Mathematics, 25.02.2020 05:25 kfloyd6046

Rothamsted Experimental Station (England) has studied wheat production since 1852. Each year many small plots of equal size but different soil/fertilizer conditions are planted with wheat. At the end of the growing season, the yield (in pounds) of the wheat on the plot is measured.
(a) Suppose for a random sample of years, one plot gave the following annual wheat production (in pounds):
4.46, 4.21, 4.40, 4.81, 2.81, 2.90, 4.93, 3.54, 4.16, 4.48, 3.26, 4.74, 4.97, 4.02, 4.91, 2.59
For this plot, the sample variance is _.
(b) Another random sample of years for a second plot gave the following annual wheat production (in pounds):
3.89, 3.81, 3.95, 4.07, 4.01, 3.73, 4.02, 3.78, 3.72, 3.96, 3.62, 3.76, 4.02, 3.73, 3.94, 4.03
For this plot, the sample variance is _.
(c) Test the claim using that the population variance of annual wheat production for the first plot is larger than that for the second plot.

Answers: 3

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Answer from: dondre54

(a) 0.653

(b) 0.0198

(c) Yes, after testing we conclude that the population variance of annual wheat production for the first plot is larger than that for the second plot.

Step-by-step explanation:

(a) We are give the sample of one year annual production of wheat (in pounds) ;

4.46, 4.21, 4.40, 4.81, 2.81, 2.90, 4.93, 3.54, 4.16, 4.48, 3.26, 4.74, 4.97, 4.02, 4.91, 2.59

For calculating sample variance, firstly we will calculate mean of the above data;

Mean of above data, X_1bar = Sum of all values ÷ n (no. of values)

= $\frac{4.46 + 4.21+4.40+.......+4.91+2.59}{16}$ = 4.074

Sample Variance, $s_1^{2}$ = $\frac{\sum (X-X_1bar)^{2} }{n-1}$ = $\frac{(4.46-4.074)^{2}+(4.21-4.074)^{2}+.........+(4.91-4.074)^{2}+(2.59-4.074)^{2} }{16-1}$ = 0.653

(b) Another sample for annual wheat production (in pounds);

3.89, 3.81, 3.95, 4.07, 4.01, 3.73, 4.02, 3.78, 3.72, 3.96, 3.62, 3.76, 4.02, 3.73, 3.94, 4.03

Mean of above data, X_2bar = Sum of all values ÷ n (no. of values)

= $\frac{3.89+3.81+3.95.......+3.94+4.03}{16}$ = 3.88

Sample Variance, $s_2^{2}$ = $\frac{\sum (X-X_2bar)^{2} }{n-1}$ = $\frac{(3.89-3.88)^{2}+(3.81-3.88)^{2}+.........+(3.94-3.88)^{2}+(4.03-3.88)^{2} }{16-1}$ = 0.0198

(c) Now, we have to test the claim that population variance of annual wheat production for the first plot is larger than that for the second plot i.e.;

Null Hypothesis, H_0 : $\sigma_1^{2} = \sigma_2^{2}$ or H_0 : $\frac{\sigma_1^{2} }{\sigma_2^{2} } = 1$

Alternate Hypothesis, H_0 : $\sigma_1^{2} \sigma_2^{2}$ or H_0 : $\frac{\sigma_1^{2} }{\sigma_2^{2} } 1$

The test statistics used here is;

$\frac{s_1^{2} }{s_2^{2} }* \frac{\sigma_1^{2} }{\sigma_2^{2}}$ ~ F_n__1-1_,n_2-1 where, n_1 = 16 and n_2 =16

Test Statistics = $\frac{0.653 }{ 0.0198}* 1$ ~ F_1_5_,_1_5

= 32.98

Since, we are not provided with any significance level so we assume it to be 5% and at this level, the F table gives critical value of 2.4282.

Since our test statistics is higher than the critical value and it falls in the rejection region so we have sufficient evidence to reject null hypothesis and conclude that the population variance of annual wheat production for the first plot is larger than that for the second plot.

Answer from: Quest

the answer is 375 miles

step-by-step explanation:

Answer from: Quest

europa2433

step-by-step explanation: