SSD = SSQ - SUM^{2}/n

= 50 - 6^{2}/3

= 50 - 12 = 38## PART 2. CALCULATIONS FROM A SAMPLE

2.1.A sample of three observations has a sum of 6 and a sum of squares equal to 50. One observation is -3. What are the other two observations?

[1]: X1 + X2 + X3 = 6[2]: X1^{2}+ X2^{2}+ X3^{2}= 50[3]: X3 = -3Use [3] in [1] and [2]:[1']: X1 + X2 = 9; X2 = 9 - X1[2']: X1^{2}+ X2^{2}= 41,[2'']: X1^{2}+ (9-X1)^{2}= 41X1^{2}+ 81 - 18 X1 + X1^{2}= 412*X1^{2}- 18*X1 -40 = 0(X1 - 4)(X1 - 5) = 0X1 = 4 or 5Say X1 = 4; then X2 = 9-X1 = 5.(Many people obtain the solution by trial-and-error.)

2.2.A sample of 10 observations has a mean of 100. The sum of 9 of the observations is 900. What is the value of the other observation?SOLUTION: Sum of all ten = 10 x mean = 10 x 100 = 1000. Sum of nine = 900

Difference = value of other observation = 100

2.3.A sample ofn= 14 has a standard deviation of 3.1.

What is the sum of squared deviations?SOLUTION:

Sample variance, sFind the value of^{2}= 3.1^{2 }= 9.61 = SSD/(n-1) = SSD/13; SSD = (9.61)(13) = 124.932.4.Consider the following table. TABLE. Distribution of Number of Magazine Subscriptions in Households ----------------------------------------------- Number of subscriptions 0 1 2 3 Number of households 10 40 30f-----------------------------------------------fsuch that the mean number of subscriptions per household is 2.0.f= ?SOLN: 2.0 = mean = sum/number

= [10(0)+40(1)+30(2)+3f] / (10+40+30+f);f= 60## PART 3. PROBABILITY: EQUALLY LIKELY CASES

A custodian is asked to rank four brands (A, B, C, D) of common household cleanser according to his preference, number 1 being the cleanser he prefers most, and so on. Suppose the custodian really has no preference among the four brands and hence all orders are equally likely to occur.

3.1.What is the probability that C is first and D is third in the ranking? C D There are two ways to fill in the blanks, A then ___ ___ ___ ___ B or B then A. The total number of possibilities 1st 2nd 3rd 4th is 4!, or 24. The prob. is 2/24 = 1/12 = .0833.

3.2.What is the probability that A is ranked either second or third? These are 2 out of 4 equally probable rankings for A; hence the prob is 2/4, or 1/2.## PART 4. PROBABILITY: COMPOUND EVENTS

4.1.A state highway department has contracted for the delivery of sand, gravel, and cement at a construction site. Due to other work commitments and labor force problems, contracting firms cannot always deliver items on the agreed delivery date. Based on past evidence, the probabilities that sand, gravel, and cement will be delivered on the promised delivery dates by the contracting firms are .3, .6 and .8, respectively. Assume that the delivery or nondelivery of one material is independent of another. Find the probability that all three materials will be delivered on time..3 x .6 x .8 = .144

## PART 5. EXPECTED VALUE OF A RANDOM VARIABLE

5.1. Consider the following probability distribution of a random variablex.-----------------------------v-3 5 10P(x = v) .2 p .8-p-----------------------------

What is the value ofpso that the expected value ofxis 5.0?5.0 =

E(x) = .2(-3) + 5p+ 10(.8-p),p= .48

## PART 6. DISCRETE DISTRIBUTIONS

6.1.Often the values 1, 2, 3, 4 and 5 are assigned to categories such as "Strongly Disagree," "Disagree Somewhat," "Neutral," "Agree Somewhat," "Strongly Agree." This question has to do with such a "scaling." Of all random variables taking the values 1, 2, 3, 4, 5, the one with P(x=1) = 1/2 and P(x=5) = 1/2 has maximum standard deviation. What is the value of this standard deviation?SOLUTION: Variance of a random variable = probability-weighted average of the squared deviations from the mean

= .5(1-3)+ .5(5-3) ^{2}=4; std.dev. = 2.

