INSTRUCTIONS - University of Manitobahome.cc.umanitoba.ca/~thavane/ASSIGN/f2801200mt.pdf ·...

PART A 005.200 TERM TEST February 28, 2001 of 16 Sections L04,L05 and L08 5:30pm - 7:00pm

INSTRUCTIONS

1. You have been provided with:

a) the examination paper,

b) a multiple choice answer sheet,

c) selected tables/Cheng & Fu Tables and formula sheet,

d) couple of blank sheets for rough work.

2. The total number of marks possible is 30.

3.

It is suggested that you first complete the questions on the examination paper by choosing the

BEST answer out of five in each case; then transfer your answers to the multiple choice answer

sheet by blackening the appropriate space with an "HB" or "F" pencil. Only one space should be

blackened; otherwise, the question will be marked wrong. The questions are of equal value. There

is no correction made for guessing; therefore, all questions should be attempted.

4. At the end of the examination period, turn in your multiple choice answer sheet . Be sure to write

your NAME, STUDENT NUMBER, and INSTRUCTOR's NAME on your MULTIPLE

CHOICE ANSWER SHEET .

5. Calculators are permitted.


1. Which of the following statements about confidence intervals is INCORRECT?

(A) If we keep the sample size fixed, the confidence interval gets wider as we increase the confidence

coefficient.

(B) A confidence interval for a mean always contains the sample mean.

(C) If we keep the confidence coefficient fixed, the confidence interval gets narrower as we increase the sample

size.

(D) If the population standard deviation increases, the confidence interval decreases in width.

(E) If the confidence intervals for two means do not overlap very much, there is evidence that the two

population means are different.

2. The diameter of ball bearings are known to be normally distributed with unknown mean and known

variance σ2. A random sample of size 25 gave a mean 2.5 cm. The 95% confidence interval had length 4

cm. Then

(A) The population variance is 5.10.

(B) The population variance is 26.03.

(C) The sample variance is 26.03.

(D) The sample variance is 5.10.

(E) The sample variance is 4.69.

3. A turkey producer knows from previous experience that profits are maximized by selling turkeys when

their average weight is 12 kilograms. Before determining whether to put all their full grown turkeys on the

market this month, the producer wishes to estimate their mean weight. Prior knowledge indicates that turkey

weights have a standard deviation of around 3 kilograms. The number of turkeys that must be sampled in

order to estimate their true mean weight to within 1 kilogram with 95% confidence is:

(A) 150

(B) 5

(C) 65

(D) 10

(E) 35


4. A random sample of 4 Herefords, each with a frame size of three (on a one-to-seven scale), gave a sample

mean weight of 452 kg and a sample standard deviation of 12 kg. A 95% confidence interval for the

average weight of all Herefords of this frame size is:

(A) (435.3, 468.7)

(B) (432.9, 471.1)

(C) (440.2, 463.8)

(D) (428.5, 475.5)

(E) (436.6, 467.4)

5. You have this year's scores on a Canadian Math Contest for a SRS of 225 grade 12 students. You plan

to test:H0: µ = 140(the previous year's mean score)

vs.Ha: µ ≠ 140

at the 5% level of significance(i.e. has the mean changed?). The standard deviation of the scores of all

grade 12 students from the previous year is known to be 30; we believe it has not changed. The

probability of Type II error under the alternative µ =135 is:

(A) .2946

(B) .7054

(C) .9922

(D) less than .0001

(E) .025

6. Refer to Question #5. The power of the test against the alternative Ha: µ = 145 is

(A) .2946

(B) .7054


(C) .9922

(D) less than .0001

(E) .025

7. DDT is an insecticide that accumulates up the food chain. Predator birds can be contaminated with quite

high levels of the chemical by eating many lightly contaminated prey. One effect of DDT upon birds is to

inhibit the production of the enzyme carbonic anhydrase which controls calcium metabolism. It is believed

that this causes egg shells to be thinner and weaker than normal and makes the eggs more prone to

breakage. (This is one of the reasons why the condor in California is near extinction.) An experiment was

conducted where 16 sparrow hawks were fed a mixture of 3 ppm dieldrin and 15 ppm DDT (a combination

often found in contaminated prey). The first egg laid by each bird was measured and the mean shell

thickness was found to be 0.19 mm with a standard deviation of 0.01 mm. A normal egg shell has a mean

thickness of 0.2 mm.

