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STA 2023 EXAM-3

Practice Problems

Mudunuru, Venkateswara Rao

STA 2023

Spring 2016

From Chapters 6, 7, & 8 SOLUTIONS

© Ven M

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© Venkateswara Rao Mudunuru Spring 2016 Exam-3 Review Solutions 1

For z values greater than 3.49, use 1.000 to approximate the area.

Standard Normal z-TABLE: P( Z < z )

z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09

0.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359

0.1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5753

0.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141

0.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517

0.4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879

0.5 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224

0.6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549

0.7 .7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852

0.8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133

0.9 .8159 .8186 .8212 .8238 .8264 .8289 .8315 .8340 .8365 .8389

1.0 .8413 .8438 .8461 .8485 .8508 .8531 .8554 .8577 .8599 .8621

1.1 .8643 .8665 .8686 .8708 .8729 .8749 .8770 .8790 .8810 .8830

1.2 .8849 .8869 .8888 .8907 .8925 .8944 .8962 .8980 .8997 .9015

1.3 .9032 .9049 .9066 .9082 .9099 .9115 .9131 .9147 .9162 .9177

1.4 .9192 .9207 .9222 .9236 .9251 .9265 .9279 .9292 .9306 .9319

1.5 .9332 .9345 .9357 .9370 .9382 .9394 .9406 .9418 .9429 .9441

1.6 .9452 .9463 .9474 .9484 .9495 .9505 .9515 .9525 .9535 .9545

1.7 .9554 .9564 .9573 .9582 .9591 .9599 .9608 .9616 .9625 .9633

1.8 .9641 .9649 .9656 .9664 .9671 .9678 .9686 .9693 .9699 .9706

1.9 .9713 .9719 .9726 .9732 .9738 .9744 .9750 .9756 .9761 .9767

2.0 .9772 .9778 .9783 .9788 .9793 .9798 .9803 .9808 .9812 .9817

2.1 .9821 .9826 .9830 .9834 .9838 .9842 .9846 .9850 .9854 .9857

2.2 .9861 .9864 .9868 .9871 .9875 .9878 .9881 .9884 .9887 .9890

2.3 .9893 .9896 .9898 .9901 .9904 .9906 .9909 .9911 .9913 .9916

2.4 .9918 .9920 .9922 .9925 .9927 .9929 .9931 .9932 .9934 .9936

2.5 .9938 .9940 .9941 .9943 .9945 .9946 .9948 .9949 .9951 .9952

2.6 .9953 .9955 .9956 .9957 .9959 .9960 .9961 .9962 .9963 .9964

2.7 .9965 .9966 .9967 .9968 .9969 .9970 .9971 .9972 .9973 .9974

2.8 .9974 .9975 .9976 .9977 .9977 .9978 .9979 .9979 .9980 .9981

2.9 .9981 .9982 .9982 .9983 .9984 .9984 .9985 .9985 .9986 .9986

3.0 .9987 .9987 .9987 .9988 .9988 .9989 .9989 .9989 .9990 .9990

3.1 .9990 .9991 .9991 .9991 .9992 .9992 .9992 .9992 .9993 .9993

3.2 .9993 .9993 .9994 .9994 .9994 .9994 .9994 .9995 .9995 .9995

3.3 .9995 .9995 .9995 .9996 .9996 .9996 .9996 .9996 .9996 .9997

3.4 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9998

Level of Significance α= 0.05 α= 0.01 α= 0.1

Critical Value for

a left-tailed test -1.645 -2.33 -1.28

a right-tailed test 1.645 2.33 1.28

a two tailed test ±1.96 ±2.58 ±1.645

Level of Confidence c Critical Value Zc

0.80 or 80% 1.28

0.90 or 90% 1.645

0.95 or 95% 1.96

0.99 or 99% 2.58

(b) Confidence Interval z-Critical

values

(c) Hypothesis testing z-Critical

values

© Ven M

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http://www.facebook.com/venkatvnv.significantly.insignificant


