Confidence interval for variance

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From users who are members of ENG2012/CME1027 practice :

Aamir Khan

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Needs to be tested

4 years, 1 month ago

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Aamir Khan 4 years, 1 month ago

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Aamir Khan 4 years, 1 month ago

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Aamir Khan 4 years, 1 month ago

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Question statement

A smartphone manufacturer claims that their new smartphone will last 24 hours on a full charge under normal usage with a standard deviation of 0.8 hours.

Testing 6 of these smartphones we see the following usage (in hours) until the battery dies:

{x} hours.

Construct a {confl}% confidence interval for the standard deviation.

Give any introductory information the student needs.

			Name	Type	Generated Value

n	integer	6
xbar	number	23.7833333333
sd	number	0.2316606714
chi_lower	number	11.07
chi_upper	number	1.145
var_lower	number	0.0242396868
var	number	0.0536666667
var_upper	number	0.2343522562
sd_lower	number	0.1556909979
sd_upper	number	0.484099428
alpha	number	0.1
confl	integer	90
x	list	List of 6 items
lower_round	number	0.156
upper_round	number	0.484

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
6

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: `integer`

→ Used by:

var_lower
var_upper
x
xbar

Advice

Parts

Gap-fill
1. Gap Gap 0. Number entry
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2. Gap Gap 1. Number entry
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Gap-fill

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

Calculate a {confl}% confidence interval $(a,b)$ for the population standard deviation:

$a=\;$Gap 0. hours $b=\;$Gap 1. hours

Enter both to 1 decimal place.

Advice

We use the $\chi^2$ tables to find the confidence interval for the population variance and then we can square root our lower and upper points of the interval to give us the confidence interval for the standard deviation.

The formula for the confidence interval for the population variance is:

\[\frac{(n-1)s^2}{\chi^2_{n-1;\alpha/2}}\le\sigma^2\le\frac{(n-1)s^2}{\chi^2_{n-1;1-\alpha/2}}\]

We first calculate the sample variance using our formula:

\[\textrm{Sample Variance} = \frac{1}{n-1}\left(\sum_{j=1}^{n}x_j^2 -n\bar{x}^2\right)=\var{var}\]

We now need to find the critical values of the $\chi^2$ distribution. Our degrees of freedom here are $n-1=\var{n}-1=\var{n-1}$.

Our confidence interval is $\var{confl}\%$ so our significance level is $\alpha=\var{alpha}$. Our critical values are then

\[\chi^2_{n-1;\alpha/2}=\chi^2_{\var{n-1};\var{alpha/2}}=\var{chi_lower}\]

and

\[\chi^2_{n-1;1-\alpha/2}=\chi^2_{\var{n-1};\var{1-alpha/2}}=\var{chi_upper}\]

Hence, inputting all of these values into our formula for the confidence interval

\[\frac{(n-1)s^2}{\chi^2_{n-1;\alpha/2}}\le\sigma^2\le\frac{(n-1)s^2}{\chi^2_{n-1;1-\alpha/2}}\]

gives the confidence intervals for the population variance $\sigma^2$

Lower value of the confidence interval $=\;\displaystyle \var{var_lower}$.

Upper value of the confidence interval $=\;\displaystyle \var{var_upper}$.

Square rooting these values gives us the confidence interval for the population standard deviation $\sigma$ (to 3 d.p.):

\[(\var{lower_round},\var{upper_round})\]

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Aamir Khan 4 years, 1 month ago

Aamir Khan 4 years, 1 month ago

Aamir Khan 4 years, 1 month ago

Error in variable testing condition

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Generated value: `integer`

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Confidence interval for variance

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Aamir Khan 4 years, 1 month ago

Aamir Khan 4 years, 1 month ago

Aamir Khan 4 years, 1 month ago

Error in variable testing condition

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Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`