CHAPTER 1

INTRODUCTION TO DATA ANALYSIS

1.2.

(10 points) One possibility:

1.4.

(10 points) No data are needed. Ideas about what affects the phenomenon under study are needed.

1.6.

(10 points) The process is not stationary. Plot of URATE versus DATE shows this:

1.10.

(a)

(4 points) Stratified plot of percent versus shift:

(b)

(6 points: 2 each) Shift 1 has highest and most consistent percentages.
Shift 2 has second highest and second most consistent percentages. Shift 3 has lowest and most variable percentages.

(5 points) Possible causes: Quality of workers
Quality of supervisors
Training

or anything else that makes sense.

1.26.

(a)

(5 points) R&R:repeatability and reproducibility.
Repeatability measures the consistency of the scale in repeated measurements of this weight by the same operator.
Reproducibility measures the variation between different operators.

(b)

(3 points) Stationarity must be checked. (2 points) Plot each operator's measurements versus time.

(c)

(5 points) Operator 2 has a much larger repeatability problem than either operator 1 or operator 3. Operator 2 should be worked with to identify and eliminate this problem.

CHAPTER 2

SUMMARIZING DATA

2.6.

(10 points) The bottom fifth cannot pull down a median. It must be the bottom half or more.

2.10.

80% (5 points); 0.15 (5 points)

2.18.

(10 points) His quote would imply that the see-saw could be balanced by putting all the weight on one side of the fulcrum.

2.24.

Skewed right (5 points), unimodal (5 points). Location: Either the mode at $\approx 16$, or the median, at 17.7. (5 points). Spread: IQR=Q₃ -Q₁=23.3-12.2=11.1. (5 points)

2.26.

The data are bimodal (5 points) with modal bars at 75-85 (mode at 80) (3 points) and 135-145 (mode at 140) (3 points). Spread of modal regions is 65-105 (3 points) and 115-165 (3 points) . Possible explanation: Data consist of two different kinds of concete. (5 points)

CHAPTER 3

DESIGNING STUDIES AND COLLECTING DATA

3.2.

(5 points) The effect of Pepsi over Coke is $$\frac{4+3+5+4+2}{5} - \frac{3+2+1+3}{4} = 3.6-2.25 = 1.35$$

3.6.

Yes it is a controlled experiment since treatments (old, new: 5 points) are assigned to experimental units (subjects: 5 points) and a response (removal of clots: 5 points) observed. (Reason it is a controlled experiment: 10 points)

3.14.

Driver-to-driver variation is included in the variation between alloys. (5 points) Better to use both alloys on each car. (5 points)

3.20.

The experimenter could have blocked by kind of worker and assigned full time workers to both types of incentive schemes and part time workers to both types of incentive schemes. (10 points)

3.24.

(a)

Yes it was a controlled experiment since treatments (aspirin/placebo) were assigned to experimental units (mice) and a response (death/not) observed. (10 points)

(b)

Not necessarily. It may be due to the drug. (10 points)

3.32.

(a)

This is a sample survey since a sample is drawn from the target population and the responses observed from the sample are used to assess the satisfaction of the target population. (5 points)

(b)

This is also a sample survey since a sample is drawn from the target population and a response measured in order to summarize responses for the population. (5 points)

(c)

This is an observational study, since the observational units are sampled, and variables are observed on each. (5 points)

(d)

This is a controlled experiment since treatments (the training program or not) are assigned to experimental units (workers) and a response observed (productivity). (5 points)

CHAPTER 4

AN INTRODUCTION TO STATISTICAL MODELING

4.14.

(a)

(10 points)

y 1 2 3 4 5 6

p_Y(y) 1/21 2/21 3/21 4/21 5/21 6/21

(b)

E(Y) = (1)(1/21) + (2)(2/21) + (3)(3/21) + (4)(4/21)+ (5)(5/21) + (6)(6/21) = 13/3 (5 points)
E(Y²) = (1)(1/21) + (4)(2/21) + (9)(3/21) + (16)(4/21) + (25)(5/21) + (36)(6/21) = 21
$\sigma^2_Y = E(Y^2) - E^2(Y) = 21 - (13/3)^2 = 20/9$ (5 points)

4.20.

(10 points) Let Y = number of weights ending in 0 or 5. If there is no rounding $Y \sim b(100, 0.2)$. The observed value of Y is y = 32. Using exact binomial calculations, $P(Y \geq 32) = 0.0031.$

Using the normal approximation,

$$P(Y \geq 32) = P(Y \geq 31.5) = P\left(\displaystyle\frac{Y... ...t{100(0.2)(0.8)}}\right) \doteq P(Z \geq 2.875) \doteq 0.0020,$$

where $Z \sim N(0,1)$.

Either method gives strong evidence against the assumption of no rounding.

4.34.

