SOLUTION: Sample size of A = 10 Sample mean of A = 29.5 Sample Standard Deviation of A = 0.7 Sample size of B = 9 Sample mean of B = 30.3 Sample Standard Deviation of B = 0.5 Norma

Algebra ->  Probability-and-statistics -> SOLUTION: Sample size of A = 10 Sample mean of A = 29.5 Sample Standard Deviation of A = 0.7 Sample size of B = 9 Sample mean of B = 30.3 Sample Standard Deviation of B = 0.5 Norma      Log On


   

Question 1058361: Sample size of A = 10
Sample mean of A = 29.5
Sample Standard Deviation of A = 0.7
Sample size of B = 9
Sample mean of B = 30.3
Sample Standard Deviation of B = 0.5
Normally distributed random variables.
Let alpha=0.05
Is the variability of A significantly different from B?
Is the mean of A significantly different from B?
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Let's start with the difference in means first.
Null Hypothesis-Ho:u%5BA%5D=u%5BB%5D or u%5BA%5D=u%5BB%5D
Alternate Hypothesis-Ha:u%5BA%5D%3C%3Eu%5BB%5D
alpha=0.05
Use unpooled standard deviation method,
t=%2829.5-30.3%29%2Fsqrt%280.7%5E2%2F10%2B0.5%5E2%2F9%29=-2.887+
C=+%28%280.7%29%5E2%2F10%29%2F%28%280.7%29%5E2%2F10%2B%280.5%29%5E2%2F9%29=0.6382+

Rounding down,
DOF=11
Checking the critical t value for DOF=11 and alpha=0.05.
t%5Bc%5D=-2.201
Since t%3Ct%5Bc%5D, reject the null hypothesis.
The sample means are significantly different.
.
.
.
With regards to the variances, are you supposed to use a particular test to check. F test?
In this case,
F=s%5B1%5D%5E2%2Fs%5B2%5D%5E2=%280.7%2F0.5%29%5E2=1.96
I used EXCEL with X=1.96,DOF%5B1%5D=9 and DOF%5B2%5D=8 to get a p value of 0.356.
Since 0.356%3E0.05, there is no significant difference between the samples using the variances.