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 percent confidence interval.

The Confidence Interval is (227.96 , 248.84)

Step-by-step explanation:

The sample size is given as n = 61

The average pressure is given as = 238.4 pounds

The standard deviation is given to be = 35 pounds

The data collected for the radial tires is considered to be normally distributed.

Since the population is normally distributed the sample is also normally distributed, therefore = =238.4 pounds

The confidence interval is given as C.I. = 98% = 0.98

Therefore the Significance Level, = 1 - 0.98 = 0.02

Critical Z value is given as =

The Confidence Interval is given by the formula CI  =

                                 Therefore C.I.  =

                                                          = 238.4 ± 10.441

                                                          = 227.96 , 248.84

jfkc

step-by-step explanation:

jhcxzbfzc

answer: idk

step-by-step explanation: