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

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.

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 = (\sum x_i )/(n)`

=>
`\= x = (373.2 +376.7 +\cdots + 383.5)/(11)`

=>
`\= x = 383.08`

Generally the sample mean for Y is mathematically represented as

`\= y = (\sum y_i )/(n)`

=>
`\= y = (395.5 +384.8 +\cdots + 398.8)/(10)`

=>
`\= y = 390.89`

Generally the standard deviation for X is mathematically represented as

`\sigma_1 = \sqrt{(\sum (x_i - \= x_1 )^2)/(n) }`

=>
`\sigma_1 = \sqrt{(( 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{(\sum (y_i - \= y )^2)/(n) }`

=>
`\sigma_2 = \sqrt{(( 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{(\sigma_1^2 )/( n_1 ) +(\sigma_2^2 )/( n_2 ) } }`

=>
`t = \frac{383.08 - 390.89}{ \sqrt{(7.63^2 )/( 11 ) +(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