Question

The tensile strength of silicone rubber is thought to be a function of curing temperature. A study was carried out in which 12 specimens of the rubber were prepared using curing temperatures of 20 C and 45 C. The accompanying data table shows the tensile strength values in megapascals. Compute the sample standard deviation in tensile strength for the samples separately for the two temperatures. Does it appear as if an increase in temperature influences the variability in tensile strength? Explain. E Click the icon to view the data on rubber tensile strengths. The standard deviation for the low-temperature group is The standard deviation for the high-temperature group is (Round to four decimal places as needed.) Does it appear as if an increase in temperature influences the variability in tensile strength? Explain. the standard deviation for the low-temperature group. Therefore, an increase i The standard deviation for the high-temperature group is temperature variability in tensile strength. X Rubber Tensile Strengths 20°C: 2,09 2,11 2.22 2,04 2.25 2,04 2,02 2,12 2.06 2.11 2.14 2.08 45°C: Enter your answer in each of the answer boxes. 241 2,13 2 58 2.04 2.35 2.02 198 2 44 203 2.44 2.21 2.07

Answer 1

Given data:

Low temperature(x1) High temperature (x2) (x1- x1 bar)^2 (x2-x2 bar)^2 S.no 20 45 1 2,09 2.41 0.0002777778 0.034225 2 2,11 2.13 1.111111E-05 0.009025 3 2.22 2.58 0.0128444444 0.126025 4 2,04 2,04 0.0044444444 0.034225 5 2,25 2,35 0.0205444444 0.015625 2,04 6 2,02 0.0044444444 0.042025 7 2,02 1,98 0.0075111111 0.060025 8 2,12 2.44 0.0001777778 0.046225 2.06 2.03 0.0021777778 0.038025 10 2.11 2,44 1.111111E-05 0.046225 0.000225 11 2,14 2.21 0.0011111111 12 2,08 2,07 0.0007111111 0.024025 SUM 25.28 26.7 0.0542667 0.4759 MEAN 2.10667 2.225

Mean of low temperature =$\frac{\sum x1}{n} = \frac{25.28}{12} = 2.10667$

Mean of High temperature = $\frac{\sum x2}{n} = \frac{26.7}{12} = 2.225$

Standard deviation of low temperature = $\sqrt{\frac{\sum (x1 - \bar{x1})^{2}}{n-1}} = \sqrt{\frac{0.054267}{12-1}} = 0.07024$

Standard deviation of low temperature = 0.0702

Standard deviation of high temperature = $\sqrt{\frac{\sum (x2 - \bar{x2})^{2}}{n-1}} = \sqrt{\frac{0.4759}{12-1}} = 0.20799$

Standard deviation of high temperature = 0.2080

The standard deviation for the high temperature group is 0.2080 greater than the standard deviation for the low temperature group. Therefore an increase in temperature increasing variability in tensile strength