Two chemical processes for manufacturing the same product are being compared under the same conditions. Yield from Process A gives an average value of 96.2 from six runs, and the estimated standard deviation of yield is 2.75. Yield from Process B gives an average value of 93.3 from seven runs, and the estimated standard deviation is 3.35. Yields follow a normal distribution. Is the difference between the mean yields statistically significant? Use the 5% level of significance, and show rejection regions for the difference of mean yields on a sketch.

Here since the number of samples is less than 40, it follows the t distribution. Mean is given by µ, standard differentiation is given by s , degrees of freedom is given by df and number of samples is given by n. Null hypothesis: µA-µB=0 Alternative hypothesis: µA-µB?0 µA=96.2 µB=93.3 sA=2.75 sB=3.35 nA=6 nB=6 dfB=5 dfB=5 t=xA-xB/s To find standard deviation, we have to apply this formula Variance=[ (5*7.56)+(5*11.22)]/10 =9