Twenty laboratory mice were randomly divided into two groups of 10. Each group w
ID: 3130178 • Letter: T
Question
Twenty laboratory mice were randomly divided into two groups of 10. Each group was fed according to a prescribed diet. At the end of 3 weeks, the weight gained by each animal was recorded. Do the data in the following table justify the conclusion that the mean weight gained on diet B was greater than the mean weight gained on diet A, at the = 0.05 level of significance? Assume normality. (Use Diet B - Diet A.)
(a) Find t. (Give your answer correct to two decimal places.)
(ii) Find the p-value. (Give your answer correct to four decimal places.)
Explanation / Answer
a)
Formulating the null and alternative hypotheses,
Ho: u1 - u2 <= 0
Ha: u1 - u2 > 0
At level of significance = 0.05
As we can see, this is a right tailed test.
Calculating the means of each group,
X1 = 14.8
X2 = 8.7
Calculating the standard deviations of each group,
s1 = 4.984420171
s2 = 2.750757471
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):
n1 = sample size of group 1 = 10
n2 = sample size of group 2 = 10
Thus, df = n1 + n2 - 2 = 18
Also, sD = 1.800308616
Thus, the t statistic will be
t = [X1 - X2 - uD]/sD = 3.388307953 [ANSWER]
where uD = hypothesized difference = 0
*****************************
ii)
Also, using p values,
p = 0.001637608 [ANSWER]
*****************************************************
Hi! If you use another method/formula in calculating the degrees of freedom in this t-test, please resubmit this question together with the formula/method you use in determining the degrees of freedom. That way we can continue helping you! Thanks!
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.