Question

The data below are yields for two different types of com 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 2148 1924 2100 2544 2170 2006 2270 1415 Type 2 2097 1912 2091 2445 2155 1927 21831425 In this example. He is the mean value of the differences d for the population of all pairs of data, where each individual difference is defined as the type 1 seed yield minus the type 2 seed yield. The 95% confidence interval is < < (Round to two decimal places as needed)

Answer 1

$\bar d$ = (51 + 12 + 9 + 99 + 15 + 79 + 87 + (-10))/8 = 42.75

s_d = sqrt(((51 - 42.75)^2 + (12 - 42.75)^2 + (9 - 42.75)^2 + (99 - 42.75)^2 + (15 - 42.75)^2 + (79 - 42.75)^2 + (87 - 42.75)^2 + (-10 - 42.75)^2)/7) = 41.647

df = 8 - 1 = 7

At 95% confidence level, the critical value is t^* = 2.365

The 95% confidence interval is

$\bar d \pm t^{*} * \frac{s_{d}}{\sqrt n}$

$= 42.75 \pm 2.365 * \frac{41.647}{\sqrt {8}}$

$= 42.75 \pm 34.82$

The 95% confidence interval is $7.93 < \mu_{d} < 77.57$ ​​​​​​