Student t-distribution Table

One-tail

area 0.250 0.125 0.100 0.075 0.050 0.025 0.010 0.005 0.0005

Two-tail

area 0.500 0.250 0.20 0.150 0.100 0.050 0.020 0.010 0.0010

d.f./c 0.500 0.750 0.800 0.850 0.900 0.950 0.980 0.990 0.999

1 1.000 2.414 3.078 4.165 6.314 12.706 31.821 63.657 636.619

2 0.816 1.604 1.886 2.282 2.920 4.303 6.965 9.925 31.599

3 0.765 1.423 1.638 1.924 2.353 3.182 4.541 5.841 12.924

4 0.741 1.344 1.533 1.778 2.132 2.776 3.747 4.604 8.610

5 0.727 1.301 1.476 1.699 2.015 2.571 3.365 4.032 6.869

6 0.718 1.273 1.440 1.650 1.943 2.447 3.143 3.707 5.959

7 0.711 1.254 1.415 1.617 1.895 2.365 2.998 3.499 5.408

8 0.706 1.240 1.397 1.592 1.860 2.306 2.896 3.355 5.041

9 0.703 1.230 1.383 1.574 1.833 2.262 2.821 3.250 4.781

10 0.700 1.221 1.372 1.559 1.812 2.228 2.764 3.169 4.587

11 0.697 1.214 1.363 1.548 1.796 2.201 2.718 3.106 4.437

12 0.695 1.209 1.356 1.538 1.782 2.179 2.681 3.055 4.318

13 0.694 1.204 1.350 1.530 1.771 2.160 2.650 3.012 4.221

14 0.692 1.200 1.345 1.523 1.761 2.145 2.624 2.977 4.140

15 0.691 1.197 1.341 1.517 1.753 2.131 2.602 2.947 4.073

16 0.690 1.194 1.337 1.512 1.746 2.120 2.583 2.921 4.015

17 0.689 1.191 1.333 1.508 1.740 2.110 2.567 2.898 3.965

18 0.688 1.189 1.330 1.504 1.734 2.101 2.552 2.878 3.922

19 0.688 1.187 1.328 1.500 1.729 2.093 2.539 2.861 3.883

20 0.687 1.185 1.325 1.497 1.725 2.086 2.528 2.845 3.850

21 0.686 1.183 1.323 1.494 1.721 2.080 2.518 2.831 3.819

22 0.686 1.182 1.321 1.492 1.717 2.074 2.508 2.819 3.792

23 0.685 1.180 1.319 1.489 1.714 2.069 2.500 2.807 3.768

24 0.685 1.179 1.318 1.487 1.711 2.064 2.492 2.797 3.745

25 0.684 1.198 1.316 1.485 1.708 2.060 2.485 2.787 3.725

26 0.684 1.177 1.315 1.483 1.706 2.056 2.479 2.779 3.707

27 0.684 1.176 1.314 1.482 1.703 2.052 2.473 2.771 3.690

28 0.683 1.175 1.313 1.480 1.701 2.048 2.467 2.763 3.674

29 0.683 1.174 1.311 1.479 1.699 2.045 2.462 2.756 3.659

30 0.683 1.173 1.310 1.477 1.697 2.042 2.457 2.750 3.646

35 0.682 1.170 1.306 1.472 1.690 2.030 2.438 2.724 3.591

40 0.681 1.167 1.303 1.468 1.684 2.021 2.423 2.704 3.551

45 0.680 1.165 1.301 1.465 1.679 2.014 2.412 2.690 3.520

50 0.679 1.164 1.299 1.462 1.676 2.009 2.403 2.678 3.496

60 0.679 1.162 1.296 1.458 1.671 2.000 2.390 2.660 3.460

70 0.678 1.160 1.294 1.456 1.667 1.994 2.381 2.648 3.435

80 0.678 1.159 1.292 1.453 1.664 1.990 2.374 2.639 3.416

100 0.677 1.157 1.290 1.451 1.660 1.984 2.364 2.626 3.390

500 0.675 1.152 1.283 1.442 1.648 1.965 2.334 2.586 3.310

1000 0.675 1.151 1.282 1.441 1.646 2.962 2.330 2.581 3.300

0.674 1.150 1.282 1.440 1.645 1.960 2.326 2.576 3.291 © Ven M

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1. Scores on exam-1 for statistics score are normally distributed with mean 70 and

standard deviation 1.5.

a. What is the probability that a student selected at random has a score

below 72 on this exam? [Ans: 0.9082]

𝑷 ( 𝒙 < 𝟕𝟐 ) = 𝑃 ( 𝑥 − 𝜇

𝜎≤ 72 − 70

1.5) = 𝑃( 𝑍 ≤ 1.33 ) = 𝟎. 𝟗𝟎𝟖𝟐

b. What is the probability that a student selected at random has a score

above 60 on this exam? [Ans: 1 ]

𝑷 ( 𝒙 > 𝟔𝟎 ) = 𝑃 ( 𝑥 − 𝜇

𝜎≤ 60 − 70

1.5) = 𝑃( 𝑍 ≤ −6.67 ) = 𝑃(𝑧 < 6.67) = 𝟏

c. What is the probability that a student selected at random has a

score between 65 and 72 on this exam? [Ans: 0.9078]

𝑷 ( 𝟔𝟓 < 𝒙 < 𝟕𝟐 ) = 𝑃 (65− 70

1.5< 𝑥 − 𝜇

𝜎<72 − 70

1.5)

= 𝑃(−3.33 < 𝑍 < 1.33 ) = 𝑃(𝑧 < 1.33)−𝑃(𝑧 < −3.33)

= 𝑃(𝑧 < 1.33) − [1 − 𝑃(𝑧 < 3.33)]

= 0.9082 − [1 − 0.9996] = 0.9082 − 0.0004 = 𝟎. 𝟗𝟎𝟕𝟖

d. What is the probability that a student selected at random has a

score between 55 and 67 on this exam? [Ans: 0.0228]

𝑷 ( 𝟓𝟓 < 𝒙 < 𝟔𝟕 ) = 𝑃 (55− 70

1.5< 𝑥 − 𝜇

𝜎<67 − 70

1.5)

= 𝑃(−10 < 𝑍 < −2 ) = 𝑃(𝑧 < −2)−𝑃(𝑧 < −10)

= [1 − 𝑃(𝑧 < 2)] − [1 − 𝑃(𝑧 < 10)]

= [1 − 0.9772] − [1 − 1] = 0.0228 − 0 = 𝟎. 𝟎𝟐𝟐𝟖

e. What score separates the lowest 4.95% of observations of the distribution?

[Ans: 67.53]

𝑷(𝒙 < 𝒂) = 𝟎. 𝟎𝟒𝟗𝟓 ⟹ 𝑷(𝒙 > 𝒂) = 𝟎. 𝟗𝟓𝟎𝟓 ⟹ 𝑃 ( 𝑧 > 𝑎 − 70

1.5) = 0.9505;

𝑁𝑜𝑤 𝑓𝑟𝑜𝑚 𝑧 − 𝑡𝑎𝑏𝑙𝑒, 𝑤𝑒 ℎ𝑎𝑣𝑒 𝑃( 𝑧 > −1.65 ) = 0.9505

𝐸𝑞𝑢𝑎𝑡𝑖𝑛𝑔 𝑎 − 70

1.5= −1.65 𝑎𝑛𝑑 𝑠𝑜𝑙𝑣𝑖𝑛𝑔 𝑓𝑜𝑟 𝑎, 𝑤𝑒 𝑔𝑒𝑡 𝒂 = 𝟕𝟎 − (𝟏. 𝟓 × 𝟏. 𝟔𝟓) = 𝟔𝟕. 𝟓𝟑 © Ven M

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f. What score separates the highest 10% of observations of the distribution?

