Question

The purchasing director for an industrial parts factory is investigating the possibility of purchasing a new type of milling machine. She determines that the new machine will be bought if there is evidence that the parts produced have a higher average breaking strength than those from the old machine. A sample of 100 parts taken from the old machine indicates a sample mean of 65 kilograms and standard deviation is 10 kilograms. Whereas a similar sample of 100 from the new machine indicates a sample mean of 72 kilograms and a standard deviation of 9 kilograms. Is there evidence that the purchasing director should buy the new machine? Test at 5% significance level. Assume the two population variances are not equal

Answer 1

Let , be the population mean for old machines and be the population mean for new machines.

The null and alternative hypothesis is ,

The test is left-tailed test.

Since, population standard deviations are not equal.

The critical value is ,

; The Excel function is , =TINV(2*0.05,196)

The test statistic is ,

Decision : Here , the value of the test statistic lies in the rejection region.

Therefore , raject Ho.

Conclusion : There is sufficient evidence that the purchasing director should buy the new machine.