Answer:
The null hypothesis is [tex]H_o : \mu_1 = \mu_2[/tex]
The alternative hypothesis is [tex]H_a : \mu_1 < \mu_2[/tex]
The test statistics is [tex]t = -2.65[/tex]
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
[tex]\= x = \frac{\sum x_i }{n}[/tex]
=> [tex]\= x = \frac{373.2 +376.7 +\cdots + 383.5}{11}[/tex]
=> [tex]\= x = 383.08 [/tex]
Generally the sample mean for Y is mathematically represented as
[tex]\= y = \frac{\sum y_i }{n}[/tex]
=> [tex]\= y = \frac{395.5 +384.8 +\cdots + 398.8}{10}[/tex]
=> [tex]\= y = 390.89 [/tex]
Generally the standard deviation for X is mathematically represented as
[tex]\sigma_1 = \sqrt{\frac{\sum (x_i - \= x_1 )^2}{n} }[/tex]
=> [tex]\sigma_1 = \sqrt{\frac{( 373.2 - 383.08)^2 +( 376.7 - 383.08)^2 +\cdots + ( 383.5 - 383.08)^2 }{11} }[/tex]
=> [tex]\sigma_1 = 7.63 [/tex]
Generally the standard deviation for Y is mathematically represented as
[tex]\sigma_2 = \sqrt{\frac{\sum (y_i - \= y )^2}{n} }[/tex]
=> [tex]\sigma_2 = \sqrt{\frac{( 395.5 - 390.89)^2 +( 384.8 - 390.89)^2 +\cdots + ( 398.8 - 390.89)^2 }{10} }[/tex]
=> [tex]\sigma_2 = 5.82 [/tex]
The null hypothesis is [tex]H_o : \mu_1 = \mu_2[/tex]
The alternative hypothesis is [tex]H_a : \mu_1 < \mu_2[/tex]
Generally the test statistics is mathematically represented as
[tex]t = \frac{\= x - \= y}{ \sqrt{\frac{\sigma_1^2 }{ n_1 } +\frac{\sigma_2^2 }{ n_2 } } }[/tex]
=> [tex]t = \frac{383.08 - 390.89}{ \sqrt{\frac{7.63^2 }{ 11 } +\frac{5.82^2 }{10} } }[/tex]
=> [tex]t = -2.65[/tex]
Generally the level of significance is [tex]\alpha = 0.05[/tex]
Generally the degree of freedom is mathematically represented as
[tex]df = n_1 + n_2 - 2[/tex]
=> [tex]df = 11 + 10 - 2[/tex]
=> [tex]df = 19[/tex]
Generally from the student t- distribution table the critical value of [tex] \alpha [/tex] at a df = 19 is
[tex]t_{\alpha ,df} =t_{0.05 ,19} = -2.093[/tex]
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
