Stats: Modeling the World Nasta Edition Grades 9-12
Stats: Modeling the World Nasta Edition Grades 9-12
3rd Edition
ISBN: 9780131359581
Author: David E. Bock, Paul F. Velleman, Richard D. De Veaux
Publisher: PEARSON
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Question
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Chapter 25, Problem 27E

(a)

To determine

To explain is there evidence that after training players can throw strikes than 60% of the time.

(a)

Expert Solution
Check Mark

Answer to Problem 27E

There is sufficient evidence to support the claim that after training players can throw strikes than 60% of the time.

Explanation of Solution

The table of number of strikes before and after is given in the question. It is also given that,

  n=20

Now, the sample mean and the standard deviation of the number of strikes are as follows:

  x¯=35+36+....+35+3220=33.15s=(3533.15)2+(3633.15)2+....+(3533.15)2+(3233.15)2201=2.3232

Let us now define the hypotheses as:

  H0:μ=60%×50=30Ha:μ>30

Now the degree of freedom is then as:

  df=n1=201=19

Thus, the value of the test statistics is then as:

  t=x¯μsn=33.15302.323220=6.064

Thus, the P-value can be calculated using table T of appendix F as:

  P<0.005

As we know that the P-value is less than the significance level then the null hypothesis is rejected. Thus, we have,

  P<0.05Reject H0

Thus, we conclude that there is sufficient evidence to support the claim that after training players can throw strikes than 60% of the time.

(b)

To determine

To explain is there evidence that the training is effective in improving a player’s ability to throw strikes.

(b)

Expert Solution
Check Mark

Answer to Problem 27E

There is no sufficient evidence to support the claim that the training is effective in improving a player’s ability to throw strikes.

Explanation of Solution

The table of number of strikes before and after is given in the question. It is also given that,

  n=20

And we will find the difference between the two samples as:

  Stats: Modeling the World Nasta Edition Grades 9-12, Chapter 25, Problem 27E

Now, the sample mean and the standard deviation of the difference are as follows:

  d¯=77+....+2+520=0.1sd=(7(0.1))2+(7(0.1))2+....+(2(0.1))2+(5(0.1))2201=3.3230

Let us now define the hypotheses as:

  H0:μd=0Ha:μd<0

Now the degree of freedom is then as:

  df=n1=201=19

Thus, the value of the test statistics is then as:

  t=d¯sd/n=0.13.3230/20=0.135

Thus, the P-value can be calculated using table T of appendix F as:

  P>0.20

As we know that the P-value is less than the significance level then the null hypothesis is rejected. Thus, we have,

  P>0.05Fail to Reject H0

Thus, we conclude that there is no sufficient evidence to support the claim that the training is effective in improving a player’s ability to throw strikes.

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