[Ans: 71.92]

𝑷(𝒙 > 𝒂) = 𝟎. 𝟏𝟎 ⟹ 𝑷(𝒙 < 𝒂) = 𝟎. 𝟗 ⟹ 𝑃 ( 𝑧 < 𝑎 − 70

1.5) = 0.9;

𝑁𝑜𝑤 𝑓𝑟𝑜𝑚 𝑧 − 𝑡𝑎𝑏𝑙𝑒, 𝑤𝑒 ℎ𝑎𝑣𝑒 𝑃( 𝑧 < 1.28 ) = 0.9


1.5= 1.28 𝑎𝑛𝑑 𝑠𝑜𝑙𝑣𝑖𝑛𝑔 𝑓𝑜𝑟 𝑎, 𝑤𝑒 𝑔𝑒𝑡 𝒂 = 𝟕𝟎 + (𝟏. 𝟓 × 𝟏. 𝟐𝟖)

= 𝟕𝟏. 𝟗𝟐

g. Find ‘a’ such that P(x > a) = 0.7357 [Ans:69.06 ]

𝑷(𝒙 > 𝒂) = 𝟎. 𝟕𝟑𝟓𝟕 ⟹ 𝑃 ( 𝑧 > 𝑎 − 70

1.5) = 0.7357;



1.5= −0.63 𝑎𝑛𝑑 𝑠𝑜𝑙𝑣𝑖𝑛𝑔 𝑓𝑜𝑟 𝑎, 𝑤𝑒 𝑔𝑒𝑡 𝒂 = 𝟕𝟎 − (𝟏. 𝟓 × 𝟎. 𝟔𝟑)

= 𝟔𝟗. 𝟎𝟔

2. Given that the POPULATION MEAN is 90 and the STANDARDDEVIATION is 8 on a

continuous NORMALLY DISTRIBUTED scale. Find the following:

(a) 𝑃( 𝑥 < 94) [Ans: 0.6915]

𝑷 ( 𝒙 < 𝟗𝟒 ) = 𝑃 ( 𝑥 − 𝜇

𝜎≤ 94 − 90

8) = 𝑃( 𝑧 ≤ 0.5 ) = 𝟎. 𝟔𝟗𝟏𝟓

(b) 𝑃(𝑥 > 86) [Ans: 0.6915]

𝑷 ( 𝒙 > 𝟖𝟔 ) = 𝑃 ( 𝑥 − 𝜇

𝜎> 86 − 90

8) = 𝑃( 𝑧 > −0.5 ) = 𝑃(𝑧 < 0.5) = 𝟎. 𝟔𝟗𝟏𝟓

(c) 𝑃(82 < 𝑥 < 94) [Ans: 0.5328]

𝑷 ( 𝟖𝟐 < 𝒙 < 𝟗𝟒 ) = 𝑃 (82 − 90

8< 𝑥 − 𝜇

𝜎<94 − 90

8)

= 𝑃(−1 < 𝑧 < 0.5 ) = 𝑃(𝑧 < 0.5)−𝑃(𝑧 < −1)

= 0.6915 − [1 − 𝑃(𝑧 < 1)] = 0.6915 − [1 − 0.8413] = 0.6915 − 0.1587 = 𝟎. 𝟓𝟑𝟐𝟖

© Ven M

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(d) 𝑃(80 < 𝑥 < 90) [Ans: 0.3944]

𝑷 ( 𝟖𝟎 < 𝒙 < 𝟗𝟎 ) = 𝑃 (80 − 90

8< 𝑥 − 𝜇

𝜎<90 − 90

8)

= 𝑃(−1.25 < 𝑧 < 0 ) = 𝑃(𝑧 < 0)−𝑃(𝑧 < −1.25)

= 0.5 − [1 − 𝑃(𝑧 < 1.25)] = 0.5 − [1 − 0.8944] = 0.5 − 0.1056 = 𝟎. 𝟑𝟗𝟒𝟒

(e) 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑎 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 𝑃(𝑥 < 𝑎) = 0.9656 [Ans: 104.56]

𝑷(𝒙 < 𝒂) = 𝟎. 𝟗𝟔𝟓𝟔 ⟹ 𝑃 ( 𝑧 < 𝑎 − 90

8) = 0.9656;



8= 1.82 𝑎𝑛𝑑 𝑠𝑜𝑙𝑣𝑖𝑛𝑔 𝑓𝑜𝑟 𝑎,

𝑤𝑒 𝑔𝑒𝑡 𝒂 = 𝟗𝟎 + (𝟖 × 𝟏. 𝟖𝟐) = 𝟏𝟎𝟒. 𝟓𝟔

(f) 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑎 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 𝑃(𝑥 > 𝑎) = 0.9418 [Ans: 77.44]

𝑷(𝒙 > 𝒂) = 𝟎. 𝟗𝟒𝟏𝟖 ⟹ 𝑃 ( 𝑧 > 𝑎 − 90

8) = 0.9418;



8= −1.57 𝑎𝑛𝑑 𝑠𝑜𝑙𝑣𝑖𝑛𝑔 𝑓𝑜𝑟 𝑎,

𝑤𝑒 𝑔𝑒𝑡 𝒂 = 𝟗𝟎 − (𝟖 × 𝟏. 𝟖𝟐) = 𝟕𝟓. 𝟒𝟒

3. Suppose 150 people are repeatedly sampled from a population in which the

standard deviation is 35. Find the standard error? [Ans: 2.858]

𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝐸𝑟𝑟𝑜𝑟, 𝑆. 𝐸. = 𝜎

√𝑛=

35

√150= 2.858

4. Suppose a standard deviation for raw scores is 14 and the sample size is 49. What is

the standard error? [Ans: 2]

𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝐸𝑟𝑟𝑜𝑟, 𝑆. 𝐸. = 𝜎

√𝑛=14

√49= 2

5. Suppose a population mean is 26 and the standard error is 0.5 based on 200

samples. What is the probability that the sample mean is above 26? [Ans:0.5]

𝑈𝑠𝑖𝑛𝑔 𝑧 = �̅� − 𝜇𝜎√𝑛⁄

𝑎𝑛𝑑 𝑡ℎ𝑒 𝑆. 𝐸. , 𝜎√𝑛⁄ = 0.5,𝑤𝑒 𝑔𝑒𝑡

𝑷(�̅� > 𝟐𝟔) = 𝑃 ( 𝑧 > 26 − 26

0.5) = 𝑃( 𝑧 > 0) = 1 − 𝑃(𝑧 < 0) = 1 − 0.5 = 0.5

© Ven M

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6. Suppose the random variable X is normally distributed with mean 80 and standard

deviation 16. What is the probability that a random variable is between 63 and 78?

