Question

A personal computer manufacturer is interested in comparing assembly times for two keyboard assembly processes. Assembly times can vary considerably from worker to worker, and the company decides to eliminate this effect by selecting a random sample of 8 workers and timing each worker on each assembly process. Half of the workers are chosen at random to use Process 1 first, and the rest use Process 2 first. For each worker and each process, the assembly time (in minutes) is recorded, as shown in Table 1 Difference Worker Process 1 Process 2 (Process 1 Process 2) 1 66 44 22 2 46 50 -4 3 68 61 7 89 72 17 5 85 86 -1 6 81 63 18 7 - 52 61 9 8 49 48 Table 1 Based on these data, can the company conclude,, at the 0.05 level of significance, that the mean assembly times for the two processes differ? Answer this question by performing a hypothesis test regarding u (which is u with a letter "d" subscript), the population mean difference in assembly times for the two processes. Assume that this population of differences (Process 1 minus Process 2) is normally distributed. Perform a two-tailed test. Then fill in the table below. Carry your intermediate computations to at least three decimal places and round your answers as specified in the table. (If necessary, consult a list of formulas.) H The null hypothesis: H The alternative hypothesis: OSO (Choose one) The type of test statistic: The value of the test statistic: (Round to at least three decimal places.) ? The two critical values at the 0,05 level of significance: and (Round to at least three decimal places.) At the 0.05 level, can the company conclude that the mean assembly times for the two processes O Yes O No

please answer neatly and correctly!

Answer 1

Answer)

Null hypothesis Ho : ud = 0

Alternate hypothesis Ha : ud not equal to 0

As the population s.d is unknown here we will use t distribution to conduct the test

Type of test = t test

Test statistics t = (mean difference)/(s.d/√n)

We just need to find the mean of the difference and s.d of the differences and n = 8

t = 6.375/(11.4635/√8) = 1.573

For n-1 = 7 dof and 0.05 alpha critical values are -2.365 and 2.365

Rejection region is reject Ho if test statistics is less than -2.365 or greater than 2.365

As 1.573 is not in rejection region

We fail to reject the null hypothesis Ho

So we do not have enough evidence to conclude that ud not equal to 0