Question

An article presents a study of the effect of the subbase thickness on the amount of surface deflection caused by aircraft landing on an airport runway. In six applications of a 160 kN load on a runway with a subbase thickness of 864 mm, the average surface deflection was 2.28 mm with a standard deviation of 0.090 mm. Find a 90% confidence interval for the mean deflection caused by a 160 kN load. Round the answers to three decimal places.

Answer 1

90% confidence interval for the mean deflection caused by a 160 KN load is [2.206 , 2.354].

Explanation:

We are given that in six applications of a 160 KN load on a runway with a sub base thickness of 864 mm, the average surface deflection was 2.28 mm with a standard deviation of 0.090 mm.

So, the pivotal quantity for 90% confidence interval for the mean deflection is given by;

P.Q. =
`(\bar X - \mu)/((s)/(√(n) ) )` ~
`t_n_-_1`

where,
`\mu` = sample average surface deflection = 2.28 mm

`\sigma` = sample standard deviation = 0.090 mm

n = sample of applications = 6

`\mu` = mean surface deflection

So, 90% confidence interval for the mean deflection,
`\mu` is ;

P(-2.015 <
`t_5` < 2.015) = 0.90

P(-2.015 <
`(\bar X - \mu)/((s)/(√(n) ) )` < 2.015) = 0.90

P(
`-2.015 * {(s)/(√(n) )` <
`{\bar X - \mu}` <
`2.015 * {(s)/(√(n) )` ) = 0.90

P(
`\bar X -2.015 * {(s)/(√(n) )` <
`\mu` <
`\bar X +2.015 * {(s)/(√(n) )` ) = 0.90

90% confidence interval for
`\mu` = [
`\bar X -2.015 * {(s)/(√(n) )` ,
`\bar X +2.015 * {(s)/(√(n) )` ]

= [
`2.28 -2.015 * {(0.090)/(√(6) )` ,
`2.28 -2.015 * {(0.090)/(√(6) )` ]

= [2.206 , 2.354]

Therefore, 90% confidence interval for the mean deflection caused by a 160 KN load is [2.206 , 2.354].

Answer 2

`2.28-2.02(0.090)/(√(6))=2.206`

`2.28+2.02(0.090)/(√(6))=2.354`

So on this case the 90% confidence interval would be given by (2.206, 2.354)

Explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

`\bar X=2.28` represent the sample mean for the sample

`\mu` population mean (variable of interest)

s=0.090 represent the sample standard deviation

n=6 represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

`\bar X \pm t_(\alpha/2)(s)/(√(n))` (1)

In order to calculate the critical value
`t_(\alpha/2)` we need to find first the degrees of freedom, given by:

`df=n-1=6-1=5`

Since the Confidence is 0.90 or 90%, the value of
`\alpha=0.1` and
`\alpha/2 =0.05`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,5)".And we see that
`t_(\alpha/2)=2.02`

Now we have everything in order to replace into formula (1):

`2.28-2.02(0.090)/(√(6))=2.206`

`2.28+2.02(0.090)/(√(6))=2.354`

So on this case the 90% confidence interval would be given by (2.206, 2.354)