Question

The quality control manager of a tire company wishes to estimate the tensile strength of a standard size of rubber used to make a class of radial tires. A random sample of 50 pieces of rubber from different production batches is subjected to a stress test. The test measures the force needed to break the rubber in pounds. According to the sample results, the average pressure is 173 pounds. The podulation variance Is known to be 1225 pounds2. Construct a 98 % confidence interval estimate of the true population mean tensile strength. Use only the appropriate formula and/or statistical table in your textbook to answer this question. Negative amounts should be indicated by a minus sign. Report your answer to 2 decimal places, using conventional rounding rules. Answer pounds s(Click to select)

Answer 1

We have given that,                              
                              
Sample mean =173 pounds                      
Population standard deviation=   35 pounds                  
Sample size=50                      
Level of significance =   1-0.98=0.02                      
Z critical value (by using Z table)=2.33                  
                              
Confidence interval formula is
$\bar{x}\pm Z*\frac{\sigma }{\sqrt{n}}$
$=173\pm 2.33*\frac{35}{\sqrt{50}}$
=(161.49,184.51)                              
                              
Therefore, 98% confidence interval estimate of true population mean tensile strength is

$161.49<\mu<184.51$