[Ans: 0.3037]

𝑷 ( 𝟔𝟑 < 𝒙 < 𝟕𝟖 ) = 𝑃 (63 − 80

16< 𝑥 − 𝜇

𝜎<78 − 80

16) = 𝑃(−1.06 < 𝑧 < −0.13 )

= 𝑃(𝑧 < −0.13) − 𝑃(𝑧 < −1.06) = [1 − 𝑃(𝑧 < 0.13)] − [1 − 𝑃(𝑧 < 1.06)]

= [1 − 0.5517] − [1 − 0.8554] = 0.4483 − 0.1446 = 𝟎. 𝟑𝟎𝟑𝟕


deviation 16. What is the probability that a random variable is between 74 and 97?

[Ans: 0.5034]

𝑷 ( 𝟕𝟒 < 𝒙 < 𝟗𝟕 ) = 𝑃 (74 − 80

16< 𝑥 − 𝜇

𝜎<97 − 80

16)

= 𝑃(−0.38 < 𝑍 < 1.06 ) = 𝑃(𝑧 < 1.06)−𝑃(𝑧 < −0.38) = [𝑃(𝑧 < 1.06)]− [1 − 𝑃(𝑧 < 0.38)]

= 0.8554 − [1 − 0.6480] = 𝟎. 𝟓𝟎𝟑𝟒


deviation 16. Find ‘a’ such that P(x>a) = 0.8888 [Ans: 60.48]

𝑷(𝒙 > 𝒂) = 𝟎. 𝟖𝟖𝟖𝟖 ⟹ 𝑃 ( 𝑧 > 𝑎 − 80

16) = 0.8888;



16= −1.22 𝑎𝑛𝑑 𝑠𝑜𝑙𝑣𝑖𝑛𝑔 𝑓𝑜𝑟 𝑎, 𝑤𝑒 𝑔𝑒𝑡 𝒂 = 𝟖𝟎 − (𝟏𝟔 × 𝟏. 𝟐𝟐) = 𝟔𝟎. 𝟒𝟖


deviation 16. Find ‘a’ such that P(x<a) = 0.9505 [Ans: 106.4]

𝑷(𝒙 < 𝒂) = 𝟎. 𝟗𝟓𝟎𝟓 ⟹ 𝑃 ( 𝑧 < 𝑎 − 80

16) = 0.9505;



16= 1.65 𝑎𝑛𝑑 𝑠𝑜𝑙𝑣𝑖𝑛𝑔 𝑓𝑜𝑟 𝑎, 𝑤𝑒 𝑔𝑒𝑡 𝒂 = 𝟖𝟎 + (𝟏𝟔 × 𝟏. 𝟔𝟓) = 𝟏𝟎𝟔. 𝟒 © Ven M

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deviation 16. Find ‘a’ such that P(x>a) = 0.9 [Ans: 59.52]

𝑷(𝒙 > 𝒂) = 𝟎. 𝟗 ⟹ 𝑃 ( 𝑧 > 𝑎 − 80

16) = 0.9;



16= −1.28 𝑎𝑛𝑑 𝑠𝑜𝑙𝑣𝑖𝑛𝑔 𝑓𝑜𝑟 𝑎, 𝑤𝑒 𝑔𝑒𝑡 𝒂 = 𝟖𝟎 − (𝟏𝟔 × 𝟏. 𝟐𝟖) = 𝟓𝟗. 𝟓𝟐


deviation 16. Find ‘a’ such that P(x<a) = 0.975 [Ans: 111.36]

𝑷(𝒙 < 𝒂) = 𝟎. 𝟗𝟕𝟓 ⟹ 𝑃 ( 𝑧 < 𝑎 − 80

16) = 0.975;



16= 1.96 𝑎𝑛𝑑 𝑠𝑜𝑙𝑣𝑖𝑛𝑔 𝑓𝑜𝑟 𝑎, 𝑤𝑒 𝑔𝑒𝑡 𝒂 = 𝟖𝟎 + (𝟏𝟔 × 𝟏. 𝟗𝟔) = 𝟏𝟏𝟏. 𝟑𝟔

12. Suppose the population value of the mean is 200, the population standard

deviation is 10. For a sample of size 36, what is the probability that the sample

mean is between 180 and 220? [Ans: 1]

𝑃( �̅� > 120 ) = 𝑃 ( �̅� − 𝜇𝜎√𝑛⁄

> 120 − 100

25√16⁄

) = 𝑃(𝑧 > 3.2)

= 1 − 𝑃(𝑧 < 3.2) = 1 − 0.9993 = 0.0007

13. Suppose the proportion of all college students who have used i-clickers in the last

two semesters is 60%. If a class of 200 students is considered who are using i-clickers,

what is the probability that the proportion of students who have used i-clickers in

the last 2 semesters is at least 56%? [Ans: 0.8749]

𝑃( �̂� ≥ 0.56 ) = 𝑃

(

�̂� − 𝑝

√𝑝𝑞𝑛

≥0.56 − 0.60

√(0.60)(0.40)200 )

= 𝑃(𝑧 ≥ −1.15) = 𝑃(𝑧 ≤ 1.15) = 0.8749

14. According to a study conducted nationwide in the year 2000, 35% of American

households own more than 2 cars. What is the probability that in a random sample

of 60 selected households at most 40% households will own 2 cars? [Ans: 0.7910]

𝑃( �̂� ≤ 0.4 ) = 𝑃

(

�̂� − 𝑝

√𝑝𝑞𝑛

≤0.4 − 0.35

√(0.35)(0.65)60 )

= 𝑃(𝑧 ≤ 0.81) = 0.791 © Ven M

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15. The daily wages of a person in the United States is normally distributed with a mean

of $100 and standard deviation of $25. If a random sample of 16 people were

surveyed, what is the probability that their average daily wages are greater than

$120? [Ans:0.0007]

𝑃( �̅� > 120 ) = 𝑃 ( �̅� − 𝜇𝜎√𝑛⁄

> 120 − 100

25√16⁄

) = 𝑃(𝑧 > 3.2)

= 1 − 𝑃(𝑧 < 3.2) = 1 − 0.9993 = 0.0007

16. Assume that flight durations from London to JFK can be described as a normal

model with mean time of 16 hours and standard deviation of 1.5 hours. Suppose a

certain airways is currently providing new service in this route with 10 new flights.

