n_251
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I just came across a past question:

"The heights of a random sample of 10 imported orchids are measured.
The mean height of the sample is found to be 20.1 cm. The heights of the orchids are normally distributed.

Given that the population standard deviation is 0.5 cm,
(a) estimate limits between which 95% of the heights of the orchids lie,"

I knew that around 95% of values lie within 2 standard deviations of the mean. But, as a general point, say I wanted to estimate limits between which x% of the values lay. Would I just do the mean plus/minus the z-value of a two tailed test multiplied by the standard deviation?

So, z=1.96 for 95%, z=2.5758 for 1%, etc.

That's what the mark scheme does but I was wondering if this works for a range of limits or only 95%?

Thanks
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n_251
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bump
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Clarity Incognito
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(Original post by n_251)
I knew that around 95% of values lie within 2 standard deviations of the mean. But, as a general point, say I wanted to estimate limits between which x% of the values lay. Would I just do the mean plus/minus the z-value of a two tailed test multiplied by the standard deviation?

So, z=1.96 for 95%, z=2.5758 for 1%, etc.

That's what the mark scheme does but I was wondering if this works for a range of limits or only 95%?

Thanks
Almost

 \bar{x} \pm (relevant number) \times \dfrac{\sigma}{\sqrt{n}}
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n_251
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(Original post by Clarity Incognito)
Almost

 \bar{x} \pm (relevant number) \times \dfrac{\sigma}{\sqrt{n}}
no that's for confidence intervals. I thought the same as you but the mark scheme disagrees.
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Clarity Incognito
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(Original post by n_251)
no that's for confidence intervals. I thought the same as you but the mark scheme disagrees.
Isn't that what we're doing here? Your confidence interval is between the limits you're talking about?
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n_251
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(Original post by Clarity Incognito)
Isn't that what we're doing here? Your confidence interval is between the limits you're talking about?
apparently not...it asks for an estimate between which 95% of the values lie
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Clarity Incognito
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(Original post by n_251)
apparently not...it asks for an estimate between which 95% of the values lie
That's confidence intervals! It's asking for the confidence limits. If I'm honest, I don't see why they have used standard deviation instead of standard error. Considering that the mean is estimated from the sample :dontknow: either bump this tomorrow or make a new thread and I won't intervene again but quote me in it as I will be interested to see the answer.
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n_251
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BUMP! Someone must know!
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Alpha95
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Did you ever come to a solution for this? I am having the exact same problem and my exam is in a few days...!!
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