S1 - Confidence intervals

I'm stuck on the second part of the question - here it is:

Rice that can be cooked in microwave ovens is sold in packets which the manufacturer claims contain a mean weight of more than 250 grams of rice.

The weight of rice in a packet may be modelled by a normal distribution.

A consumer organisation’s researcher weighed the contents, x grams, of each of a random sample of 50 packets. Her summarised results are:

\bar{x}=251.1

and

\sum(x-\bar{x})^2=184.5

(a) Show that, correct to two decimal places,

s=1.94

, where

s^2

denotes the unbiased estimate of the population variance.
.. Which I've already shown

(b) Construct a 96% confidence interval for the mean weight of rice in a packet, giving the limits to one decimal place.

So in this part, I know that the 96% equals 0.96 in probability. However, I need to find the z score so I can use the confidence interval formula, but I don't know how to find it. I thought I should use z = x-μ divided by σ but it doesn't make sense. So I looked at the z score tables for 0.96 - this gives me 0.83147, which isn't right because the mark scheme says that z=2.0537. How do I find z? Many thanks in advance.

Reply 1

Hopple

18

I'm not sure how you'd get z to so many decimal places (unless you have super tables), but there are two things I think you've missed. Firstly is you want 0.98 rather than the 0.96 you mention, can you think why? And the other is I think you've looked for 0.96 within the table, i.e. you've set Z=0.96, rather than looking for what Z gives the probability 0.96 (and again, all those 0.96s should be 0.98 :tongue:

).

Reply 2

Prestoria

OP

9

Original post by Hopple

I'm not sure how you'd get z to so many decimal places (unless you have super tables), but there are two things I think you've missed. Firstly is you want 0.98 rather than the 0.96 you mention, can you think why? And the other is I think you've looked for 0.96 within the table, i.e. you've set Z=0.96, rather than looking for what Z gives the probability 0.96 (and again, all those 0.96s should be 0.98 :tongue:

).

Oh, you're right - I checked 0.98 in the tables and it gives 2.0537 (the formula book from AQA gives to 4 decimal places). Just me being careless.. thanks very much!

Reply 3

Jasprar

Sorry, Why is it 0.98 and not 0.96? Thanks. :smile:

Reply 4

maggiehodgson

14

I have to say that I don't understand why BUT whatever the confidence interval % is, you subtract it from 100, divide the result by 2 and add it back on to the original percentage. No idea why.

So if you wanted 98% CI you would look up 99 in the table and for 99% CI you'd look up 99.5.

If anyone knows the reason why I'd be pleased to hear.

Reply 5

Music99

4

Original post by maggiehodgson

I have to say that I don't understand why BUT whatever the confidence interval % is, you subtract it from 100, divide the result by 2 and add it back on to the original percentage. No idea why.

So if you wanted 98% CI you would look up 99 in the table and for 99% CI you'd look up 99.5.

If anyone knows the reason why I'd be pleased to hear.

Because you are going either way, you want the upper and lower part of the interval so you divide by 2 :smile:

.

Reply 6

maggiehodgson

14

Thanks for trying to help but I'm so unclued-up about statistics that what you've told me means nothing to me. But what I'll do now is go and read it up bearing in mind what you've said and see if I can cobble it together.

TSR is the best.

Reply 7

davros

16

Original post by maggiehodgson

Thanks for trying to help but I'm so unclued-up about statistics that what you've told me means nothing to me. But what I'll do now is go and read it up bearing in mind what you've said and see if I can cobble it together.

TSR is the best.

I haven't got a drawing package to hand, so here's a wordy explanation:

think about the Normal distribution and imagine marking out an interval on the x-axis centred on the mean which is going to be your 96% confidence interval for some measurement - in other words, you want to be 96% confident that your desired value lies in that interval.

That leaves you with a 4% "non-confident" bit left over, and because of the symmetry of the Normal distribution that means you have a 2% "bit" at the left before the start of the confidence interval, and a 2% "bit" at the right after the end of the interval.

Now forget about the confidence interval for a minute and pretend you're asked to work out that "2% bit" on the right. This is the same as 1 (which is total probability) minus the 98% bit to the left of it (which in other words is 98% cumulative probability which you can find in a table).

So, you can derive the "2% right bit" from a 98% cumulative probability value, and now you need to double this to account for the bit at the left hand too. This gives us our "area" that lies outside the 96% confidence interval.