What’s the probability that the mean flight duration of these flights will be more

than 17.3 hours? [Ans: 0.0031]

𝑃( �̅� > 17.3 ) = 𝑃 ( �̅� − 𝜇𝜎√𝑛⁄

> 17.3 − 16

1.5√10⁄

) = 𝑃(𝑧 > 2.74)

= 1 − 𝑃(𝑧 < 2.74) = 1 − 0.9969 = 0.0031

17. Sixty wind speed readings gave an average speed of 1105 cm/sec. From long term

experience, it can be assumed that the standard deviation is 240 cm/sec. Find and

interpret a 90% confidence interval for the parameter of interest. [Ans: (1054.03,

1155.97)]

𝐺𝑖𝑣𝑒𝑛: 𝑛 = 60; 𝑚𝑒𝑎𝑛, �̅� = 1105 ⇒ 𝐶𝑙𝑒𝑎𝑟𝑙𝑦 𝑃𝑎𝑟𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝑖𝑠 𝜇

𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛, 𝜎 = 240,⇒ 𝜎 − 𝑘𝑛𝑜𝑤𝑛 ⇒ 𝑧 − 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛

𝑐 = 90% = 0.90 ⇒ 𝑧𝑐 = 𝑧0.90 = 1.645

The Point Estimate of Mean µ is �̅� = 1105

The 90% Confidence Interval is (�̅� − 𝐸, �̅� + 𝐸),𝑤ℎ𝑒𝑟𝑒 𝐸 = 𝑧𝑐𝜎

√𝑛= 1.645 (

240

√60) = 50.97

Therefore, 90% CI for mean µ is,

(�̅� − 𝐸, �̅� + 𝐸) = (1105 − 50.97, 1105 + 50.97) = (1054.03, 1155.97)

© Ven M

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18. Given a random sample of 24 measurements recorded a sample mean 35 and

with a standard deviation 5. What is the maximal margin of error E? Construct a 90%

confidence interval for parameter of interest. [Ans: (33.251,36.749)]

𝐺𝑖𝑣𝑒𝑛: 𝑛 = 24; 𝑚𝑒𝑎𝑛, �̅� = 35 ⇒ 𝐶𝑙𝑒𝑎𝑟𝑙𝑦 𝑃𝑎𝑟𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝑖𝑠 𝜇 𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛, 𝑠 = 5,⇒ 𝜎 − 𝑢𝑛𝑘𝑛𝑜𝑤𝑛 𝐴𝑁𝐷 𝑛 < 30 ⇒ 𝑡 − 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛

𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓 𝑓𝑟𝑒𝑑𝑜𝑚, 𝑣 = 𝑛 − 1 = 24 − 1 = 23 𝑎𝑛𝑑 𝑐 = 90% = 0.90 ⇒ 𝑡𝑣,𝑐 = 𝑡23,0.90 = 1.714


The 90% Confidence Interval is (�̅� − 𝐸, �̅� + 𝐸),𝑤ℎ𝑒𝑟𝑒 𝐸 = 𝑡𝑣,𝑐𝑠

√𝑛= 1.714 (

5

√24) = 1.749

Therefore, 90% CI for mean µ is,

(�̅� − 𝐸, �̅� + 𝐸) = (35 − 1.749, 35 + 1.749) = (𝟑𝟑. 𝟐𝟓𝟏, 𝟑𝟔. 𝟕𝟒𝟗)

19. Given a random sample of 50 measurements of carbon in air from a region on one

day recorded a sample mean of 210 with a standard deviation of 55. What is the

maximal margin of error E? Construct a 95% confidence interval for the parameter

of interest. [Ans: (194.755, 225.245)]

𝐺𝑖𝑣𝑒𝑛: 𝑛 = 50; 𝑚𝑒𝑎𝑛, �̅� = 210 ⇒ 𝐶𝑙𝑒𝑎𝑟𝑙𝑦 𝑃𝑎𝑟𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝑖𝑠 𝜇 𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛, 𝑠 = 55,⇒ 𝜎 − 𝑢𝑛𝑘𝑛𝑜𝑤𝑛 𝑩𝑼𝑻 𝒏 > 𝟑𝟎 ⇒ 𝑧 − 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛

𝑐 = 95% = 0.95 ⇒ 𝑧𝑐 = 1.96


The 95% Confidence Interval is (�̅� − 𝐸, �̅� + 𝐸),𝑤ℎ𝑒𝑟𝑒 𝐸 = 𝑧𝑐𝜎

√𝑛= 1.96 (

55

√50) = 15.245

Therefore, 95% CI for mean µ is, (�̅� − 𝐸, �̅� + 𝐸) = (210 − 15.245, 210 + 15.245)

= (𝟏𝟗𝟒. 𝟕𝟓𝟓, 𝟐𝟐𝟓. 𝟐𝟒𝟓)

20. In a random sample of 519 students, it was found that 234 were seniors. Compute

and interpret a 99% confidence interval for the total number of seniors in the

campus. [Ans: (0.395,0.507)]

𝐺𝑖𝑣𝑒𝑛: 𝑛 = 519; 𝑟 = 234 ⇒ 𝐶𝑙𝑒𝑎𝑟𝑙𝑦 𝑃𝑎𝑟𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝑖𝑠 𝑝𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛 𝑝

�̂� =𝑟

𝑛=234

519= 0.451

𝐴𝑠𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛𝑠: 𝐶𝑙𝑒𝑎𝑟𝑙𝑦 ℎ𝑖𝑔ℎ𝑒𝑟 𝑠𝑡𝑢𝑑𝑖𝑒𝑠 𝑐𝑎𝑡𝑒𝑔𝑜𝑟𝑦 = 234 > 5 𝑎𝑛𝑑