The null and alternate hypotheses, to test the belief, are:

(A) H0: µ = 0.2 vs Ha : µ < 0.2

(B) H0: µ < 0.2 vs Ha: µ = 0.2

(C) H0: x = 0.2 vs Ha: x < 0.2

(D) H0: x = 0.19 vs Ha: x = 0.2

(E) H0: µ = 0.2 vs Ha: µ ≠ 0.2


8. (Refer to question 7) The value of the test statistic is:

(A) -1.00

(B) -4.00

(C) 0.01

(D) 1.96

(E) 1.75

9. (Refer to question 7) The null hypothesis will be rejected (α=0.05) if the test statistic is less than: (note

that if the rejection region is two sided, only one side has been shown)

(A) +2.1314

(B) +1.7531

(C) +1.6450

(D) -1.6450

(E) -1.7531

10. A Canadian railway company claims that its trains block crossings for no more that 5 minutes per train

on the average. The actual times (minutes) that 10 randomly selected trains block crossings were:

10.4 9.7 6.5 9.5 8.8 11.2 7.2 10.5 8.2 9.3

giving x = 9.130 and s2 = 2.209.

In testing this claim, at the significance level of 0.05 and assuming that the crossing times are

normally distributed, the value of the test statistic and the critical value are, respectively:

(A) 5.91 and -2.2622

(B) 8.79 and -1.8331

(C) 5.91 and -1.8331

(D) 8.79 and -2.2622

(E) 2.78 and -1.96


11. A drug company claims it has developed a new blood pressure drug that can reduce, after one week ofmedication, a patients blood pressure by more than 10 units, the current average for a competing drug. To test the drug companys claim, it is decided to test H0: µ = 10 vs. Ha: µ > 10 by recording the bloodpressure medication for 36 patients after treatment with the drug. The null hypothesis is to be rejected ifx > 13. The standard deviation of reductions in blood pressure is known to be 9. Assuming normality forblood pressure readings, the probability of a type I error is:

(A) 0.05

(B) 0.0228

(C) 0.1056

(D) 0.9772

(E) 0.8749


12. Refer to Question 11. What is the power of the test under the alternative µ = 17.5?

(A) 0.4013

(B) 0.9987

(C) 0.05

(D) 0.4801

(E) 0.0013

13. A drug antibiotic manufacturer randomly sampled 12 different locations in a fermentation vat to determine

the average potency for a batch of antibiotic being prepared. The data was entered into JMPin with the

following output:

MomentsMeanStd DevStd Error MeanUpper 95% MeanLower 95% MeanNSum Weights

8.96667 0.10731 0.03098 9.03485 8.89849

12.00000 12.00000

Using this information, it is possible to test whether µ is 9.5. At the 5% level of significance, one would:

(A) Fail to reject H0: µ = 9.5 is inside the appropriate confidence interval.

(B) Reject H0: µ = 9.5 is outside the appropriate confidence interval.

(C) Reject H0: µ = 9.5 is inside the appropriate confidence interval.

(D) Fail to reject H0: µ = 9.5 is outside the appropriate confidence interval.

(E) Not make any conclusion; information provided is not sufficient.

14. A researcher wished to test the effect of the addition of extra calcium to yogurt on the "tastiness" of yogurt.

A collection of 20 adult volunteers was randomly divided into two groups of 10 subjects each. Group 1


tasted yogurt containing the extra calcium. Group 2 tasted yogurt from the same batch as group 1 but

without the added calcium. Both groups rated the flavor on a scale of 1 to 10, 1 being "very unpleasant" and

10 being "very pleasant". The mean rating for a sample from group 1 was 6.5 with a standard deviation0.5. The mean rating for a sample from group 2 was 7.0 with a standard deviation 2.0 Let µ1 and µ2represent the mean ratings we would observe for the entire population represented by the volunteers if all

members of this population tasted, respectively, the yogurt with and without the added calcium. Using theabove data, what is a 90% confidence interval for µ1 - µ2 ?


