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

`32.9-2.365(4.9)/(√(8))=28.80`

`32.9+2.365(4.9)/(√(8))=37.00`

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)(s)/(√(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(4.9)/(√(8))=28.80`

`32.9+2.365(4.9)/(√(8))=37.00`