1. Due to differences in the environment, the masses of a certain species of sma
ID: 3150820 • Letter: 1
Question
1. Due to differences in the environment, the masses of a certain species of small animal are believed to be greater in Region A than in Region B. It is known that the masses in both regions are normally distributed, with masses in Region A having a standard deviation of 0.04 kg and masses in Region B having a standard deviation of 0.09 kg. To test the theory, random samples are taken: 60 animals from Region A had a mean mass of 3.03 kg and 50 animals from Region B had a mean mass of 3.00 kg.
Does this provide evidence, at the 1% level that the animals of this species in Region A have a greater mass than Region B?
2. The same physical fitness test was given to a group of 100 scouts and to a group of 144 guides. The maximum score was 30. The guides obtained a mean score of 26.81 and the scouts obtained a mean score of 27.53. Assuming that the fitness scores are normally distributed with a common population standard deviation of 3.48, test at the 5% level of significance whether the guides did not do as well as the scouts in the fitness test.
3. An investigation was carried out to assess the effects of adding certain vitamins to the diet. A group of two-week old rats was given a vitamin supplement in their diet for a period of one month, after which time their masses were noted. A control group of rats of the same age was fed on an ordinary diet and their masses were also noted after one month.
The results are summarized in the table:
No. in sample
Mean
Std Dev.
With vitamin supplement
64
89.6 g
12.96 g
Without vitamin supplement
36
83.5 g
11.41 g
Treating the samples as large samples from normal distributions with the same population variance, test at the 5% level whether the results provide evidence that rats given the vitamin supplement have a greater mass, at age six weeks, than those not given the vitamin supplement.
No. in sample
Mean
Std Dev.
With vitamin supplement
64
89.6 g
12.96 g
Without vitamin supplement
36
83.5 g
11.41 g
Explanation / Answer
Let A be the mass , in kg, of an animal in region A and let the population mean be u1. Then A~N(u1,.04^2).
Let B be the mass , in kg, of an animal in region B and let the population mean be u2. Then B~N(u2,.09^2).
H1:u1-u2=0(there is no difference in the masses between the regions)
H2:u1-u2>0 (the animals in Region Ahave greater mass)
A-B~N(u1-u2,a1/n1+a2/n2)with n1=60;n2=50
if H1 is true then u1-u2=0
so A-B~N(0,.04^2/60+.09^2/50)
Use a one tailed test at the 1% level.
the critical z value is 2.326, so reject H1 if z>2.326
where z=x1-x2/sqrt(.04^2/60+.09^2/50)
=3.03-3.00/.0137
=2.184
since z<2.326, do not reject H1.
there is no evidence , at the 1% level that the animals of this species in Region A have a greater mass than those in Region B.
3.
Let A be the mass , in gms, of rat with vitamin and let the wieght mean be u1. Then A~N(u1,.12.96^2).
Let B be the mass , in gms, of rat without vitamin and let the population mean be u2. Then B~N(u2,.11.41^2).
H1:u1-u2=0(there is no difference in the masses between the rats)
H2:u1-u2>0 (the rats with vitamin have greater mass)
A-B~N(u1-u2,a1^2/n1+a2^2/n2)with n1=64;n2=36
if H1 is true then u1-u2=0
so A-B~N(0,12.96^2/64+11.41^2/36)
Use a one tailed test at the 5% level.
the critical z value is 1.645, so reject H1 if z>1.645
where z=x1-x2/sqrt(12.96^2/64+11.41^2/36)
=64-36/sqrt(2.62+3.6)
=28/2.49
=11.2
since z>1.645, we reject H1.
there is evidence ,at the 5% level , that the rats given the vitamin supplement have a greater mass than the rats not given the supplement
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