(A) −.5±1.383(0.5)2

10+ 2( )2

10

(B) −.5±1.833(0.5)2

10+ 22

10

(C) −.5±1.734(0.5)2

10+ 22

10

(D) −.5±1.8330.510

+ 210

(E) −.5±1.734 2.1251

10+ 1

10

15. Refer to question 14. If the researcher assumes that the population variances are equal then the 90%

confidence for µ1 - µ2 is:

(A) −.5±1.383(0.5)2

10+ 2( )2

10

(B) −.5±1.833(0.5)2

10+ 22

10

(C) −.5±1.734(0.5)2

10+ 22

10

(D) −.5±1.833 2.1251

10+ 1

10

(E) −.5±1.734 2.1251

10+ 1

10

16. A researcher wished to compare the average amount of time spent in extracurricular activities by high

school students in a suburban school district with that in a school district of a large city. The researcher

obtained a SRS of 10 high school students in a large suburban school district and found the mean time

spent in extracurricular activities per week to be 6 hours with a standard deviation of 3 hours. The

researcher also obtained an independent SRS of 12 high school students in a large city school district and

found the mean time spent in extracurricular activities per week to be 4 hours with a standard deviation of 1


hour. Let µ1 and µ2 represent the mean amount of time spent in extracurricular activities per week by the

populations of all high school students in the suburban and city school districts, respectively. Using the

above data, suppose the researcher wished to test whether the time spent on extracurricular activities is the

same in the two types of districts. What is the P-value for the test?

(A) larger than .10

(B) between .05 and .10

(C) between 025 and .05

(D) between .01 and .025

(E) below .01

17. You wish to compare a new method of teaching reading to "slow learners" to the current standard method.

You decide to base this comparison on the results of a reading test given at the end of a learning period of 6

months. Of a random sample of 20 slow learners, 8 are taught by the new method and 12 are taught by the

standard method. All 20 children are taught by qualified instructors under similar conditions for a 6 month

period. The results of the reading test at the end of the period are summarized below:

New Method

Score

Standard Method

Score

x1= 76.9 x2 = 72.7

s1 = 4.85 s2 = 6.35

Assuming the underlying populations are normally distributed with equal variance, the P-value to test the

hypothesis that the mean scores are the same for the two methods is:

(A) below 0.01

(B) between 0.01 and .0.025

(C) between 0.025 and 0.05

(D) between 0.05 and 0.1

(E) between 0.1 and 0.2


18. An experiment is conducted to compare the starting salaries of male and female college graduates who find

jobs. Pairs are formed by choosing a male and female with the same major and grade point average.

Suppose a random sample of 10 pairs is formed in this manner and the starting annual salary of each

person is recorded. To test whether there is evidence that the mean starting salaries for males and females

are different, the appropriate test is:

(A) pooled two sample t test with 18 d.f.

(B) conservative two sample t test with 9 d.f.

(C) conservative two sample t test with 18 d.f.

(D) paired t test with 19 d.f.

(E) paired t test with 9 d.f.


19. Wild horse populations on federal lands have been protected since 1971. Since that time, the populations

have grown large and need to be managed and kept to a supportable size. Management of the mustang

population has been a controversial issue; one common method is periodic removal of the horses.

Researchers were curious if a new method would work better. In 1985, twelve bands of horses were

rounded up, and the male horses in each band were treated. The number of foals in each band for three

years was recorded. Year 1 was prior to treatment, year 2 was the year the treatment was applied, and year 3

was one year after treatment. The number of bands, mean number of foals per band, along with the standard

deviations, are given below.

Year # of bands Means(xi ) St. Dev. (si )

1 14 15.25 7.10

2 12 15.64 14.14

3 12 15.25 14.03

The researchers did an ANOVA F test of the data and obtained the following results.

Analysis of Variance for Foals

� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �

Year 0.68

Error 5321.71

Total 37

The approximate P-value of the test of equality of means is:

(A) 0.0045

(B) below 0.01

(C) between 0.01 and .0.025

(D) between 0.025 and 0.05

(E) larger than .05


20. Refer to Question 19. The estimate of the population common variance is:

A) 5321.71

B) 0.68

C) 1.36

D) 152.05

E) 0.0045


21. Refer to Question 19. Assuming the population means are µ1, µ2, and µ3, and the correspondingvariances are σ1

2,σ22,and σ3

2 . The appropriate null and alternative hypotheses for theAnalysis Variance are:

A) H0: x1 = x2 = x3 vs. Ha: x1 ≠ x2 ≠ x3

B) H0: µ1 = µ2 = µ3 vs. Ha: µ1 ≠ µ2 ≠ µ3C) H0: µ1 = µ2 = µ3 vs. Ha: not all the µi's are equal

D) H0: µ1 = xi , i =1,2,3 vs. Ha: µ1 ≠ xi for at least one i = 1, 2, 3

E) H0: σ1 = σ2 = σ3 vs. Ha: σ1 ≠ σ2 ≠ σ3

22. The length of time to complete recovery was recorded for patients subjected to two different surgical

procedures. The patients were randomly assigned to a procedure. The results are:

Procedure 1 Procedure 2

Sample Size 21 23

Sample Mean 7.3 8.9

Sample Variance 0.54 2.22

Do the data present sufficient evidence to indicate a difference in mean recovery time for the two surgical

procedures? The test statistic and the appropriate degrees of freedom are:

(A)7.3 − 8.9

0.2916

21+

4.9284

23

with 20 d.f.

