Solution to From a random sample of 10 pigs, fed on diet A, the increases in weight … - Sikademy
Author Image

Archangel Macsika

From a random sample of 10 pigs, fed on diet A, the increases in weight in a certain period were 10,6,16,17,13,12,8,14,15,9 lbs. For another random sample of 12 pigs fed on diet B, the increases in the same period were 7,12,22,15,12,14,18,8,21,23,10,17 lbs. Test whether diets A and B differ significantly as regards their effect on increases in weight? (t % at d. f. 20 is 2.08

The Answer to the Question
is below this banner.

Can't find a solution anywhere?

NEED A FAST ANSWER TO ANY QUESTION OR ASSIGNMENT?

Get the Answers Now!

You will get a detailed answer to your question or assignment in the shortest time possible.

Here's the Solution to this Question

Diet A

n_1=10\\\bar x_1={\sum x\over n_1}={120\over10}=12

s_1^2={\sum x^2-{(\sum x)^2\over n_1}\over n_1-1}={1560-1440\over9}=13.33333

Diet B

n_2=12\\\bar x_2={\sum x\over n_2}={179\over12}=14.92

s_2^2={\sum x^2-{(\sum x)^2\over n_2}\over n_2-1}={2989-2670.083\over11}=28.9924245

Before we test for the difference in means we have to test their variability using F-test.

We test,

H_0:\sigma_1^2=\sigma_2^2\\vs\\H_1:\sigma_1^2\not=\sigma_2^2

The test statistic is,

F_c={s_2^2\over s_1^2}={28.9924245\over13.3333333}=2.1744

The table value is,

F_{\alpha,n_1-1,n_2-1}=F_{0.05,11,9}= 3.102485 and we reject the null hypothesis if F_c\gt F_{\alpha,n_1-1,n_2-1}

Since F_c=2.1744\lt F_{0.05,11,9}= 3.102485, we fail to reject the null hypothesis and conclude that the means for both diets are equal.


Now,

The hypothesis tested are,

H_0:\mu_1=\mu_2\\vs\\H_1:\mu_1\not=\mu_2 

The test statistic is,

t_c={(\bar x_1-\bar x_2)\over \sqrt{sp^2({1\over n_1}+{1\over n_2})}}

where sp^2 is the pooled sample variance given as,

sp^2={(n_1-1)s_1^2+(n_2-1)s_2^2\over n_1+n_2-2}={(9\times13.3333333)+(11\times28.9924245)\over20}={438.91667\over20}=21.9458335

Therefore,

t_c={(12-14.91667)\over \sqrt{21.95({1\over 10}+{1\over 12})}}={2.91667\over2.0058}=1.4541(4dp)

t_c is compared with the table value at \alpha=0.05 with n_1+n_2-2=10+12-2=20 degrees of freedom.

The table value is,

t_{{0.05\over2},20}=t_{0.025,20}= 2.085963

The null hypothesis is rejected if t_c\gt t_{0.025,20}.

Since, t_c=1.4541\lt t_{0.025,20}=2.085963, we fail to reject the null hypothesis and conclude that there is no sufficient evidence to show that the two sample means for diet A and diet B differ significantly regarding their effect on weight increase at 5% level of significance.


Related Answers

Was this answer helpful?

Join our Community to stay in the know

Get updates for similar and other helpful Answers

Question ID: mtid-4-stid-46-sqid-2428-qpid-898