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, 2098 1931 2053 2415 2207 2004 2221 1592 Type 2, 2065 1962 2061 2448 2145 1969 2146 1491 In this​ example, mu 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.)

Answer 1

Type 1, 2098 1931 2053 2415 2207 2004 2221 1592

Type 2, 2065 1962 2061 2448 2145 1969 2146 1491

The difference d between type 1 seed yield minus the type 2 seed yield is,

33, -31, -8, -33, 62, 35, 75, 101

Mean difference = (33 -31 -8 -33 + 62 + 35 + 75 + 101)/ 8 = 29.25

Standard deviation of difference , d is,

sd = sqrt[{(33 - 29.25)² + (-31 - 29.25)² + (-8 - 29.25)² + (-33 - 29.25)² + (62 - 29.25)² + (35 - 29.25)² + ( 75 - 29.25)² + (101 - 29.25)² }/ 7 ]

= 49.64661

Standard error of the mean difference = sd / = 49.64661 / = 17.55273

Degree of freedom = n - 1 = 8 - 1 = 7

t value for 95% confidence interval and degree of freedom = 7 is 2.36

95% confidence interval of the mean difference =

(29.25 - 2.36 * 17.55273, 29.25 + 2.36 * 17.55273)

= (-12.17, 70.67)

As, the 95% confidence interval contains the value 0, there is no significantg evidence in farmer​ Joe's claim that type 1 seed is better than type 2​ seed.