(B)7.3 − 8.9

0.54

21+

2.22

23

with 42 d.f.


(C)7.3 − 8.9

(21)(0.54) + (23)(2.22)

21+ 23 − 2

1

21+

1

23

with 42 d.f.

(D)7.3 − 8.9

(20)(0.54) + (22)(2.22)

21+ 23 − 2

1

11+

1

14

with 42 d.f.

(E)7.3 − 8.9

0.54

21+

2.22

23

with 20 d.f.

23. A physician wants to compare the blood pressures of six patients before and after treatment with a drug.

The blood pressures are as follows:

Patient Before Drug After Drug Difference

1 161 140 21

2 158 148 10

3 170 150 20

4 155 140 15

The physician wants to test if there is a significant change of the blood pressure before and after taking the

drug at 0.05 level of significance. The absolute value of the test statistic and the absolute critical value of the

test are respectively:

(A) 3.257 and 1.96

(B) 2.761 and 3.1824

(C) 5.066 and 3.1824

(D) 6.514 and 3.1824

(E) 6.514 and 1.96


24. The infamous researcher, claims to have found a drug that causes people to grow taller. The coach of the

Basketball team at Brandon University has expressed interest but demands evidence. Ten people are

randomly selected from students at Brandon, their heights measured, the drug administered, and 2 hours

later their heights remeasured. The results were as follows:

Pre-Drug 68 69 74 78 70 66 71 70 71 65

Post-Drug 70 69 75 78 73 69 72 73 72 66

Person 1 2 3 4 5 6 7 8 9 10

Using the proper test statistic, an appropriate decision rule for the hypotheses H0:Drug has no effect versus

Ha: Drug increases height at (α = .05) will be

(A) Reject H0 if the test statistic is > 1.96

(B) Reject H0 if the test statistic is > 1.645

(C) Reject H0 if the test statistic is > 1.83

(D) Reject H0 if the test statistic is > 1.73

(E) Reject H0 if the test statistic is > 2.10


Table entry for z is the area under the standard normal curve to the left of z.

