Question
Pitcher 1 Pitcher 2
87 82
86 92
82 70
84 96
83 89
81 84
85 84
93 80
86 81
85 89
84 86
92 72
83 77
84 87
80 89
87 93
88 78
87 81
79 82
82 87
82 81
87 84
80 88
88 93
90 80
85 79
86 87
87 74
86 78
85 80
85 83
88 79
84 95
83 81
88 89
87 91
94 93
83 91
88 92
86 84
85 83
84 84
82 87
89 87
85 80
80 83
83 88
79 88
85 80
86 93

83. FILE Fastballs. The manager of a minor league baseball team wants to estimate the average fastball speed of two pitchers.

A. Use the data file to complete this problem.

B. Construct 95% confidence intervals for the mean speed for each pitcher.

C. Explain why the widths of the two intervals are different.

D. Who is the more consistent pitcher in terms of the speed of their fastball? Explain.

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Answer #1

Solution

[Note: Final answers are given first. Detailed Working follow at the end.]

Back-up Theory

100(1 - α) % Confidence Interval for μ, when σ is not known is: M ± MoE,where

MoE = (tn- 1, α /2)s/√n wherein

M = sample mean,

tn – 1, α /2 = upper (α/2)% point of t-distribution with (n - 1) degrees of freedom,

s = sample standard deviation and

n = sample size.

Now to work out the solution,

B. 95% confidence intervals for the mean speed for pitcher 1: [84.02, 86.09] Answer 1

    95% confidence intervals for the mean speed for pitcher 2: [82.99, 86.37] Answer 2

C. From the formula for CI given above, it is clear that the width of the interval is solely determined by the MoE.

Since tn – 1, α /2 and n are the same for both pitchers, the above further, boils down to only the s, the sample standard deviation, which is more for Pitcher 2 than for Pitcher 1

That explains why the widths of the two intervals are different. Answer 3

D. Consistency is measured by the standard deviation, the relation being less the standard deviation more the consistency. Sample standard deviation being more for Pitcher 2 than for Pitcher 1,

Pitcher 1 is the more consistent pitcher in terms of the speed of their fastball. Answer 4

Detailed Working

n

50

50

Mean - M

85.16

84.68

SD - s

3.2972

5.9365

tn – 1, α /2

2.0096

2.0096

√n

7.0711

7.0711

MoE

0.9371

1.6872

Lower Bound

84.2229

82.9928

Upper Bound

86.0971

86.3672

DONE

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