Twenty laboratory mice were randomly divided into two groups of 10. Each group w
ID: 3315064 • 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.)
(b) State the appropriate conclusion.
a: Reject the null hypothesis, there is significant evidence that diet B had a greater weight gain.
b: Reject the null hypothesis, there is not significant evidence that diet B had a greater weight gain.
c: Fail to reject the null hypothesis, there is significant evidence that diet B had a greater weight gain
d: Fail to reject the null hypothesis, there is not significant evidence that diet B had a greater weight gain.
Diet A 9 9 6 8 13 12 14 5 13 12 Diet B 20 6 20 15 17 14 21 12 22 23Explanation / Answer
we have to calculate above proble considering the two-sample t- test
A t-test is used when you're looking at a numerical variable - for example, height - and then comparing the averages of two separate populations or groups (e.g., males and females).
Requirements
Null Hypothesis
H0: u1 - u2 > 0, where u1 is the mean of first population and u2 the mean of the second.
As above, the null hypothesis tends to be that there is grater difference between the means of the one to another populations; or, more formally, that the difference is zero
calculation and t-statistica as followes
Difference Scores Calculations
Treatment 1
N1: 10
df1 = N - 1 = 10 - 1 = 9
M1: 17
SS1: 254
s21 = SS1/(N - 1) = 254/(10-1) = 28.22
Treatment 2
N2: 10
df2 = N - 1 = 10 - 1 = 9
M2: 10.1
SS2: 88.9
s22 = SS2/(N - 1) = 88.9/(10-1) = 9.88
T-value Calculation
s2p = ((df1/(df1 + df2)) * s21) + ((df2/(df2 + df2)) * s22) = ((9/18) * 28.22) + ((9/18) * 9.88) = 19.05
s2M1 = s2p/N1 = 19.05/10 = 1.9
s2M2 = s2p/N2 = 19.05/10 = 1.9
t = (M1 - M2)/(s2M1 + s2M2) = 6.9/3.81 = 3.53
conclusin:
The t-value is 3.53498. The p-value is .001183. The result is significant at p < .05.
the clain is correct
thanks
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