𝑡ℎ𝑒 𝑟𝑒𝑠𝑡 𝑜𝑓 𝑡ℎ𝑒𝑚(𝑜𝑡ℎ𝑒𝑟 𝑐𝑎𝑡𝑒𝑔𝑜𝑟𝑦) 𝑎𝑟𝑒 519 − 234 = 285 > 5 𝑎𝑛𝑑 𝑐 = 99% = 0.99

⇒ 𝑧𝑐 = 𝑧0.99 = 2.58 © Ven M

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The Point Estimate of Proportion p is = �̂� = 0.451

The 99% Confidence Interval is (�̂� − 𝐸, �̂� + 𝐸),𝑤ℎ𝑒𝑟𝑒 𝐸 = 𝑧𝑐√𝑝�̂�

𝑛

𝐸 = 𝑧𝑐 × √�̂��̂�

𝑛= 2.58 × √

0.451 × 0.549

519= 0.056

Therefore, 99% CI for proportion p is,

(�̂� − 𝐸, �̂� + 𝐸) = (0.451 − 0.056, 0.451 + 0.056) = (𝟎. 𝟑𝟗𝟓, 𝟎. 𝟓𝟎𝟕)

21. Find the sample size necessary to determine the proportions within 5% at the 10%

level of significance if no preliminary estimate of proportion is known. [Ans: 271]

Since NO preliminary estimate of proportion p is given, we use the following formula for

estimation of sample size, n.

𝑛 =1

4(𝑧𝑐𝐸)2

𝐺𝑖𝑣𝑒𝑛 𝛼 = 10% ⇒ 𝑐 = 90% = 0.90, 𝑎𝑛𝑑 ℎ𝑒𝑛𝑐𝑒 𝑧0.90 = 1.645 𝑎𝑛𝑑 𝐸 = 5% = 0.05

𝑛 =1

4(𝑧𝑐𝐸)2

=1

4(1.645

0.05)2

= 270.60 ≃ 𝟐𝟕𝟏

22. A child psychologist wants to estimate the mean age at which a child learns to talk.

Find the sample size necessary for a 95% confidence level to ensure that the

sample mean is within 5 weeks for the mean age at which a child learns to talk.

Assume standard deviation as 10 weeks. [Ans: 16]

Here we are given with the details of confidence level, c=95%; Maximal Margin

of error, E = 5; and σ =10. We are asked to calculate the Sample size, n.

𝐴𝑙𝑠𝑜 𝑧𝑐 = 𝑧0.95 = 1.96; 𝐹𝑜𝑟 𝑆𝑎𝑚𝑝𝑙𝑒 𝑠𝑖𝑧𝑒 𝑒𝑠𝑡𝑖𝑚𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑀𝑒𝑎𝑛𝑠, 𝑤𝑒 ℎ𝑎𝑣𝑒

𝑛 = (𝑧𝑐𝜎

𝐸)2

= (1.96 × 10

5)2

= 15.36 ≈ 𝟏𝟔

© Ven M

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23. A cloth manufacturing company claims that the mean tearing strength of their

curtain fabric is 120 pounds. A government inspection agency conducted tests on

100 curtain lengths revealing a mean strength of 115 pounds with a standard

deviation of 20 pounds. At the 10% level of significance, does the governments’

information support the manufacturer’s claim? [Ans: Reject H0]

a. Details given in the problem: 𝐺𝑖𝑣𝑒𝑛, 𝑛 = 100, �̅� = 115, 𝛼 = 0.1,

𝜎 = 20, 𝜇 = 120.𝐻𝑒𝑟𝑒 𝜎 𝑖𝑠 𝑢𝑛𝑘𝑛𝑜𝑤𝑛 𝐵𝑈𝑇 𝑛 = 100 > 30,

𝑤𝑒 𝑤𝑖𝑙𝑙 𝑢𝑠𝑒 𝑧 − 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛

b. State the Null hypothesis, Alternate hypothesis and what tail test is this:

Here 𝐻0: 𝜇 = 120 𝑣𝑠 𝐻1: 𝜇 ≠ 120; 𝑇ℎ𝑖𝑠 𝑖𝑠 𝑎 𝑡𝑤𝑜 𝑡𝑎𝑖𝑙 𝑡𝑒𝑠𝑡.

c. Find the Critical Statistic(Table value) and Illustrate: zα/2 = ±1.645

d. State and Compute test Statistic:

𝑧 =�̅�−𝜇𝜎

√𝑛

=115−120

20

√100

= −2.5

e. Compute p-value:

𝑝 = 2𝑃(|𝑍| > 𝑧) = 2𝑃(𝑍 > |−2.5|) = 2[1 − 𝑃(𝑍 < 2.5)] = 2[1 − 0.9938] = 0.0124

f. Conclusion using critical method:

𝑆𝑖𝑛𝑐𝑒 𝑡ℎ𝑒 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 𝑡𝑒𝑠𝑡 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐, 𝑧 = −2.5 𝑓𝑎𝑙𝑙𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑟𝑒𝑗𝑒𝑐𝑡𝑖𝑜𝑛 𝑟𝑒𝑔𝑖𝑜𝑛,

𝑤𝑒 𝑟𝑒𝑗𝑒𝑐𝑡 𝑡ℎ𝑒 𝑛𝑢𝑙𝑙 ℎ𝑦𝑝𝑜𝑡ℎ𝑒𝑠𝑖𝑠.

g. Conclusion using p-value method:

