Answer to Question #299476 in Statistics and Probability for jay

Question

A certain book manufacturing company presents the time it takes for sheets from a press to be folded, gathered,

sewn, tipped on end sheets, and bound. The following data represent samples of books of two book production

companies and the processing time for these jobs. The processing time refers to the time in days starting from

the time the books come off the press to the time they are packed in cartons. Assume the data values came from

normally distributed populations with unequal variances and use 5% level of significance.

Company 1 5.42 5.32 16.25 10.45 21.5

Company 2 9.46 11.36 16.62 12.6 15.5 18.7 10.75 14.2

Is there sufficient evidence to conclude that Company 1 is more efficient than Company 2 if the processing

time is a measure of efficiency?

Answer 1

Company 1

$n_1=5\\\bar x_1={\sum x\over n_1}={ 58.94\over 5}=11.788\\s_1^2={\sum x^2-{(\sum x)^2\over n_1}\over n_1-1}={ 893.1938-694.78472\over4}=49.60227$

Company 2

$n_2=8\\\bar x_2={\sum x\over n_2}={ 109.19\over 8}=13.64875\\s_2^2={\sum x^2-{(\sum x)^2\over n_2}\over n_2-1}={ 1560.668-1490.30701\over7}=10.05157$

Hypotheses,

$H_0:\mu_1=\mu_2\\vs\\H_1:\mu_1\gt\mu_2$

The test statistic is,

$t_c={\bar x_1-\bar x_2\over \sqrt{{s_1^2\over n_1}+{s_2^2\over n_2}}}={11.788-13.64875\over \sqrt{{49.60227\over 5}+{10.05157\over8}}}={-1.85075\over 3.34318713}=-0.5566$

The critical value is the t distribution table value at $\alpha =0.05$ with $v$ degrees of freedom.

$v={({49.60227\over 5}+{10.05157\over 8})^2\over ({49.60227\over5})^2+({10.05157\over8})^2}= 5.0313\approx 5$

Now,

$t_{\alpha, v}=t_{0.05,5}=2.015$

Reject the null hypothesis if $|t_c|\gt t_{0.05,5}$

Since $|t_c|=0.5536\lt t_{0.05,5}=2.015,$ we fail to reject the null hypothesis and conclude that there is no sufficient evidence to show that Company 1 is more efficient than Company 2 if the processing time is a measure of efficiency at 5% level of significance.