z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

–3.4 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0002–3.3 0.0005 0.0005 0.0005 0.0004 0.0004 0.0004 0.0004 0.0004 0.0004 0.0003–3.2 0.0007 0.0007 0.0006 0.0006 0.0006 0.0006 0.0006 0.0005 0.0005 0.0005–3.1 0.0010 0.0009 0.0009 0.0009 0.0008 0.0008 0.0008 0.0008 0.0007 0.0007–3.0 0.0013 0.0013 0.0013 0.0012 0.0012 0.0011 0.0011 0.0011 0.0010 0.0010–2.9 0.0019 0.0018 0.0018 0.0017 0.0016 0.0016 0.0015 0.0015 0.0014 0.0014–2.8 0.0026 0.0025 0.0024 0.0023 0.0023 0.0022 0.0021 0.0021 0.0020 0.0019–2.7 0.0035 0.0034 0.0033 0.0032 0.0031 0.0030 0.0029 0.0028 0.0027 0.0026–2.6 0.0047 0.0045 0.0044 0.0043 0.0041 0.0040 0.0039 0.0038 0.0037 0.0036–2.5 0.0062 0.0060 0.0059 0.0057 0.0055 0.0054 0.0052 0.0051 0.0049 0.0048–2.4 0.0082 0.0080 0.0078 0.0075 0.0073 0.0071 0.0069 0.0068 0.0066 0.0064–2.3 0.0107 0.0104 0.0102 0.0099 0.0096 0.0094 0.0091 0.0089 0.0087 0.0084–2.2 0.0139 0.0136 0.0132 0.0129 0.0125 0.0122 0.0119 0.0116 0.0113 0.0110–2.1 0.0179 0.0174 0.0170 0.0166 0.0162 0.0158 0.0154 0.0150 0.0146 0.0143–2.0 0.0228 0.0222 0.0217 0.0212 0.0207 0.0202 0.0197 0.0192 0.0188 0.0183–1.9 0.0287 0.0281 0.0274 0.0268 0.0262 0.0256 0.0250 0.0244 0.0239 0.0233–1.8 0.0359 0.0351 0.0344 0.0336 0.0329 0.0322 0.0314 0.0307 0.0301 0.0294–1.7 0.0446 0.0436 0.0427 0.0418 0.0409 0.0401 0.0392 0.0384 0.0375 0.0367–1.6 0.0548 0.0537 0.0526 0.0516 0.0505 0.0495 0.0485 0.0475 0.0465 0.0455–1.5 0.0668 0.0655 0.0643 0.0630 0.0618 0.0606 0.0594 0.0582 0.0571 0.0559–1.4 0.0808 0.0793 0.0778 0.0764 0.0749 0.0735 0.0721 0.0708 0.0694 0.0681–1.3 0.0968 0.0951 0.0934 0.0918 0.0901 0.0885 0.0869 0.0853 0.0838 0.0823–1.2 0.1151 0.1131 0.1112 0.1093 0.1075 0.1056 0.1038 0.1020 0.1003 0.0985–1.1 0.1357 0.1335 0.1314 0.1292 0.1271 0.1251 0.1230 0.1210 0.1190 0.1170–1.0 0.1587 0.1562 0.1539 0.1515 0.1492 0.1469 0.1446 0.1423 0.1401 0.1379–0.9 0.1841 0.1814 0.1788 0.1762 0.1736 0.1711 0.1685 0.1660 0.1635 0.1611–0.8 0.2119 0.2090 0.2061 0.2033 0.2005 0.1977 0.1949 0.1922 0.1894 0.1867–0.7 0.2420 0.2389 0.2358 0.2327 0.2296 0.2266 0.2236 0.2206 0.2177 0.2148–0.6 0.2743 0.2709 0.2676 0.2643 0.2611 0.2578 0.2546 0.2514 0.2483 0.2451–0.5 0.3085 0.3050 0.3015 0.2981 0.2946 0.2912 0.2877 0.2843 0.2810 0.2776–0.4 0.3446 0.3409 0.3372 0.3336 0.3300 0.3264 0.3228 0.3192 0.3156 0.3121–0.3 0.3821 0.3783 0.3745 0.3707 0.3669 0.3632 0.3594 0.3557 0.3520 0.3483–0.2 0.4207 0.4168 0.4129 0.4090 0.4052 0.4013 0.3974 0.3936 0.3897 0.3859–0.1 0.4602 0.4562 0.4522 0.4483 0.4443 0.4404 0.4364 0.4325 0.4286 0.4247–0.0 0.5000 0.4960 0.4920 0.4880 0.4840 0.4801 0.4761 0.4721 0.4681 0.4641


Table entry for z is the area under the standard normal curve to the left of z

z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

0.0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.53590.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.57530.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.61410.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.65170.4 0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.68790.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.72240.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.75490.7 0.7580 0.7611 0.7642 0.7673 0.7704 0.7734 0.7764 0.7794 0.7823 0.78520.8 0.7881 0.7910 0.7939 0.7967 0.7995 0.8023 0.8051 0.8078 0.8106 0.81330.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.8289 0.8315 0.8340 0.8365 0.83891.0 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.86211.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770 0.8790 0.8810 0.88301.2 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 0.8980 0.8997 0.90151.3 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.91771.4 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.93191.5 0.9332 0.9345 0.9357 0.9370 0.9382 0.9394 0.9406 0.9418 0.9429 0.94411.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 0.95451.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 0.9608 0.9616 0.9625 0.96331.8 0.9641 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 0.9699 0.97061.9 0.9713 0.9719 0.9726 0.9732 0.9738 0.9744 0.9750 0.9756 0.9761 0.97672.0 0.9772 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812 0.98172.1 0.9821 0.9826 0.9830 0.9834 0.9838 0.9842 0.9846 0.9850 0.9854 0.98572.2 0.9861 0.9864 0.9868 0.9871 0.9875 0.9878 0.9881 0.9884 0.9887 0.98902.3 0.9893 0.9896 0.9898 0.9901 0.9904 0.9906 0.9909 0.9911 0.9913 0.99162.4 0.9918 0.9920 0.9922 0.9925 0.9927 0.9929 0.9931 0.9932 0.9934 0.99362.5 0.9938 0.9940 0.9941 0.9943 0.9945 0.9946 0.9948 0.9949 0.9951 0.99522.6 0.9953 0.9955 0.9956 0.9957 0.9959 0.9960 0.9961 0.9962 0.9963 0.99642.7 0.9965 0.9966 0.9967 0.9968 0.9969 0.9970 0.9971 0.9972 0.9973 0.99742.8 0.9974 0.9975 0.9976 0.9977 0.9977 0.9978 0.9979 0.9979 0.9980 0.99812.9 0.9981 0.9982 0.9982 0.9983 0.9984 0.9984 0.9985 0.9985 0.9986 0.99863.0 0.9987 0.9987 0.9987 0.9988 0.9988 0.9989 0.9989 0.9989 0.9990 0.99903.1 0.9990 0.9991 0.9991 0.9991 0.9992 0.9992 0.9992 0.9992 0.9993 0.99933.2 0.9993 0.9993 0.9994 0.9994 0.9994 0.9994 0.9994 0.9995 0.9995 0.99953.3 0.9995 0.9995 0.9995 0.9996 0.9996 0.9996 0.9996 0.9996 0.9996 0.99973.4 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9998


