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 2162 2050 2164 2526 2148 2030 2218 1451 Type 2 2045 1964 2079 2455 2106 1943 2184 1465 In this? example, mu Subscript d 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 nothingless thanmu Subscript dless than nothing. ?(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? A. Because the confidence interval includes ?zero, there is sufficient evidence to support farmer? Joe's claim. B. Because the confidence interval only includes positive values and does not include ?zero, there is sufficient evidence to support farmer? Joe's claim. C. Because the confidence interval only includes positive values and does not include ?zero, there is not sufficient evidence to support farmer? Joe's claim. D. Because the confidence interval includes ?zero, there is not sufficient evidence to support farmer? Joe's claim. Click to select your answer(s).

Answer 1

(a)

Type 1 (X)            Type 2 (Y)               d=X-Y

2162                     2045                     117

2050                    1964                      86

2164                    2079                     85

2526                    2455                     71

2148                   2106                       42

2030                    1943                    87

2218                    2184                    34

1451                    1465                    - 14

From the d values, the following statistics are calculated:

n = 8

= 508/8 = 63.5

s_d = 41.0192

SE = s_d/

= 41.0192/ = 14.5026

= 0.05

ndf = 8 - 1 = 7

From Table, critical values of t = 2.3646

Confidence interval:

t SE

= 63.5 (2.3646 X 14.5026)

= 63.5 34.2929

=(29.21 , 97.79)

95% Confidence Interval is:

29.21 < < 97.79

(b)

Correct option:

B. Because the confidence interval only includes positive values and does not include zero, there is sufficient evidence to support farmer Joe's claim.