Mathematics: 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.

Add the cost of pansies to 15 and the cost of trees to 8. Then add them together.Step-by-step explanation:That s what I said and I got it correct so there you go, hope this helps....

A pharmaceutical company is interested in testing

Ответ:

loyaltyandgood

03.02.2022

Mathematics

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

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