Question

A pharmaceutical company is interested in testing the effect of humidity on the weights of pills sold in a new aluminum package. Let X and Y denote the weight of a pill in the old aluminum package and in the new aluminum package (respectively) after the packaged pill has spent one week in chamber kept at 30 °C and 100 % humidity. Define a null and alternates hypothesis to test whether the old aluminum package pills weigh less than the new aluminum package pills following the humidity-chamber treatment. The following random samples of X yielded the following weights in milligrams:
373.2 376.7 381.6 382.1 388.7 384.0
397.9 389.8 385.1 371.3 383.5
and the following random sample of Y yielded the following weights in milligrams:
395.5 384.8 383.5 386.5 394.8
391.6 397.7 384.0 391.7 398.8
Define a critical region with a significance level of alpha = 0.05 and calculate the value of the test statistic. Accept or reject the null hypothesis and then state a scientific conclusion about the packaged-pill humidity experiment.

Answer 1

Answer:

The null hypothesis is  $H_o : \mu_1 = \mu_2$

The alternative hypothesis is   $H_a : \mu_1 < \mu_2$

The test statistics is   $t =  -2.65$

Reject the null hypothesis

There is  sufficient evidence to conclude that the weight of the old aluminium package pills is less than the new aluminium pills.

Step-by-step explanation:

From the question we are told that

The data for X is  373.2 , 376.7  , 381.6 , 382.1  , 388.7 ,  384.0 ,  397.9 , 389.8 ,385.1 , 371.3 , 383.5

The data for Y is 395.5 ,  384.8 , 383.5  , 386.5  ,394.8

, 391.6  , 397.7 , 384.0 , 391.7 , 398.8

Generally the sample mean for X is mathematically represented as

          $\= x = \frac{\sum x_i }{n}$

=>       $\= x = \frac{373.2 +376.7 +\cdots + 383.5}{11}$

=>       $\= x = 383.08$

Generally the sample mean for Y is mathematically represented as

          $\= y = \frac{\sum y_i }{n}$

=>       $\= y = \frac{395.5 +384.8 +\cdots + 398.8}{10}$

=>       $\= y = 390.89$

Generally the standard deviation for  X is mathematically represented as

     $\sigma_1 = \sqrt{\frac{\sum (x_i - \= x_1 )^2}{n} }$

=>   $\sigma_1 = \sqrt{\frac{( 373.2 - 383.08)^2 +( 376.7 - 383.08)^2 +\cdots + ( 383.5 - 383.08)^2 }{11} }$

=> $\sigma_1 = 7.63$  

Generally the standard deviation for  Y is mathematically represented as

     $\sigma_2 = \sqrt{\frac{\sum (y_i - \= y )^2}{n} }$

=>   $\sigma_2 = \sqrt{\frac{( 395.5 - 390.89)^2 +( 384.8 - 390.89)^2 +\cdots + ( 398.8 - 390.89)^2 }{10} }$

=> $\sigma_2 = 5.82$  

The null hypothesis is  $H_o : \mu_1 = \mu_2$

The alternative hypothesis is   $H_a : \mu_1 < \mu_2$

Generally the test statistics is mathematically represented as

     $t = \frac{\= x - \= y}{ \sqrt{\frac{\sigma_1^2 }{ n_1 } +\frac{\sigma_2^2 }{ n_2 } } }$

=>   $t = \frac{383.08 - 390.89}{ \sqrt{\frac{7.63^2 }{ 11 } +\frac{5.82^2 }{10} } }$

=>   $t = -2.65$

Generally the level of significance is $\alpha = 0.05$

Generally the degree of freedom is mathematically represented as

      $df = n_1 + n_2 - 2$

=>   $df = 11 + 10 - 2$

=>   $df = 19$

Generally from the student t- distribution table the critical value of   $\alpha$ at a df = 19 is      

    $t_{\alpha ,df} =t_{0.05 ,19} = -2.093$

Now comparing the critical value  obtained with test statistic calculated we see that the critical value is the region of the calculated test statistic( i.e  between  -2.65 and  2.65) hence we reject the null hypothesis

Therefore there sufficient evidence to conclude that the old aluminium package pills weight is less than the new aluminium pills