# Stt 825 Fall, 01 homework #2

 Дата канвертавання 26.04.2016 Памер 22.8 Kb.
STT 825 Fall, 01 HOMEWORK #2 SOLUTIONS Due on Monday, September 24
Note late homework policy on the course syllabus. Computer use is encouraged.
1. (8 points) Text problem #8 (Chapter 2).

To convert data back to “raw” form, enter 0 (28 times), 1 (4 times), 2 (3 times),…, 10 (once), to get a sample of 50 measurements.

(a) Minitab output Histogram

Histogram of number N = 50
Midpoint Count

0 28 ****************************

1 4 ****

2 3 ***

3 4 ****

4 4 ****

5 2 **

6 1 *

7 0

8 2 **

9 1 *

10 1 *

(b) Descriptive Statistics: number

Variable N Mean Median TrMean StDev SE Mean

number 50 1.780 0.000 1.409 2.682 0.379

## Sample mean= 1.78 SE(mean) = (.9685)(.379) = .367

c. No the sample is highly skewed with a big spike at zero and n=50 isn’t that large.

d. = 28/50 = .56, 95% CI is .56  (1.96)(.9685)(.0709) which gives .56  .135.

e. Why is the normal o.k. to use for part (d)?

n= 28 and n(1-) = 32, both greater than 5, so n=50 is o.k. (large enough).
f. Give the sampling units and define yi.

SU = faculty member yi = number of refereed publications in (a), yi = 1 if faculty has publication, 0 if not in part (d). (Either (a) or (d) answer is accepted).
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2. (5 points) Text problem #10 (Chapter 2).

Note that N is not given, and we assume that n/N is close to zero so that FPC is close to 1.
a. .175  (1.96) [ .175(.825)/999 ]1/2 which gives .175  .024

b. YES
c. Give the sampling units and define yi.

SU = entries, yi =1 from south, 0 if not from south.

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3. (13 points) Text problem #12 (Chapter 2). Use 95% confidence.

Use e=.04 to get no = (1.96)2 (.5)(.5)/ (.04)2 = 600.25, then get n* for the different cities:
Buckeye n* = 534 (or 535)

Gilbert n* = 594 (or 595)

Gila Bend n*= 445 (or 446)

Phoenix n* = 600

Tempe n* = 598
The FPC makes a difference in Buckeye and Gila Bend, but doesn’t affect the others much.

b. Suppose we believe that the proportion in Buckeye is at least .65. Plan the sample size (for Buckeye only) to achieve margin of error 4 percentage points.

Use p = .65 to plan which gives no = (1.96)2 (.65)(.35)/(.04)2 = 546.2275, then plug into n* to get

n* = 491
4. (9 points) A storage facility contains 1500 shelves (equal sizes), and the problem is to estimate the mean value of items per shelf and the total value of the items in storage. For planning sample size, a pilot sample (SRS) of 10 shelves was taken, and the total value of the items on each of the 10 shelves was recorded as

\$31, 40, 53, 58, 60, 67, 74, 77, 82, 89.

Use this pilot sample to get values of the parameters for planning the sample size.

## Descriptive stats for the pilot sample are

Variable N Mean Median TrMean StDev SE Mean

value 10 63.10 63.50 63.88 18.43 5.83
a. Give the sampling units and define yi.

SU = shelf yi=total value of all items on the ith shelf.
b. What sample size is needed to estimate the mean value per shelf to within \$3.00 of the actual mean value with 95% confidence?

Absolute error e = \$3.00. Use S=18.43 or 18.5 is o.k.

no = (1.96)2 (18.5)2/32=146.09 which gives n*=133 or 134.
c. What sample size is needed to estimate the mean value per shelf to within 4% of the actual mean value with 95% confidence?

Relative error e = .04. Use CV = 18.43/63.10 = .292 or .30 is o.k. This gives

no = (1.96)2(.3)2/(.04)2 = 216.09 which then gives n*=189.
d. What sample size is needed to estimate the total value of items in storage to within \$4500 of the actual total value with 95% confidence?

D = \$4500, gives e= D/N = \$3.00, so we get the same answer as in (b), n*=133 or 134.
e. What sample size is needed to estimate the total value of items in storage to within 4% of the actual total value with 95% confidence?

D = .04, relative error, gives same answer as (d), n* = 189.

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5. (6 points) Consider the storage facility in #4, but now suppose we want to estimate the number of items in storage that are more than 6 months old. Data from the pilot sample was

12, 18, 07, 08, 15, 18, 21, 13, 08, 25 items on 10 sampled shelves were older than 6 months.

Descriptive stats:

Variable N Mean Median TrMean StDev SE Mean

number 10 14.50 14.00 14.13 6.02 1.90
a. Give the sampling units and define yi.

SU = shelf, yi = number of items older than 6 months.
b. Use the sample of size 10 to compute a 95% confidence interval for the number of items in storage which are older than 6 months.

CI for total t. Compute a CI for the mean number of items older than 6 months (per shelf) then convert to the total by multiplying by 1500.

## CI for mean: 14.50  (t.025, 9)(.99666)(1.90) which gives 14.50  4.30 (using t = 2.262)

Multiplying by 1500 gives the answer: 21,750 6450.
c. Comment on the validity of your confidence level in (b).

n=10 is very small for using the normal unless we have a normal population. And there’s no reason to assume that the population of counts is approximately normal. Thus, the 95% claim on the reported interval is suspect.

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6. (5 points) A company has freight bills issued during a 10 day period and ranging in value from \$1.10 to \$14.00. In (a) - (e) below, give the value of S for planning the sample size.

a = \$1.10, b=\$14.00 b-a=\$12.90
a. The bill amounts are approximately uniformly distributed.

S2 = (b-a)2/12= 13.87 which gives S = 3.72
b. The bill amounts are approximately normally distributed.

S= (b-a)/6 = 2.15
c. The bill amounts are approximately J-shaped distributed.

S2 - (b-a)2/18 = 9.245 which gives S = 3.04.
d. The bill amounts are approximately unimodal, symmetrically distributed.

S2 = (b-a)2/24 = 6.93 which gives S = 2.63.
e. No distributional shape for the bill amounts is assumed.

S2 = (b-a)2/4 = 41.6 which gives S = 6.45.

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7. (4 points) For problem #6, a manager used information from a similar company to find S=\$4.20 to use for planning the sample size. The manager recommended a sample of size 271 to achieve the accuracy of  .50 with 95% confidence. The "boss" noted that the sample size of 271 was larger than the total number of bills in the population. What did the manager do wrong? How did the manager get 271?

no = (1.96)2(4.20)2/(.5)2 = 271.06 , so the manager did not correct for a finite population.

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