multiplied by 2 gives 4.262 which the software has now accepted!

Thanks man!

No problem, I have a test in statistics at the end of this semester so I'm glad this stuff is making sense to me