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 61 pieces of rubber from different production batches is subjected to a stress test. The test measures the force (in pounds) needed to break the rubber. According to the sample results, the average pressure is 238.4 pounds with a population standard deviation of 35 pounds. Determine the 98% confidence interval.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 238.4

Population standard deviation = = 35

Sample size = n = 61

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

Z_/2 = Z_0.01 = 2.326

Margin of error = E = Z_/2* ( /n)

= 2.326 * ( 35 / 61 )

= 10.42

At 98% confidence interval estimate of the population mean is,

- E < < + E

238.4 - 10.42 < < 238.4 + 10.42

227.98 < < 248.82

( 227.98 , 248.82)