Respuesta :
Answer:
The 99% confidence interval is (228.035, 234.233).
Step-by-step explanation:
The sample size selected to compute the 95% confidence interval for the true average natural frequency (Hz) of delaminated beams of a certain type is
n = 5
The sample size is very small and the population standard deviation is also not known.
So, we will use t-interval for the confidence interval.
The (1 - α)% confidence interval for the true mean is:
[tex]CI=\bar x\pm t_{\alpha/2, (n-1)}\times \frac{s}{\sqrt{n}}[/tex]
The 95% confidence interval for the true average natural frequency (Hz) of delaminated beams of a certain type is:
(Upper limit, Lower limit) = (229.266, 233.002).
Compute the value of sample mean as follows:
[tex]\bar x=\frac{UL+LL}{2}=\frac{233.002+229.266}{2}=231.134[/tex]
The critical value of t for α = 0.05 and n = 5 is:
[tex]t_{\alpha/2, (n-1)}=t_{0.05/2, (5-1)}=t_{0.025,4}=2.776[/tex]
*Use a t-table.
Compute the value of sample standard deviation as follows:
[tex]\frac{UL-LL}{2}=t_{\alpha/2, (n-1)}\times \frac{s}{\sqrt{n}}\\\frac{233.002-229.266}{2}=2.776\times\frac{s}{\sqrt{5}}\\s=1.505[/tex]
The critical value of t for α = 0.01 and n = 5 is:
[tex]t_{\alpha/2, (n-1)}=t_{0.01/2, (5-1)}=t_{0.005,4}=4.604[/tex]
*Use a t-table.
Construct the 99% confidence interval as follows:
[tex]CI=\bar x\pm t_{\alpha/2, (n-1)}\times \frac{s}{\sqrt{n}}\\=231.134\pm 4.604\times \frac{1.505}{\sqrt{5}}\\=231.134\pm3.099\\=(228.035, 234.233)[/tex]
Thus, the 99% confidence interval is (228.035, 234.233).
