Question

5) A certain company is interested to compare two drill bits, A and B. The company is currently using drill bit A but is investigating a possibility of switching to drill bit B because it is slightly cheaper. However, the duration of the manufacturing cycle is critical to the company, and it will consider switching to drill bit B only if such a change will result in a shorter manufacturing cycle. To make an informed decision, the company collects seven measurements of the time necessary to drill a hole in a steel plate of specified thickness using each of the drill bits. The observed sample mean and sample standard deviation for drill bit A are 20.1 sec and 1.02 sec, respectively, and for drill bit B the same parameters are 19.8 and 0.97 sec, respectively. Should the company switch to drill bit B? Test at a 5% significance level.

Answer 1

given

n1=n2=7

so here we have to test

$H0: \mu_A -\mu_B \leq 0 ,H1: \mu_A -\mu_B>0$

first, we find the pooled standard deviation

$S_{pooled}=\sqrt{\frac{(n1-1)S^2_A +(n2-1)S^2_B}{n1+n2-2}}=\sqrt{\frac{(7-1)1.02^2+(7-1)0.97^2}{7+7-2}}$

=0.9953

so

test statistics is given by

$t=\frac{\bar{x}_A-\bar{x}_B -0}{S_{pooled}\sqrt{\frac{1}{n1}+\frac{1}{n2}}}=\frac{20.1-19.8}{0.9953\sqrt{\frac{1}{7}+\frac{1}{7}}}=\frac{0.3}{0.5320}$

=0.5639

df =n1+n2-2 =7+7-2 =12

now

P value =P(t>0.5639) =0.2916

Since P value is greater than level of significance hence we failed to reject the null hypothesis so that means there is not sufficient evidence to the claim that B need shorter manufacturing cycle so company shouldnt switch to B