# In test of new surgical procedure five respected surgeons in

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```					Paired t test

Classical Approach

Let 1 and  2 denote the average amount quoted by the two companies Progressive Car
Insurance and GEICO Mutual Insurance respectively.
Here we want to test the null hypothesis H 0 : 1   2 against the alternative hypothesis
H a : 1   2 .

Let xi and y i denote the amount quoted by the two companies Progressive Car Insurance
and GEICO Mutual Insurance respectively for the i-th family. Define di  xi  yi . Let
1 n                                   1 n
d   di be the mean and sd                   (di  d )2 be the sample standard deviation of
n i 1                              n  1 i 1
the differences d i .
The test statistic for testing H0 is
d
t           t( n1) , a Student’s t distribution with (n-1) degrees of freedom.
sd n
Here n = 15, d = -246.33 and s d = 546.959
246.33
Thus the calculated value of t               = -1.74427
546.959 15
Here the level of significance is α = 0.10

Since α = 0.10, from the student’s t able with (n-1) = (15-1) = 14 degrees of freedom, the
critical value is t = 1.761

Thus the critical region is |t| > 1.761

Since the calculated |t| value = 1.74427 < 1.761 we fail to reject the null hypothesis H 0 .
So we may conclude that at the 0.10 significance level, there is no difference in the
amount quoted.

P-Value Approach

Let 1 and  2 denote the average amount quoted by the two companies Progressive Car
Insurance and GEICO Mutual Insurance respectively.
Here we want to test the null hypothesis H 0 : 1   2 against the alternative hypothesis
H a : 1   2 .
Let xi and y i denote the amount quoted by the two companies Progressive Car Insurance
and GEICO Mutual Insurance respectively for the i-th family. Define di  xi  yi . Let
1 n                                   1 n
d     
n i 1
di be the mean and sd              (di  d )2 be the sample standard deviation of
n  1 i 1
the differences d i .
The test statistic for testing H0 is
d
t           t( n1) , a Student’s t distribution with (n-1) degrees of freedom.
sd n
Here n = 15, d = -246.33 and s d = 546.959
246.33
Thus the calculated value of t             = -1.74427
546.959 15

Here the level of significance is α = 0.10

The p-value of the test is given by
p-value = P[|t| > 1.74427] = 0.1030

Since the p-value is greater than 0.10, the level of significance, we fail to reject the null
hypothesis H 0 .
So we may conclude that at the 0.10 significance level, there is no difference in the
amount quoted.

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