Question

The data below are yields for two different types of corn seed that were used on adjacent plots of land. Assume that the data are simple random samples and that the differences have a distribution that is approximately normal. Construct a 95% confidence interval estimate of the difference between type 1 and type 2 yields. What does the confidence interval suggest about farmer Joe's claim that type 1 seed is better than type 2 seed? Type 1 2029 1968 2120 2469 2135 2037 2212 1407 Type 2 2069 1974 2071 2499 2102 1923 2149 1439 In this example, He is the mean value of the differences d for the population of all pairs of data, where each individual difference d is defined as the type 1 seed yield minus the type 2 seed yield. The 95% confidence interval is - 27.10 < Ho < 64.85 (Round to two decimal places as needed.) What does the confidence interval suggest about farmer Joe's claim that type 1 seed is better than type 2 seed? O A. Because the confidence interval only includes positive values and does not include zero, there is not sufficient evidence to support farmer Joe's claim. O B. Because the confidence interval only includes positive values and does not include zero, there is sufficient evidence to support farmer Joe's claim. OC. Because the confidence interval includes zero, there is not sufficient evidence to support farmer Joe's claim. OD. Because the confidence interval includes zero, there is sufficient evidence to support farmer Joe's claim.

Answer 1

• We construct the following table:

  	Type I	Type II	d
  	2029	2069	-40
  	1968	1974	-6
  	2120	2071	49
  	2469	2499	-30
  	2135	2102	33
  	2037	1923	114
  	2212	2149	63
  	1407	1439	-32
  Total	16377	16226	151

  
  
  The 95% Confidence Interval is given by:
  $(\overline{d}-\frac{s_d}{\sqrt{n}}*t_{n-1;0.025},\overline{d}+\frac{s_d}{\sqrt{n}}*t_{n-1;0.025})$
  
  Now,
  $\overline{d}=\frac{\sum_{i=1}^8d_i}{8}=\frac{151}{8}=18.875$
  $s_d=\sqrt{\frac{\sum_{i=1}^{8}(d_i-\overline{d})^2}{8-1}}=54.98684908$
  $t_{n-1;0.025}=t_{7;0.025}=2.36462$
  
  Putting in the given values, we get,
  
  (18.875- 54.98684908 + 2.36462, 18.875 + 8 54.98684908 8 *2.36462)
  
  
  
  Hence, the 95% Confidence Interval is
  
  
• 95% Confidence Interval is actually the acceptance region of the test based on the sample. Since the acceptance region includes 0, this means that there is no signifcant evidence for the claim.
  
  Hence, option C) is the correct option. That is,
  Because the Confidence Interval includes zero, there is sufficient evidence to support farmer Joe's claim.
  

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