Question

A firm is manufacturing two brands, A and B, of battery cells. It claims that the average life of brand A cells is more than that of brand B cells by 100 hrs, the variances of the two brands being the same. To maintain this standard, two independent samples of 12 cells of each brand are selected on the 20th of every month and a t value of the difference of sample means computed. The firm is satisfied with its claim if the computed t value falls between ±t0.025. A sample of 12 cells of brand A gives a mean life of 1200 hrs and variance of 49 hrs, and that of 12 cells of brand B gives a mean life of 1095 hrs and variance of 64 hrs. Comment on the outcome of the sample results

Answer 1

The first step is to state the null hypothesis and an alternative hypothesis.                        H0: μA – μB = 100 Hrs;        H1 : μA – μB ≠ 100.The significance level is 0.025. Using sample data,l conduct a two-sample t-test of the null hypothesis.Ho Accept: if  - t 0.025 <= t <= t0.025     and  Ho Reject:  if  t < -t0.025 and t > t0.025In this case ±t0.025 = +- 2.0739  ( Here,  dof =  22)