Question

Have you ever wondered, when you weigh yourself on a scale, how the scale is calibrated? How is it known that 150 pounds on that scale is the same as 150 pounds on another? Scales are calibrated by comparison to standard weights. Of course, millions of instruments require calibration and they cannot all be compared to the same weight. Rather, there is a single definitive weight, and a hierarchy of other weights that are compared to it. The U.S. National Bureau of Standards maintains a set of standard weights that are an important link in this chain. In Statistics, Freedman et al report the results of a set of definitive measurements by the Bureau of one 10 gram standard weight, done sometime in 1962-3. The first measurement was 9.999591. The error is very slight, on the order of what a grain of salt weights. Several other measurements also came in just slightly below 10 grams. To make interpretation easier, the Bureau chose to measure not "grams," but, rather "micrograms below 10 grams." A microgram is one-millionth of a gram. So, instead of 9.999591, the measurement was 409. Here is a sample of 20 such measurements: 409, 400, 406, 399, 402, 406, 401, 403, 401, 403, 398, 403, 407, 402, 401, 399, 400, 401, 405, 402 Based on this sample, you would estimate that this standard weight falls short by _____ micrograms. The variation between one measurement and another is probably due to __________ . Do a bootstrap simulation to determine how much variability there might be in your estimate in part (a). Produce a confidence interval. (This should be done on computer. If you have so far been totally unsuccessful in getting any of the software programs to work, you may do ten simulations with actual slips of paper and a hat, report the results, and describe how you would use the results of 1000 such simulations). (OPTIONAL) Calculate this interval using statistical software and a standard formula approach.

Answer 1

Solution

Answers are first given. Detailed working is given at the end.

Part (a)

Point estimate for the AVERAGE duration of baseball games = 2:57 ANSWER 1.

Confidence interval around the point estimate; [Since confidence level is not stipulated, the oft-used level of 95% is assumed.]

95% Confidence Interval for mean, μ, is: Xbar ± (1.96s)/√n where

Xbar = sample mean = 2:57,

2.228 is upper 2.5% point of t- distribution with 10 [i.e., (n – 1)] degrees of freedom,

s = sample standard deviation = 0:25.9436 and

n = sample size = 11.

Substituting the values, 95% Confidence Interval = [2:40, 3:14] ANSWER 2

Part (b)

Separate point estimates,

National League games: 2:58 ANSWER 3

American League games: 2:55 ANSWER 4

Part (c)

Comment on the difference between the two:

There is a difference of just 3 minutes, which cannot be really considered statistically significant. Also, figures for National League games has one value for 11 innings which is naturally higher and could be the possible reason for the marginal difference. ANSWER 5

Part (d)

The population that we are making an inference to:

All games played by the two leagues. ANSWER 6

It reasonable to assume that it is Normally-shaped. However, the sample size is a bit too small to make this assertion valid. ANSWER 7

Comment on the validity of the sampling method.

It is a random sample. However, just one season may not reflect the population of all games. From that view point, the sample is not representative. At least two seasons would be the bare minimum. ANSWER 8

Calculations Details

Game	N	A
1	29	62
2	33	83
3	53	67
4	56	20
5	110	41
6	68
n	6	5
Sum	349	273
Average	58.16667	54.6
(N+A)
n	11
Sum	622
Average	56.54545

NOTE [Going beyond]

Since the point of concern is the television view time, the above analysis is fine.

However, the fact that one of the figures for League A is based on 11 innings makes these estimates non-standard given that standard innings is 10.

The adjusted figures are given below:

Figures Adjusted for 10 innings
[One figure of 3:50(N) is for 11 innings.
This figure is adjusted for 10 innings as (230 min x 10)/11 =
209 min = 3:29] N only
n	6
Sum	12:328
Average	2:54.6667
(N+A)
n	11
Sum	22:601
Average	2:54.5454