In: Statistics and Probability
Random samples of batting averages from the leaders in both leagues prior to the All-Star break are shown. At the 0.05 level of significance, can a difference be concluded?
National
|
0.320 |
0.353 |
0.335 |
0.312 |
0.319 |
0.348 |
0.326 |
American
|
0.354 |
0.326 |
0.318 |
0.322 |
0.353 |
0.340 |
0.314 |
Perform each of the following steps. Assume the variables are normally distributed and the variances are unequal. Use μ1 for the National League average.
A) Find the critical values.
B) Compute the test value.
C) Reject or do not reject the hypothesis.
D) is there or is there not enough evidence to support the claim.
1)
Ho : µ1 - µ2 = 0
Ha : µ1-µ2 ╪ 0
Level of Significance , α =
0.05
Degree of freedom, DF= n1+n2-2 =
11
t-critical value , t* =
2.201 (excel formula
=t.inv(α/2,df)
2)
Sample #1 ----> 1
mean of sample 1, x̅1= 0.330
standard deviation of sample 1, s1 =
0.015
size of sample 1, n1= 7
Sample #2 ----> 2
mean of sample 2, x̅2= 0.332
standard deviation of sample 2, s2 =
0.017
size of sample 2, n2= 7
difference in sample means = x̅1-x̅2 =
0.3304 - 0.3 =
0.00
pooled std dev , Sp= √([(n1 - 1)s1² + (n2 -
1)s2²]/(n1+n2-2)) = N/A
std error , SE = Sp*√(1/n1+1/n2) =
0.0086
t-statistic = ((x̅1-x̅2)-µd)/SE = (
-0.0020 - 0 ) /
0.01 = -0.2335
3)
Decision: | t-stat | < | critical value |, so, Do
not Reject Ho
There is not enough evidence that batting averages from the leaders
in both leagues is different.
Please let me know in case of any doubt.
Thanks in advance!
Please upvote!