[AShiraz]

So on page 518 the solution states that c = 2.58 when I see it as 2.756.

v = n-1 = 29
p = 1-0.99/2 = 0.005

c = 2.756?

p.s.Also the forum gives difficulties with logging in. It takes you to some other page and then makes you go all over the place. Had to log in several times.

[jr.larsen]

It looks like a normal distribution has been used to find c.

c represents the number of standard deviations above the mean your value should be so that the probability of being below that value is 0.995.

If you scan the normal table at the back of the book for 0.995, you'll find z is around 2.58.

Page 504 also has a table on the topic.

[AShiraz]

It looks like a normal distribution has been used to find c.

c represents the number of standard deviations above the mean your value should be so that the probability of being below that value is 0.995.

If you scan the normal table at the back of the book for 0.995, you'll find z is around 2.58.

Page 504 also has a table on the topic.

Ok I see it is normal because n >=30 (or 30). If it was less than 30 then I would have used t distribution.

Thanks!