2012 2013 CBA3_3

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Bill Foster

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6 years, 4 months ago

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Bill Foster commented 6 years, 4 months ago

Should state in part c) that the answers are to be input to 2 decimal places.

Calculations are correct.

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Bill Foster 6 years, 4 months ago

Gave some feedback: Ready to use

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Newcastle University Mathematics and Statistics 9 years, 3 months ago

Created this.

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Question statement

$X_1,\;X_2,\;\dots, X_n$ is a random sample from $\operatorname{Exponential}(\lambda)$.

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

mle	number	5.00
r11	number	0.2
tulim	number	6.5495160535
per	integer	95
n	integer	40
mu	integer	6
tol	number	0.01
ulim	number	6.55
inf	number	1.60
z	number	1.96
llim	number	3.45
tllim	number	3.4504839465

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
precround(1/r11,2)

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`

5.00

← Depends on:

→ Used by:

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
  Add
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. $\hat{\lambda}$ for $\lambda$ here:

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

Advice

The maximum likelihood estimator (MLE) for $\lambda$ is given by

\[\operatorname{MLE} = \frac{1}{\bar{x}}=\frac{1}{\var{r11}}=\var{mle}\] to 2 decimal places.

The expected information in this case is:

$\displaystyle \operatorname{I}(\lambda)=\frac{n}{\hat{\lambda}^2}=n\bar{x}^2=\var{n}\times\var{r11}^2=\var{inf}$ to 2 decimal places.

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

\[a=\frac{1}{\bar{x}}-z\sqrt{\frac{1}{n\bar{x}^2}},\;\;b=\frac{1}{\bar{x}}+z\sqrt{\frac{1}{n\bar{x}^2}},\;\;\;z=\var{z}\]

Calculating to 2 decimal places gives:

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

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

Give a worked solution to the whole question.

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Part	Test	Passed?
Gap-fill
Gap-fill		Hasn't run yet
Number entry
Number entry		Hasn't run yet
Gap-fill
Gap-fill		Hasn't run yet
Number entry
Number entry		Hasn't run yet
Gap-fill
Gap-fill		Hasn't run yet
Number entry
Number entry		Hasn't run yet
Number entry
Number entry		Hasn't run yet

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Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

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Bill Foster commented 6 years, 4 months ago

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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Gap-fill

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2012 2013 CBA3_3

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Bill Foster commented 6 years, 4 months ago

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`