Q6. How does Aaron Rodgers, quarterback for the 2011 Super Bowl Champion Green B

Question

Q6. How does Aaron Rodgers, quarterback for the 2011 Super Bowl Champion Green Bay Packers, compare to Drew Brees, quarterback for the 2010 Super Bowl winners, the New Orleans Saints? The table below shows the number of completed passes for each athlete during the 2010 NFL football season. Aaron Rodgers Drew Brees 19 21 719 1525 27 37 25 28 34 34 27 19 12 22 29 30 27 35 33 27 26 18 21 29 22 24 23 21 24 Do the data indicate that there is a difference in the average number of completed passes for the two quarterbacks? Test using -005. a) Do the hypothesis testing by using critical value approach (by hand) and using p. value approach (by SPSS. b) Construct a 95% confidence interval for the difference in the average number of completed passes for the two quarterbacks. Does the confidence interval confirm your conclusion in part a)?

Explanation / Answer

We are given the following sample

Aaron rogers 19,21,7,19,15,25,34,27,19,12,22,27,26,18,21

Calculating mean and standard deviation from the given data

Mean,1 = 20.8

Standard deviation,s1 =6.63

Total,n1= 15

We are given another sample

Drew Brees   27,37,25,28,34,29,30,27,35,33,29,22,24,23,21,24

Calculating mean and standard deviation from the given data

Mean ,2 = 28

Standard deviation, s2=4.83

Total n2= 16

= 0.05

PART A

Hypothesis testing

CRITICAL VALUE APPROACH

#step 1: The test is

               Null hypothesis, H=1-2=0

             Alternative hypothesis, H=1-20

#step 2: since the two samples are independent and both are less than 30(as n1=15 and n2=16) we perform t-test

    So the test statistic is

T=[(1-2)-D]/(s2p(1/n1-1/n2)

Where, S2p = [(n1-1)s21 + n(n2-1)s22]/(n1+n2-2)

Has students t-distribution with Df=n1+n2-1 = 15+16-2=29

Substituting the data

We get,

    Sp2 = 32.17

   And     T = -3.53

#step 3: since the symbol in H is () this is a two-tailed test

therefore there are two critical values, ±t2=±t0.025

from the table for t-distribution

   t0.025=2.045

    so, The rejection region is (,2.045][2.045,)

#step 4: the test statistic does falls in the rejection region. The decision is to reject H0. In the context of the problem our conclusion is:

The data provides sufficient evidence, at the 5% level of significance, to conclude that there is a difference in the average number of completed passes for the two quarterbacks.

P VALUE APPROACH

#step 1: The test is

                Null hypothesis, H=1-2=0

             Alternative hypothesis, H=1-20

#step 2: since the two samples are independent and both are less than 30 we perform t-test

    So the test statistic is

T=[(1-2)-D]/(s2p(1/n1-1/n2)

        Where, S2p = [(n1-1)s21 + n(n2-1)s22]/(n1+n2-2)

T Has students t-distribution with Df=n1+n2-1 = 15+16-2=29

Substituting the data

We get, S2p = 32.17

And T = -3.53

   #step 3: for the test statistic T = -3.53 having students t-distribution with degree of freedom = 29

            The P-Value is .001409

#step 4: Since p=0.001409<0.05= , the decision is to reject H0. In the context of the problem our conclusion is:

The data provides sufficient evidence, at the 5% level of significance, to conclude that there is a difference in the average number of completed passes for the two quarterbacks

PART B

Since the samples are small and independent

The formula for confidence interval is

     ( 1-2 )±t/2(s2p(1/n1+1/n2)

Where S2p = [(n1-1)s21 + n(n2-1)s22]/(n1+n2-2)

Substituting the values

the confidence interval is -7.2±4.17

The 95% confidence interval is [-11.37,-3.03]

We are 95% confident that the difference in the average number of completed passes lies in the interval [-11.37,-3.03]. Because the interval contains negative values

The statement in the context of the problem is that we are 95% confident that the average number of completed passes by Aaron rogers is between 3.03 units and 11.37 units lower than the average number of completed passes by Drew Brees .

This proves that there is a difference between the average number of completed passes between the two players which confirms our conclusion in part a.

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