Question

In a sample of 100 pigs from a large population the following gains in weight (kg) during a 50 day interval were recorded:
										36	23	25	21	28	17	35	32	39	30
7	31	24	26	47	30	30	19	39	22
29	36	43	21	34	57	33	36	26	44
41	19	23	41	11	41	45	33	33	33
13	35	18	26	42	30	33	18	26	31
37	34	22	40	37	18	40	14	43	28
30	42	49	27	15	31	29	29	12	16
48	27	28	20	30	46	19	53	29	24
17	21	25	35	42	31	34	38	20	38
30	26	39	24	33	32	27	25	30	30

b. What's the prob of randomly selecting a pig that added at least 44kg to its weight during the test? How does this predicted number (predicted proportion/percent) compare with the actual number? *Hint: remember there are 100 total samples*

c. What's the probability that a pig would increase no less than 10kg and no more than 47kg?

d. Construct a 99% confidence interval for this data.

Answer 1

Sample size, n =100

Sample mean, = =30.26

Sample standard deviation, s = =9.687

b.

Assuming normal distribution,

Z =(X - ​​​​​​)/s =(44 - 30.26)/9.687 =1.4184

The probability of at least 44kg =P(X 44) =P(Z 1.4184) =0.078 =7.8%

Actual probability of at least 44 kg =8/100 =0.08 =8%

Thus, the predicted percent and the actual percent are very close to each other (almost same).

c.

At X =10, Z =(10 - 30.26)/9.687 = -2.0915

At X =47, Z =(47 - 30.26)/9.687 =1.7281

The probability that a pig would increase no less than 10kg and no more than 47kg =P(10 X 47) =P(-2.0915 Z 1.7281) =0.9398

d.

At a 99% confidence level, for a two-tailed case, the critical value of Z is: Z^* =2.58

Standard Error, SE =s/ =9.687/ =9.687/10 =0.9687

Margin of Error, MoE =Z^*(SE) =2.58(0.9687) =2.50

99% confidence interval for the population mean increase in weight, for the given data is: =MoE =30.26 2.50 =(30.26 - 2.50, 30.26 + 2.50) =[27.76kg, 32.76kg].