Answer to Question #269464 in Statistics and Probability for Priyanka

Question

A company has been selling production machinery, the existing design is model A and a new model has just come out, model B. the designer claims that model B produces more efficiently than model A. However the general manager of the company wants this to be proved in actual usage by companies to whom the machinery is supplied. Therefore it was arranged to have several customers try model B and model A under precisely similar conditions for determining how many units each model produce in an hour. The trial was conducted in seven customers’ plants. However one trial of model A had to be discarded. This left the management with six trials of model A, to compare with seven trials of model B.

 

Model A

176

180

181

184

187

190

 

Model B

185

187

187

190

192

194

195

 

 

On the basis of the above sample information will you consider model B superior than model A?

Answer 1

$H_0:m_b>m_a$

$H_1:m_b ≤m_a$ , where $m_a,m_b$ - population mean of model A and model B respectively

Since we have small samlpe sizes, then it is more accurate to use t-statistic instead of z-score

Test statistic is calculated next way

$t={\frac {x_b-x_a} {\sqrt{(n-1)*s_a^2+(k-1)*s_y^2}}}*\sqrt{{\frac {nk(n+k-2)} {n+k}}}$ , where $x_a,x_b$ - sample means for models A and B respectively, $s_x,s_y-$ sample standard deviations of samples A and B respectively, n and k is the sample sizes of A and B respectively

Lets calculate required terms

n = 6

k = 7

$x_a={\frac {176+...+190} 6}={\frac {1098} 6}=183$

$x_b={\frac {185+...+195} 7}={\frac {1330} 7}=190$

$s_a^2={\frac {(176-183)^2+...+(190-183)^2} 6}={\frac {128} 6}=21.3$

$s_a^2={\frac {(185-190)^2+...+(195-190)^2} 7}={\frac {88} 7}=12.6$

So, our test statistic is equal to

$t={\frac {190-183} {\sqrt{(5*21.3+6*12.6}}}*\sqrt{{\frac {7*6(7+6-2)} {7+6}}} ={\frac 7 {13.49}}*5.96=3.09$

According to the form of null hypothesis, if $t>Cr$ then we should accept it, where Cr - critical value

Critical value depends on the significance level, which is not given. Mostly common used significance levels is 0.9, 0.95, 0.99, so we can make desicion on every of them. With the increasing of significance level the critical value also increases, so, for example, if we accept the null hypothesis with 0.99 level of significance we automatically accept it on 0.95. Lets calculate for a = 0.99 level of significance. Since we are runnin one - tailed test, then critical value is such value, that

$P(T(n+k-1)>Cr)=1-a=0.01$ , where T(n+k-1) is Student's T-criteria with n+k-1 degrees of freedom

$P(T(12)>Cr)=0.01\implies Cr=2.681$ - that value was taken from the tables of t-criteria, they can be easily found on internet

We see that t > Cr, which means we accept the null hypothesis on the 0.99 significance level(and automatically on 0.95, 0.9 too)

There are significant statistical evidence that model B is better than model A