Table t Distribution Critical Values �

� �

pdf 0.100 0.050 0.025 0.015 0.010 0.005 0.00251 3.0777 6.3137 12.7062 21.2051 31.8210 63.6559 127.32112 1.8856 2.9200 4.3027 5.6428 6.9645 9.9250 14.08923 1.6377 2.3534 3.1824 3.8961 4.5407 5.8408 7.45324 1.5332 2.1318 2.7765 3.2976 3.7469 4.6041 5.59755 1.4759 2.0150 2.5706 3.0029 3.3649 4.0321 4.77336 1.4398 1.9432 2.4469 2.8289 3.1427 3.7074 4.31687 1.4149 1.8946 2.3646 2.7146 2.9979 3.4995 4.02948 1.3968 1.8595 2.3060 2.6338 2.8965 3.3554 3.83259 1.3830 1.8331 2.2622 2.5738 2.8214 3.2498 3.689610 1.3722 1.8125 2.2281 2.5275 2.7638 3.1693 3.581411 1.3634 1.7959 2.2010 2.4907 2.7181 3.1058 3.496612 1.3562 1.7823 2.1788 2.4607 2.6810 3.0545 3.428413 1.3502 1.7709 2.1604 2.4358 2.6503 3.0123 3.372514 1.3450 1.7613 2.1448 2.4149 2.6245 2.9768 3.325715 1.3406 1.7531 2.1315 2.3970 2.6025 2.9467 3.286016 1.3368 1.7459 2.1199 2.3815 2.5835 2.9208 3.252017 1.3334 1.7396 2.1098 2.3681 2.5669 2.8982 3.222418 1.3304 1.7341 2.1009 2.3562 2.5524 2.8784 3.196619 1.3277 1.7291 2.0930 2.3457 2.5395 2.8609 3.173720 1.3253 1.7247 2.0860 2.3362 2.5280 2.8453 3.153421 1.3232 1.7207 2.0796 2.3278 2.5176 2.8314 3.135222 1.3212 1.7171 2.0739 2.3202 2.5083 2.8188 3.118823 1.3195 1.7139 2.0687 2.3132 2.4999 2.8073 3.104024 1.3178 1.7109 2.0639 2.3069 2.4922 2.7970 3.090525 1.3163 1.7081 2.0595 2.3011 2.4851 2.7874 3.078226 1.3150 1.7056 2.0555 2.2958 2.4786 2.7787 3.066927 1.3137 1.7033 2.0518 2.2909 2.4727 2.7707 3.056528 1.3125 1.7011 2.0484 2.2864 2.4671 2.7633 3.047029 1.3114 1.6991 2.0452 2.2822 2.4620 2.7564 3.038030 1.3104 1.6973 2.0423 2.2783 2.4573 2.7500 3.029831 1.3095 1.6955 2.0395 2.2746 2.4528 2.7440 3.022132 1.3086 1.6939 2.0369 2.2712 2.4487 2.7385 3.014933 1.3077 1.6924 2.0345 2.2680 2.4448 2.7333 3.008234 1.3070 1.6909 2.0322 2.2650 2.4411 2.7284 3.002035 1.3062 1.6896 2.0301 2.2622 2.4377 2.7238 2.996136 1.3055 1.6883 2.0281 2.2595 2.4345 2.7195 2.990537 1.3049 1.6871 2.0262 2.2570 2.4314 2.7154 2.985338 1.3042 1.6860 2.0244 2.2546 2.4286 2.7116 2.980339 1.3036 1.6849 2.0227 2.2524 2.4258 2.7079 2.975640 1.3031 1.6839 2.0211 2.2503 2.4233 2.7045 2.971241 1.3025 1.6829 2.0195 2.2483 2.4208 2.7012 2.967042 1.3020 1.6820 2.0181 2.2463 2.4185 2.6981 2.963043 1.3016 1.6811 2.0167 2.2445 2.4163 2.6951 2.959244 1.3011 1.6802 2.0154 2.2428 2.4141 2.6923 2.955545 1.3007 1.6794 2.0141 2.2411 2.4121 2.6896 2.952146 1.3002 1.6787 2.0129 2.2395 2.4102 2.6870 2.9488


