S2 question Sampling

swagmister

Could I have help with 6a please?

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(edited 7 years ago)

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Reply 1

Kevin De Bruyne

Original post by swagmister

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Could I have help with 6a please?

Unless you've misread the value from the table I see nothing wrong with your working

Reply 2

swagmister

Original post by SeanFM

Unless you've misread the value from the table I see nothing wrong with your working

The answers say it's 0.155 I tried 6b which was a similar question but also got the wrong answer :s-smilie:

Reply 3

Kevin De Bruyne

Original post by swagmister

The answers say it's 0.155 I tried 6b which was a similar question but also got the wrong answer :s-smilie:

Beats me, sorry. I agree with your answer :smile:

Reply 4

swagmister

Original post by SeanFM

Beats me, sorry. I agree with your answer :smile:

Maybe the textbook is wrong? I'll ask my teacher tomorrow, thanks for taking a look anyway 👍

Reply 5

NotNotBatman

Shouldn't there be a continuity correction, since it's a binomial distribution?

Original post by swagmister

Could I have help with 6a please?

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Reply 6

swagmister

Original post by NotNotBatman

Shouldn't there be a continuity correction, since it's a binomial distribution?

Idk that would make sense but I didn't use it in any of the other questions that similar and got the right answer

Reply 7

NotNotBatman

Original post by swagmister

Idk that would make sense but I didn't use it in any of the other questions and got the right answer

Try it again with a continuity correction and see if you get the answer in the textbook.

Reply 8

swagmister

Original post by NotNotBatman

Try it again with a continuity correction and see if you get the answer in the textbook.

I got 0.9292

Reply 9

swagmister

Original post by NotNotBatman

Try it again with a continuity correction and see if you get the answer in the textbook.

Do you not need a continuity correction because the means and sampling mean aren't discrete?

Reply 10

NotNotBatman

Original post by swagmister

I got 0.9292

You'd need to apply the continuity correction to the total score (mean multiplied by number of throws) and then divide by 70.

Reply 11

swagmister

Original post by NotNotBatman

You'd need to apply the continuity correction to the total score (mean multiplied by number of throws) and then divide by 70.

So ((35/12)+0.5)/70??

(edited 7 years ago)

Reply 12

NotNotBatman

Original post by swagmister

So ((32/15)+0.5)/70??

Where's 32 and 15 coming from?

You do 3.3 x 70 = 231, then apply the continuity correction and divide by 70, then you find your standardized z value.

Reply 13

swagmister

Original post by NotNotBatman

Where's 32 and 15 coming from?

You do 3.3 x 70 = 231, then apply the continuity correction and divide by 70, then you find your standardized z value.

Oh I was taught to add or minus 0.5 onto x where eg. P(X<x), I got 0.1723 this time :s-smilie:

Reply 14

NotNotBatman

Original post by swagmister

Oh I was taught to add or minus 0.5 onto x where eg. P(X<x), I got 0.1723 this time :s-smilie:

In this case it's different because it's a mean.

You've done the continuity correction the wrong way around; it's less than x, not less than or equal to, so subtract 0.5.

230.5/70 = 3.29...

Reply 15

swagmister

Original post by NotNotBatman

In this case it's different because it's a mean.

You've done the continuity correction the wrong way around; it's less than x, not less than or equal to, so subtract 0.5.

230.5/70 = 3.29...

Oh right I haven't learnt that yet, oh yeah, I got 0.1585

Reply 16

swagmister

Nevermind I read the tables wrong

Got 0.155 now!

How come you use the continuity correction in Q6 but not Q5?

Original post by NotNotBatman

In this case it's different because it's a mean.

You've done the continuity correction the wrong way around; it's less than x, not less than or equal to, so subtract 0.5.

230.5/70 = 3.29...

Reply 19

NotNotBatman