2012 2013 CBA3_1

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Bill Foster 6 years, 4 months ago

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Newcastle University Mathematics and Statistics 9 years, 3 months ago

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Name	Status	Author	Last Modified
2012 2013 CBA3_1	Ready to use	Newcastle University Mathematics and Statistics	20/11/2019 14:51
Find the confidence interval and mean of a normal distribution	draft	Xiaodan Leng	10/07/2019 22:59

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Question statement

$X_1,\;X_2,\;\dots, X_n$ is a random sample from $\operatorname{N}(\mu,\var{sd}^2)$ .

Let $\bar{x}$ denote the mean of this sample and for this exercise we have $n=\var{n},\; \bar{x}=\var{r11}$

Give any introductory information the student needs.

			Name	Type	Generated Value

r11	number	239.2
tulim	number	242.4634705453
per	integer	99
n	integer	90
mu	integer	239
tol	number	0.01
sd	integer	12
ulim	number	242.46
inf	number	0.63
z	number	2.58
llim	number	235.94
tllim	number	235.9365294547

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
precround(mean(repeat(normalsample(mu,sd),n)),1)

JME function reference

Description

Describe what this variable represents, and list any assumptions made about its value.

Can an exam override the value of this variable?

Generated value: `number`

239.2

← Depends on:

→ Used by:

tllim
tulim

Statement
Advice
"Part a)" → "Unnamed gap" - Minimum accepted value
"Part a)" → "Unnamed gap" - Maximum accepted value

Parts

Part a) Gap-fill
1. Gap Number entry
  Add
Part b) Gap-fill
1. Gap Number entry
  Add
Part c) Gap-fill
1. Gap Gap 0. Number entry
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2. Gap Gap 1. Number entry
  Add

Gap-fill

Ask the student a question, and give any hints about how they should answer this part.

Enter the m.l.e. $\hat{\mu}$ for $\mu$ here:

$\hat{\mu}=\;$?Unnamed gap

Advice

The maximum likelihood estimator (m.l.e) is, in this case, the sample mean i.e. $\hat{\mu}=\var{r11}$.

The expected information in this case is $\displaystyle \frac{n}{\sigma^2}$ where $\sigma=\var{sd}$ is the standard deviation of the sampled normal distribution.

Hence $\displaystyle I(\mu)=\frac{\var{n}}{\var{sd^2}}=\var{inf}$ to 2 decimal places.

The $\var{per}$% confidence interval in this case is given by $(a,b)$ where:

\[a=\bar{x}-z\sqrt{\frac{\sigma^2}{n}},\;\;b=\bar{x}+z\sqrt{\frac{\sigma^2}{n}},\;\;\;z=\var{z}\]

Calculating to 2 decimal places gives:

$a=\var{llim},\;\;\;b=\var{ulim}$.

Hence a $\var{per}$% confidence interval for the mean is given by $(\var{llim},\var{ulim})$.

Give a worked solution to the whole question.

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Part	Test	Passed?
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Gap-fill		Hasn't run yet
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Taxonomy: mathcentre

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Bill Foster 6 years, 4 months ago

Newcastle University Mathematics and Statistics 9 years, 3 months ago

Error in variable testing condition

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Generated value: `number`

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2012 2013 CBA3_1

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Bill Foster 6 years, 4 months ago

Newcastle University Mathematics and Statistics 9 years, 3 months ago

Error in variable testing condition

Error

Warning

Generated value: number

Parts

Gap-fill

Generated value: `number`