Question

Researchers wanted to compare the GRE scores of the students who were trained in an academy and the students who didn't receive any training. One group of 80 students, who had the training, had a mean score of 315. Another group of 120students, who had no training, had a mean score of 305. Assume the population standard deviations for the scores for the students who take the training and the students who don't take the training are 25 and 20, respectively. Find the lower bound of the 95% confidence interval for the difference in population scores between students who had the training and students who didn't have the training. Let the students who had the training be the first sample and let the students who didn't have the training be the second sample. Assume that both the population distributions are normally distributed. Assume the samples are random and independent. Round your answer to two decimal places.

Answer 1

Level of Significance , α = 0.05

mean of sample 1,               x̅1=315

standard deviation of sample 1,s1 = 25

size of sample 1,                     n1=80

mean of sample 2,               x̅2=305

standard deviation of sample 2,s2 = 20

size of sample 2,                     n2=120

difference in sample means = x̅1-x̅2   = 10

DF = min(n1-1 , n2-1 )=79

t-critical value = t α/2 = 1.990 (excel formula =t.inv(α/2,df)

std error , SE = √(s1²/n1+s2²/n2) = 3.339

margin of error, E = t*SE = 6.6452

difference of means = x̅1-x̅2   = 10

confidence interval is

Interval Lower Limit=(x̅1-x̅2) - E   = 3.35

Interval Upper Limit=(x̅1-x̅2) + E   =16.65