Question

A random sample of n = 8 E-glass fiber test specimens of a certain type yielded a sample mean interfacial shear yield stress of 32.9 and a sample standard deviation of 4.9. Assuming that interfacial shear yield stress is normally distributed, compute a 95% CI for true average stress. (Give answer accurate to 2 decimal places.)

Answer 1

Answer:

$32.9-2.365\frac{4.9}{\sqrt{8}}=28.80$    

$32.9+2.365\frac{4.9}{\sqrt{8}}=37.00$    

Step-by-step explanation:

Information given

$\bar X=32.9$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

s=4.9 represent the sample standard deviation

n=8 represent the sample size  

Confidence interval

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

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (1)

the degrees of freedom are given by:

$df=n-1=8-1=7$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, and the critical value for this cae would be $t_{\alpha/2}=2.365$

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

$32.9-2.365\frac{4.9}{\sqrt{8}}=28.80$    

$32.9+2.365\frac{4.9}{\sqrt{8}}=37.00$