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.
A) $\left[ \begin{array} { l l }
227.96 & 248.82
\end{array} \right]$
B) $[ 236.64 \quad 240.16 ]$
C) $\left[ \begin{array} { l l }
233.63 & 243.17
\end{array} \right]$
D) $\left[ \begin{array} { l l }
229.21 & 247.59
\end{array} \right]$

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 percent confidence interval.
A) $\left[ \begin{array} { l l } 
227.96 & 248.82
\end{array} \right]$
B) $[ 236.64 \quad 240.16 ]$
C) $\left[ \begin{array} { l l } 
233.63 & 243.17
\end{array} \right]$
D) $\left[ \begin{array} { l l } 
229.21 & 247.59
\end{array} \right]$

Quizplus · Accepted Answer

The correct answer is A because the 98% confidence interval is 227.96 to 248.82. This is calculated using the formula:CI = x̄ ± z * (σ/√n)where x̄ is the sample mean, z is the z-score for the desired confidence level, σ is the population standard deviation, and n is the sample size.In this case, x̄ = 238.4, z = 2.326 (for a 98% confidence level), σ = 35, and n = 61. Plugging these values into the formula gives us:CI = 238.4 ± 2.326 * (35/√61)CI = 238.4 ± 20.86CI = [227.96, 248.82]

The Quality Control Manager of a Tire Company Wishes to Estimate