47 1.2998 1.6779 2.0117 2.2380 2.4083 2.6846 2.945648 1.2994 1.6772 2.0106 2.2365 2.4066 2.6822 2.942649 1.2991 1.6766 2.0096 2.2351 2.4049 2.6800 2.939750 1.2987 1.6759 2.0086 2.2338 2.4033 2.6778 2.937060 1.6706 2.0003 2.2990 2.5044 2.6603 2.9146 3.156280 1.6641 1.9901 2.2844 2.4860 2.6387 2.8870 3.1220

100 1.6602 1.9840 2.2757 2.4751 2.6259 2.8707 3.10181000 1.6464 1.9623 2.2448 2.4366 2.5807 2.8133 3.0310z* 1.2816 1.6449 1.9600 2.1701 2.3263 2.5758 2.8070

80% 90% 95% 97% 98% 99% 99.5%

Confidence level C


Selected Formulae for 005.200

1. (yi − y)2

i =1

n

∑ = yi2

i =1

n

∑ − 1

nyi

i =1

n

∑

2

= yi2

i =1

n

∑ − ny2

2. sy = 1

n −1(yi − y)2

i =1

n

∑

3. r = 1

n -1

x i − x

sx

yi − y

sy

∑ =x iyi − 1

nx i∑( ) yi∑( )∑

(n −1) sx sy

4. b = rsy

sx

=n x iyi − x i∑( ) yi∑( )∑

n x i2 − x i∑( )2

∑ a = y − b x

5. If X has a binomial distribution with parameters n and p, then the mean of X is np and the variance of X isnp(1-p).

6. The sampling distribution of p^ has a mean of p and a standard deviation of p(1− p)

n.

7. The sampling distribution of x_ has a mean of µ and a standard deviation of

σn

.

8. Z = x − µ0σ / n

x ± z* σn

n = z*σm

2

9. Z =√p − p0

p0(1− p0)

n

√p ± z∗ √p(1− √p)

n n = z∗

m

2

p∗(1- p∗)

10. t = x − µ0s/ n

x_ ± t*

s

n


11. t = x1 − x2 − (µ1 − µ2)

sp1

n1

+ 1

n2

x1 − x2( ) ± t * sp1

n1

+ 1

n2

sp2 = (n1 −1)s1

2 + (n2 −1)s22

n1 + n2 − 2 with df = n1 + n2 − 2 and σ1

2 = σ22

12. t = x1 − x2 − (µ1 − µ2)

s12

n1

+ s22

n2

x1 − x2( ) ± t *s1

2

n1

+ s22

n2

with df = smaller of n1 −1 and n2 −1

13. Z =√p1 − √p2

sp

sp = √p(1− √p)1

n1

+ 1

n2

√p = x1 + x2

n1 + n2

(√p1 − √p2) ± z* sD sD =√p1(1− √p1)

n1

+√p2(1− √p2)

n2

14. Binomial Probability Distribution:

P(X = k) = Error! ) pError! k = 0, 1, ... , n

15. χ2 = (Observed count − Expected count)2

Expected countover all cells

∑

INSTRUCTIONS - University of Manitobahome.cc.umanitoba.ca/~thavane/ASSIGN/f2801200mt.pdf ·...

Documents

Transcript of INSTRUCTIONS - University of Manitobahome.cc.umanitoba.ca/~thavane/ASSIGN/f2801200mt.pdf ·...