2. To estimate yield of a wheat variety Tianning, 10 samples are colleceu nom u
ID: 3329329 • Letter: 2
Question
2. To estimate yield of a wheat variety Tianning, 10 samples are colleceu nom u places. The yields of these 9 samples are: 12, 15, 17, 18, 18, 18, 20, 20, 25 kg/plot. Please calculate the mean, median, mode, standard deviation, CV, and the standard error of mean (10 points). 8 19 3. A random variable follows a binomial distribution with parameters n-10 and p-0.5. What's the probability P(Y4) (10 points)? 4. Tree diameters for a certain tree species are normally distributed with a mean of 20 cm and a standard deviation of 5 cm. 1) What is the probability that the diameter of a randomly chosen tree will be between 16 cm and 23? 2) Suppose we take a sample of n-5 trees. What is the probability that the average of the 5 diameters will be between 16 and 23 cm? (20 points) 5. The yields of 9 wheat variety Tianning samples are: 12, 15, 17, 18, 18, 18, 20, 20, 25 kg/plot. 1). Test the hypothesis that the mean of the population is 20g, using alpha-0.05. 2)Construct 95% confidence interval for population mean (20 points). 6. Abuse of substance containing toluene (e.g., glue) can produce various neurological symptoms. In one study, investigators measured the concentrations of various chemicals in the brains of rats that had been exposed to a toluene-laden atmosphere, and also in unexposed control rats. The concentrations of a chemical norepinephrine (NE) for six toluene-exposed rats and 5 control rats, are given in the following table. Test the hypothesis that there is no difference in NE of rats under toluene-exposed and control rats (20 points). NE concentration Toluene exposed rat 543 523 431 635 564 549 Control rat 535 385 502 412 387Explanation / Answer
Q.2 Mean of the sample = x/n= (12 + 15 + 17 + 18 + 18 + 18 + 20 + 20 + 25)/9= 18.11 kg/plot
Median = 18 kg/plot
mode = 18 kg/plot [ highest frequency]
standard deviation s= sqrt [(x - x0)2 /n-1] = 3.856 box/ plot
standard error of the mean = s/ sqrt(n) = 3.856/ sqrt(9) = 1.195 box/ plot
Q.3 n = 10 and p = 0.5
P(X = 4) = 10C4 (0.5)10 = 0.2051
Q.4 Mean = 20 cm
Standard deviation =5 cm
Let say diameter of the treee is X cm
(1) Pr( 16 cm < X < 23 cm ; 20 cm; 5 cm) = Pr(X < 23 cm) - Pr( X < 16 cm) = Pr(Z2 ) - Pr(Z1)
We will find Z - values here
Z2 = (23 - 20)/5 = 0.6 and Z1 = (16 - 20)/ 5 = -0.8
Pr( 16 cm < X < 23 cm ; 20 cm; 5 cm) = (0.6) - (-0.8)
where is standard normal cumulative distribution function. Looking for Z - table for given values.
= 0.7257 - 0.2119 = 0.5138
(b) Now we take a sample of 5 trees.
so standard error of the mean = 5/ sqrt(5) = 2.236
estimated sample mean = population mean = 20 cm
Pr( 16 cm < X < 23 cm ; 20 cm; 2.236 cm) = Pr(X < 23 cm) - Pr( X < 16 cm) = Pr(Z2 ) - Pr(Z1)
We will find Z - values here
Z2 = (23 - 20)/2.236 = 1.34 and Z1 = (16 - 20)/ 2.236 = -1.79
Pr( 16 cm < X < 23 cm ; 20 cm; 5 cm) = (1.34) - (-1.79)
where is standard normal cumulative distribution function. Looking for Z - table for given values.
= 0.9099 - 0.0367 = 0.8732
Q.5 Test statistic
t = (18.11 - 20)/ 1.195 = - 1.58 [ all values is derived and calculaed in part 2]
tcr for dF = 8 and alpha = 0.05
tcr = 2.306
so l t l < 2.306 so we shall not reject the null hypothesis and can conclude that the mean of the population is 2g.
95% confidence interval for population mean = xbar +- tcr (s/sqrt(n)
= 18.11 +- 2.306 * 1.195
= (15.354, 20.866)
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