𝐶𝑙𝑒𝑎𝑟𝑙𝑦 𝑝 − 𝑣𝑎𝑙𝑢𝑒 = 0.0124 < 0.1(= 𝛼),𝑤𝑒 𝑟𝑒𝑗𝑒𝑐𝑡 𝑡ℎ𝑒 𝑛𝑢𝑙𝑙 ℎ𝑦𝑝𝑜𝑡ℎ𝑒𝑠𝑖𝑠

2.58 -2.58 1.645 -1.645

© Ven M

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24. A car company wishes to test the claim that at an average of 55 mph, the stopping

distance for auto brakes is 130 ft. In a random sample of 19 individuals that were

tested for their breaking distance were recorded a sample mean of 144 ft. with a

sample standard deviation of 20 ft. At the 10% level of significance, test the claim

that the stopping distance for a car traveling at 55 mph is more than the stated 130

ft. [Ans: Reject H0]

a. Details given in the problem: 𝐺𝑖𝑣𝑒𝑛, 𝑛 = 19, �̅� = 144, 𝛼 = 0.1,

𝑠 = 20, 𝜇 = 130.𝐻𝑒𝑟𝑒 𝜎 𝑖𝑠 𝑢𝑛𝑘𝑛𝑜𝑤𝑛 𝐴𝑁𝐷 𝑛 = 19 < 30,

𝑤𝑒 𝑤𝑖𝑙𝑙 𝑢𝑠𝑒 𝑡 − 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑤𝑖𝑡ℎ (𝑛 − 1)𝑑. 𝑓. = 18 𝑑. 𝑓.


Here 𝐻0: 𝜇 = 130 𝑣𝑠 𝐻1: 𝜇 > 130; 𝑇ℎ𝑖𝑠 𝑖𝑠 𝑎 𝑜𝑛𝑒(𝑟𝑖𝑔ℎ𝑡) 𝑡𝑎𝑖𝑙 𝑡𝑒𝑠𝑡.

c. Find the Critical Statistic(Table value) and Illustrate: t18,0.1 = 1.33


𝑡 =�̅� − 𝜇𝑠

√𝑛

=144 − 130

20

√19

= 3.05

e. Compute p-value:

0.0005 < 𝑝 − 𝑣𝑎𝑙𝑢𝑒 < 0.005


𝑆𝑖𝑛𝑐𝑒 𝑡ℎ𝑒 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 𝑡𝑒𝑠𝑡 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐, 𝑧𝑡 = 3.05 𝑓𝑎𝑙𝑙𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑟𝑒𝑗𝑒𝑐𝑡𝑖𝑜𝑛 𝑟𝑒𝑔𝑖𝑜𝑛, 𝑤𝑒 𝑟𝑒𝑗𝑒𝑐𝑡 𝑡ℎ𝑒 𝑛𝑢𝑙𝑙 ℎ𝑦𝑝𝑜𝑡ℎ𝑒𝑠𝑖𝑠.


𝐶𝑙𝑒𝑎𝑟𝑙𝑦 𝑝 − 𝑣𝑎𝑙𝑢𝑒 < 0.1(= 𝛼), 𝑤𝑒 𝑟𝑒𝑗𝑒𝑐𝑡 𝑡ℎ𝑒 𝑛𝑢𝑙𝑙 ℎ𝑦𝑝𝑜𝑡ℎ𝑒𝑠𝑖𝑠

2.58 -2.58 1.33

0.0005 0.005 α = 0.1

p-value © Ven M

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25. A SAT tutoring center claims that the average Math scores of their students in SAT is

500. A small sample of 40 students was considered and has given a sample mean

score of 521.5 with a sample standard deviation of 120.5. At 1% level of significance

do you have sufficient evidence to prove that the average Math score on SAT is

different than 500. [Ans: We fail to Reject H0]

a. Details given in the problem: 𝐺𝑖𝑣𝑒𝑛, 𝑛 = 40, �̅� = 521.5, 𝛼 = 0.01, 𝑠 =

120.5, 𝜇 = 500.𝐻𝑒𝑟𝑒 𝜎 𝑖𝑠 𝑢𝑛𝑘𝑛𝑜𝑤𝑛 BUT 𝑛 = 40 > 30

⟹ 𝑤𝑒 𝑤𝑖𝑙𝑙 𝑢𝑠𝑒 𝑧 − 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑡𝑟𝑒𝑎𝑡𝑖𝑛𝑔 𝑠 ≈ 𝜎


Here 𝐻0: 𝜇 = 500 𝑣𝑠 𝐻1: 𝜇 ≠ 500; 𝑇ℎ𝑖𝑠 𝑖𝑠 𝑎 𝑡𝑤𝑜 𝑡𝑎𝑖𝑙 𝑡𝑒𝑠𝑡.

c. Find the Critical Statistic(Table value) and Illustrate: zα/2 = ±2.58


𝑧 =�̅�−𝜇𝜎

√𝑛

=521.5−500

120.5

√40

= 1.13

e. Compute p-value:

𝑝 = 2𝑃(|𝑍| > 𝑧) = 2𝑃(𝑍 > |1.13|) = 2[1 − 𝑃(𝑍 < 1.13)] = 2[1 − 0.8708] = 0.2584


𝑆𝑖𝑛𝑐𝑒 𝑡ℎ𝑒 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 𝑡𝑒𝑠𝑡 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐, 𝑧 = 1.13 𝐷𝑂𝐸𝑆 𝑁𝑂𝑇 𝑓𝑎𝑙𝑙 𝑖𝑛 𝑡ℎ𝑒 𝑟𝑒𝑗𝑒𝑐𝑡𝑖𝑜𝑛 𝑟𝑒𝑔𝑖𝑜𝑛,

𝑤𝑒 𝑓𝑎𝑖𝑙 𝑡𝑜 𝑟𝑒𝑗𝑒𝑐𝑡 𝑡ℎ𝑒 𝑛𝑢𝑙𝑙 ℎ𝑦𝑝𝑜𝑡ℎ𝑒𝑠𝑖𝑠.


𝐶𝑙𝑒𝑎𝑟𝑙𝑦 𝑝 − 𝑣𝑎𝑙𝑢𝑒 = 0.2584 > 0.01(= 𝛼), 𝑤𝑒 𝑓𝑎𝑖𝑙 𝑡𝑜 𝑟𝑒𝑗𝑒𝑐𝑡 𝑡ℎ𝑒 𝑛𝑢𝑙𝑙 ℎ𝑦𝑝𝑜𝑡ℎ𝑒𝑠𝑖𝑠.