PART 7. BINOMIAL DISTRIBUTIONS

7.1.Suppose a binomial distribution has a mean of 6 and a variance of 3. Then what are the values of the parameters n and p of the distribution? mean = np = 6; 3 = variance = npq = 6; q = 1/2, p = 1/2, n = 12

## PART 8. SAMPLING FROM A FINITE POPULATION

8.1. A sample of n = 400 is to be drawn (without replacement) from a population of N = 2000 with a standard deviation of $4000. What is the standard deviation of the sample mean?Var(y) = [pop.var./n][(N-n)/(N-1)] = [4000

^{2}/400][(2000-400)/(2000-1)] = [4000^{2}/400](1600/1999) _ SD(y) = (4000/20)*sqrt(1600/1999) = 200*sqrt(.8004) = 200*.895 = $178.93

## PART 9. NORMAL DISTRIBUTIONS: PERCENTILES

9.1. What is the 95th percentile of the standard normal distribution? 95th pctile = value exceeded by only 5% of the distribution: z = 1.645 14. What is the 75th percentile of the standard normal of the standard normal distribution?z = 0.6745

## PART 10. RANDOM SAMPLING FOR MEASURED CHARACTERISTICS; THE NORMAL DISTRIBUTION

In quality control, samples are selected from a production line and various quality characteristics are measured in order to check that the process is "in control." Suppose that a bottling process is intended to fill bottles with, on average, 21 fluid ounces of beverage. Variation around this mean follows the normal distribution with a standard deviation of 0.5 fluid ounces.

10.1.If a technician samples 25 bottles (when the process is "in control") and measures the amount of beverage in each, what is the probability that the sample average (for the 25 bottles) will exceed 21.2 fluid ounces?s.e. mean = 0.5/sqrt(25) = 0.1 oz.; z = (21.2-21)/0.1 = +2.00, P(x>2.00) = .0228

## PART 11. NORMAL APPROXIMATION TO THE BINOMIAL

Statistics released by the National Highway Traffic Safety Administration and the National Safety Council show that on an average weekend night, 1 out of every 10 drivers on the road is drunk. If 400 drivers are randomly checked next Saturday night, what is the probability that the number of drunk drivers will be11.1. More than 49?

mean = np = 400(.1) = 40; variance = npq = 400(.1)(.9) = 36; SD = sqrt(36) = 6. z = (49.5-40)/6 = 9.5/6 = 1.583, P(z > 1.583) = .057

11.2. Exactly 40? z = (40.5-40)/6 = .5/6 = 1/12 = .0833 P(0 < z < .0833) = .033 P(-.0833 < z < .0833) = 2(.033) = .066

11.3.Exactly 49 ? z = (49.5-40)/6 = 9.5/6 = 1.583, P(z > 1.583) = .057

z = (48.5-40)/6 = 8.5/6 = 1.4167, P(z > 1.4167) = .078

P(exactly 49) is approx. .078-.057 = .021 .

## PART 12. INTERVAL ESTIMATION OF A BINOMIAL PROBABILITY

In a random sample of n = 625 persons, three hundred (300) favor Paul Parrot for President.

12.1.What is the estimate of the standard deviation of the sampling distribution of the sample proportion ("p-hat")?Sample proportion = 300/625 = .48 Var = pq/n, est'd by .48(.52)/625 = .000399; est of SD is sqrt of this, or .01998, very close to .02.

PART 13. SAMPLE SIZE DETERMINATION FOR A CONFIDENCE INTERVALSuppose that GMAT scores have a known standard deviation of 100. A sample of scores of UIC MBA students is to be taken to estimate the mean in that population. Determine how large a sample is required to form a 95% confidence interval with a margin of error (half-width) of 25 points.