(a)

(10 points) If Y is the width, $P(0.50 < Y < 0.51) = \displaystyle\int^{0.51}_{0.50} 50y\;dy = 0.2525$.Of the next 1000, we estimate $1000 \times 0.2525 = 252.5$ will have widths between 0.50 and 0.51.

(b)

(10 points) It is $$\frac{(50)(0.51)}{(50)(0.49)} = 1.0408$$ times as likely.

4.44.

(10 points) Let Y = area covered. Want t such that $0.005 = P(Y < t) = P\left(Z < \frac{t-250}{5}\right)$. This means $\frac{t-250}{5}= -2.58$ or t = 237.1

4.50.

(10 points) Let Y = number of heads. If fair, $Y \sim b(1000, 0.5).$ Observed y = 560. $P(Y \geq 560) = 0.000083$ exactly, or using the normal approximation.

$P(Y \geq 560) = P(Y \geq 559.5) \doteq P\left(Z \geq \frac{559.5-500}{\sqrt{1000(0.5)(0.5)}}\right) = P(Z \geq 3.76) = 0.000085$.

Either way, this presents strong evidence against the assumption that the coin is fair.

CHAPTER 5

INTRODUCTION TO INFERENCE:ESTIMATION AND PREDICTION

5.6.

(10 points) Width of interval:$$\frac{2z_{0.995} \sigma}{\sqrt{n}}$$.
For $n = 100: \displaystyle\frac{2z_{0.995}\sigma}{10}$. For $n = 10000:\,\,\displaystyle\frac{2z_{0.995}\sigma}{100}$Interval for n = 100 is 10 times as wide (or:difference is:$z_{0.995}\sigma\left(\frac{1}{5} - \frac{1}{50} \right) = \frac{(2.5758)9}{50} \sigma = 0.4636\sigma)$

5.10.

Let $\hat{p}_1 =$ sample proportion of males who prefer cookie $A = \frac{94}{179} = 0.525~ (n_1 = 179)$
$\hat{p}_2 =$ sample proportion of females who prefer cookie $A = \frac{39}{66} = 0.591~ (n_2 = 66)$
$y_1 = 94,\;\; n_1 - y_1 = 85,\;\; y_2 = 39,\;\; n_2 - y_2 = 27 \geq 10$ (Assumptions met) (5 points)
95% confidence interval for difference between two population proportions: (5 points)

There is not enough evidence to conclude that the proportion of females who prefer cookie A is greater than the proportions of males since 0 is in the interval. (5 points)

5.14.

$\overline{y} = 126,\;\; s = 16.$

(a)

99% confidence interval for $\mu$ is $\overline{y} \pm t_{15,0.995} \cdot \frac{s}{\sqrt{16}}= 126 \pm(2.9467) \cdot \frac{16}{4}= (114.2,137.8)$. (5 points) As this interval contains 120, it is not possible to conclude $\mu \gt 120$. (5 points)

(b)

99% prediction interval is $\overline{y} \pm t_{15,0.995} \cdot s\sqrt{1+\frac{1}{16}}= 126\pm (2.9467)(16) \sqrt{1 + \frac{1}{16}} =(77.4, 174.6).$ (5 points) We estimate with 99% confidence that the fire endurance of the next wall tested will be in the range 77.4 - 174.6. (5 points) 99% confidence means in repeated samplng, 99% of all 99% prediction intervals so computed will contain the fire endurance of the next wall. (5 points)

(c)

This is a tolerance interval. (5 points) It is of the form $\overline{y} \pm ks$, where k = 3.421. So the interval is $126 \pm (3.421)(16) = (71.3, 180.7)$. (5 points)

(d)

The formula is $\overline{y}_1 - \overline{y}_2 \pm \hat{\sigma}(\overline{y}_1 - \overline{y}_2)t_{\nu,0.995}$ where $\nu$ is the greatest integer less than or equal to

$$\left(\frac{s^2_1}{n_1} + \frac{s^2_2}{n_2} \right)^2\left/... ...[\frac{(16^2/16)^2}{15} + \frac{(9^2/25)^2}{24}\right] = 21.$$

$$\hat{\sigma}(\overline{y}_1-\overline{y}_2) = \sqrt{\frac{s... ...2_2}{n_2}} = \sqrt{\frac{16^2}{16} + \frac{9^2}{25}} = 4.39.$$

Interval is (5 points)

$$126 - 116 \pm (4.39)(t_{21,0.995}) = 10 \pm (4.39)(2.8314) = (-2.43, 22.43).$$

Since the inteval contains , we cannot conclude there is a difference in mean fire endurance times for the two types of walls. (5 points)

5.30.

(a)

Take differences of the Port A and Port B observations and analyze as paired data. (5 points)

(b)

If correlation decreases sufficiently with distance, use the confidence interval for independent populations. (5 points)

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