Question

A sample of 50 cotton threads shows the mean breaking strength to be 220 grams and the sample standard deviation to be 12 grams. What is the 95% confidence interval for the standard deviation of breaking strength of the thread.

Answer 1

Answer:

$220-2.01\frac{12}{\sqrt{50}}=216.589$    

$220+2.01\frac{12}{\sqrt{50}}=223.411$    

The confidence interval for this case would be (216.589, 223.411)

Step-by-step explanation:

Information given

$\bar X=220$ represent the sample mean

$\mu$ population mean

s=12 represent the sample standard deviation

n=50 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:

$df=n-1=50-1=49$

The Confidence level is 0.95 or 95%, the significance is $\alpha=0.05$ and $\alpha/2 =0.025$, the critical value for this case is $t_{\alpha/2}=2.01$

And replacing we got:

$220-2.01\frac{12}{\sqrt{50}}=216.589$    

$220+2.01\frac{12}{\sqrt{50}}=223.411$    

The confidence interval for this case would be (216.589, 223.411)