2.58 -2.58 2.58 -2.58

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NORMAL APPROXIMATION TO BINOMIAL PROBLEMS

26. In economics, of all new products put on the market, 60% failed and are taken

off the market within 2 years. Suppose a store introduces 80 new products. Use

the normal approximation for the binomial distribution what is the probability

that within 2 years 57 or more will fail with correction for continuity? [Ans: 0.0262]

𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛: 𝐴𝑠𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛𝑠: 𝑛𝑝 > 5 𝑎𝑛𝑑 𝑛𝑞 > 5. 𝐺𝑖𝑣𝑒𝑛 𝑛 = 80, 𝑝 = 0.60, 𝑞 = 1 – 𝑝 = 0.40

𝑛𝑝 = (80) (0.60) = 48 > 5 𝑎𝑛𝑑 𝑛𝑞 = (80) (0.40) = 32 > 5

⟹ 𝐴𝑠𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛𝑠 𝑎𝑟𝑒 𝑠𝑎𝑡𝑖𝑠𝑓𝑖𝑒𝑑. 𝐻𝑒𝑛𝑐𝑒 𝑤𝑒 𝑐𝑎𝑛 𝑝𝑟𝑜𝑐𝑒𝑒𝑑 𝑢𝑠𝑖𝑛𝑔 𝑁𝑜𝑟𝑚𝑎𝑙 𝑎𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑖𝑜𝑛 𝑡𝑜 𝐵𝑖𝑛𝑜𝑚𝑖𝑎𝑙.

𝑁𝑜𝑤, 𝜇 = 𝑛𝑝 = (80)(0.60) = 48; 𝜎 = √𝑛𝑝𝑞 = √80 (0.60)(0.40) = 4.38

WITH CORRECTION: 𝑃( 𝑥 ≥ 57 ) = 𝑃(𝑥 ≥ 56.5) = 𝑃 (𝑥 − 𝜇

𝜎 ≥

56.5 − 48

4.38)

= 𝑃 (𝑍 ≥ 1.94) = 1 − 𝑃(𝑍 < 1.94) = 1 − 0.9738 = 𝟎. 𝟎𝟐𝟔𝟐

EXTRA INFORMATION−WITHOUT CORRECTION: 𝑃( 𝑥 ≥ 57 ) = 𝑃 ( 𝑥 − 𝜇

𝜎 ≥

57 − 48

4.38)

= 𝑃( 𝑍 ≥ 2.05) = 1 − 𝑃(𝑍 < 2.05) = 1 − 0.9798 = 0.0202




that within 2 years fewer than 52 will fail without correction for continuity and

with correction for continuity? [Ans: Without: 0.7517 & With: 0.7881]

𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛: 𝐴𝑠𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛𝑠: 𝑛𝑝 > 5 𝑎𝑛𝑑 𝑛𝑞 > 5. 𝐺𝑖𝑣𝑒𝑛 𝑛 = 80, 𝑝 = 0.60, 𝑞 = 1 – 𝑝 = 0.40

𝑛𝑝 = (80) (0.60) = 48 > 5 𝑎𝑛𝑑 𝑛𝑞 = (80) (0.40) = 32 > 5


𝑁𝑜𝑤, 𝜇 = 𝑛𝑝 = (80)(0.60) = 48; 𝜎 = √𝑛𝑝𝑞 = √80 (0.60)(0.40) = 4.38

WITH CORRECTION: 𝑃( 𝑥 < 52 ) = 𝑃(𝑥 ≤ 51) = 𝑃(𝑥 ≤ 51.5) = 𝑃 (𝑥 − 𝜇

𝜎 ≤

51.5 − 48

4.38)

= 𝑃 (𝑍 ≤ 0.80) = 𝟎. 𝟕𝟖𝟖𝟏

WITHOUT CORRECTION: 𝑃( 𝑥 < 52 ) = 𝑃(𝑥 ≤ 51) = 𝑃(𝑥 ≤ 51) = 𝑃 (𝑥 − 𝜇

𝜎 ≤

51 − 48

4.38)

= 𝑃( 𝑍 ≤ 0.68) = 𝟎. 𝟕𝟓𝟏𝟕 © Ven M

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that within 2 years more than 62 will fail without correction for continuity? [Ans:

0.0003]

𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛: 𝐴𝑠𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛𝑠: 𝑛𝑝 > 5 𝑎𝑛𝑑 𝑛𝑞 > 5.

𝐺𝑖𝑣𝑒𝑛 𝑛 = 80, 𝑝 = 0.60, 𝑞 = 1 – 𝑝 = 0.40

𝑛𝑝 = (80) (0.60) = 48 > 5 𝑎𝑛𝑑 𝑛𝑞 = (80) (0.40) = 32 > 5


𝑁𝑜𝑤, 𝜇 = 𝑛𝑝 = (80)(0.60) = 48; 𝜎 = √𝑛𝑝𝑞 = √80 (0.60)(0.40) = 4.38

WITHOUT CORRECTION: 𝑃( 𝑥 > 62 ) = 𝑃(𝑥 ≥ 63) = 𝑃 (𝑥 − 𝜇

𝜎 ≥

63 − 48

4.38)

= 𝑃 (𝑍 ≥ 3.42) = 1 − 𝑃(𝑍 < 3.42) = 1 − 0.9997 = 𝟎. 𝟎𝟎𝟎𝟑

EXTRA INFORMATION−WITH CORRECTION: 𝑃( 𝑥 > 62) = 𝑃( 𝑥 ≥ 63 )

= 𝑃 ( 𝑥 − 𝜇

𝜎 ≥

62.5 − 48

4.38)

= 𝑃( 𝑍 ≥ 3.31) = 1 − 𝑃(𝑍 < 3.31) = 1 − 0.9995 = 𝟎. 𝟎𝟎𝟎𝟓

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Chapter-8: Hypothesis Testing Problem Template for the Exam-3

a. Details given in the problem:

b. Assumptions (if any):

c. Null, Alternate Hypotheses and the tail test?

d. Find the Critical Statistic(Table value) and Illustrate:

e. State and Compute test Statistic:

f. Compute p-value:

g. Conclusion using critical method:

h. Conclusion using p-value method:

Critical Statistic:

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