13.1.The required sample size is about ?25 =

B= 1.96 SD(y) = (1.96)(100)/sqrt(n); sqrt(n) = (1.96)(100)/25 = 7.84; n = 61.47 (use 61 or 62)

## PART 14. OBSERVED SIGNIFICANCE LEVEL:

p-VALUE14.1.In a test of the null hypothesis that the mean is 18 versus alternatives that the mean is greater than 18, a sample of n = 100 observations gave a mean of 19.5 and standard deviation of 6.00. What is the p-value (i.e., the achieved level of significance)?

z= (19.5 - 18)/0.6 = +2.5,p-value = .006

## PART 15. CONFIDENCE INTERVAL FOR A DIFFERENCE BETWEEN TWO MEANS

Lifetimes of two types of batteries were compared. Summary statistics are given in the table. Note that the means are given in hours and the standard deviations are given in minutes.

TABLE. Statistics from samples of battery lifetimesn mean std.dev.--------------------------------------------------Battery A 64 7 hr 30 minBattery B 100 6 hr 15 min----------------------------------------------------In what follows, do not pool the variances since, judging from the ratio of 2 between the two sample standard deviations, it appears that the population variances differ. Also, use the normal distribution (rather than t) due to the large sample sizes.15.1. What is the estimate of the standard deviation of the difference between sample means?

Var(diff of uncorrelated variables) = sum of variances = Var(mean1) + Var(mean2) = 30

^{2}/64 + 15^{2}/100 = 16.3125 SD(mean1-mean2) = sqrt(16.3125) = 4.039 min.15.2. What is the 95% confidence interval for the difference between means? 95% C.I. = (diff-B,diff+B), where diff = mean1-mean2 = 7 hr. - 6 hr.= 1 hr. = 60 min., and B = 1.96(4.039 min.) = 7.9 min., so 95% C.I. = (52.1, 67.9) min.

## PART 16. CHI-SQUARE GOODNESS OF FIT TEST

16.1.The numbers of accidents per day were recorded in a plant for 100 days. The data are tabulated below. Compute the value of chi-square for testing the hypothesis that the following data came from the distribution (.4, .3, .2, .1) over the values, 0, 1, 2, 3 or more.

Number of accidents | 0 | 1 | 2 | more than 2 |

Number of days with this many accidents | 45 | 25 | 20 | 10 |

Hypothesized prob of this no. of accidents | .4 | .3 | .2 | .1 |

The value of the chi-square test statistic is ?

O 45 25 20 10

E 40 30 20 10

O-E 5 -5 0 0

(O-E)^{2}25 25 0 0

(O-E)Value of chi-sq test statistic = .625 + .833 = 1.46^{2}/E 0.625 0.833 0 0

[The number of degrees of freedom is 3 and the *p*-value is .6915, which is much
greater than conventional cut-off values such as .10 or .05 and
as such indicates a satisfactory fit.]

OWN A CAR?

| Yes No |

_________

___|_______ |___

Full time | 16 1 | 17

EMPLOYMENT

Part time | 68 15 | 83

None | 50 19 | 69

____|________|_____

| 134 35 | 169

17.1. Of all 169 people, what percentage
are employed full time and own a car? 16/169 = 9.47% 25. Of
those who are employed full time, what percentage own a car?

16/17 = 94%

17.2. Of those who own a car, what percentage are employed full time?

16/134 = 11.9%

17.3. What is the number of degrees of freedom associated
with the chi-square test statistic for this table? (r-1)(c-1) = 2
28. Compute the value of the chi-square test statistic. (Answer:
4.586)

MTB > chisquare c1 c2

Expected counts are printed below observed counts

C1 C2 Total

-----------------------

1 16 1 17

13.48 3.52

2 68 15 83

65.81 17.19

3 50 19 69

54.71 14.29

-----------------------

Total 134 35 169

ChiSq = 0.471 + 1.805 + 0.073 + 0.279 + 0.405 + 1.552 = 4.586

df = 2

1 cells with expected counts less than 5.0

MTB > cdf 4.586; SUBC> chisq 2. 4.5860 0.8990

MTB > let k1 = 1-.8990

MTB > print k1 K1 0.101000

**18.1. ** If the speed (S) is 65, then the predicted
gasoline mileage (G) is ?

pred.val. of ln G is 7.216 - 1.073 ln S = 7.216 - 1.073(4.174) = 2.737; pred.val. of G is exp(2.737) = 15.44 mpg