Respuesta :
Answer:
[tex]2.28-2.02\frac{0.090}{\sqrt{6}}=2.206[/tex]
[tex]2.28+2.02\frac{0.090}{\sqrt{6}}=2.354[/tex]
So on this case the 90% confidence interval would be given by (2.206, 2.354)
Step-by-step 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".
[tex]\bar X=2.28[/tex] represent the sample mean for the sample
[tex]\mu[/tex] 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:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:
[tex]df=n-1=6-1=5[/tex]
Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], 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 [tex]t_{\alpha/2}=2.02[/tex]
Now we have everything in order to replace into formula (1):
[tex]2.28-2.02\frac{0.090}{\sqrt{6}}=2.206[/tex]
[tex]2.28+2.02\frac{0.090}{\sqrt{6}}=2.354[/tex]
So on this case the 90% confidence interval would be given by (2.206, 2.354)
Answer:
90% confidence interval for the mean deflection caused by a 160 KN load is [2.206 , 2.354].
Step-by-step 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. = [tex]\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\mu[/tex] = sample average surface deflection = 2.28 mm
[tex]\sigma[/tex] = sample standard deviation = 0.090 mm
n = sample of applications = 6
[tex]\mu[/tex] = mean surface deflection
So, 90% confidence interval for the mean deflection, [tex]\mu[/tex] is ;
P(-2.015 < [tex]t_5[/tex] < 2.015) = 0.90
P(-2.015 < [tex]\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.015) = 0.90
P( [tex]-2.015 \times {\frac{s}{\sqrt{n} }[/tex] < [tex]{\bar X - \mu}[/tex] < [tex]2.015 \times {\frac{s}{\sqrt{n} }[/tex] ) = 0.90
P( [tex]\bar X -2.015 \times {\frac{s}{\sqrt{n} }[/tex] < [tex]\mu[/tex] < [tex]\bar X +2.015 \times {\frac{s}{\sqrt{n} }[/tex] ) = 0.90
90% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X -2.015 \times {\frac{s}{\sqrt{n} }[/tex] , [tex]\bar X +2.015 \times {\frac{s}{\sqrt{n} }[/tex] ]
= [ [tex]2.28 -2.015 \times {\frac{0.090}{\sqrt{6} }[/tex] , [tex]2.28 -2.015 \times {\frac{0.090}{\sqrt{6} }[/tex] ]
= [2.206 , 2.354]
Therefore, 90% confidence interval for the mean deflection caused by a 160 KN load is [2.206 , 2.354].
