Course STA0: Statistics and Probability Lecture No to 5 Multiple Choice Questions:. Statistics deals with: a) Observations b) Aggregates of facts*** c) Individuals d) Isolated ites. A nuber of students in a class is an exaple of: a) Finite Population*** b) Infinite Population c) Hypothetical Population d) None of these. A coponent bar chart in which a) Each bar is divided into two or ore sections*** b) The value of a coon variable in the for of grouped bars c) The horizontal or vertical bars of equal with and length proportional to the values d) None of these 4. What is the value of class ark for the given class 0-5? a) 4 b).5***=0+5/=.5 c) d) 5 5. The extreely positive skewed curve is also known as:: a) Frequency curve b) U-shaped curve c) J-shaped curve d) Reverse J-shaped curve*** Nuerical Questions
Q: Differentiate between Qualitative and Quantitative Variables. A variable whose characteristics can be easured nuerically is called Quantitative Variable for exaple age, weight, incoe or nuber of children. A variable whose characteristics cannot be easured nuerically is called Qualitative Variable for exaple gender, education, intelligence or eye-color. Q: Differentiate between Priary data and Secondary data. Data that have been originally collected (raw data) and have not undergone any statistical treatent are called Priary data. Data that have undergone any statistical treatent at least once i.e. the data have been collected, classified, tabulated or presented in soe for for a certain purpose are called Secondary data. Q: Define Multiple Bar Chart. A ultiple bar chart consists of two or ore characteristics corresponding to the values of the coon variable in for of grouped bars, whose lengths are proportional to the values where each bar is shaded with different colors to show their identification. Q4: Arrange the data given below in an array and construct a frequency distribution, using a class interval of, nuber of classes is 5 and the sallest observation is 09. Indicate the class boundaries and class liits clearly. 6, 7, 85, 4, 68, 54, 8, 7,, 6, 68, 69, 75, 59, 5, 58, 6, 85, 88, 7, 5, 5, 6, 68, 9, 7, 9, 76, 9, 57, 46, 4, 8,, 09, 66, 7, 68, 4, 76. Class Liits Frequency Class Boundaries 9 8.5.5 4 7.5 4.5 5 47 6 4.5 47.5
48 60 7 47.5 60.5 6 7 60.5 7.5 74 86 5 7.5 86.5 87 99 86.5 99.5 40 Q5:Draw a Histogra and a Frequency Polygon for the following distribution. Daily wages (Rs.) No. of eployees 4 6 6 8 8 0 0 4 4 6 8 05 9 7 Daily wages No. of wages Mid points 4 6 5 6 8 7 8 0 8 9 0 05 4 9 4 6 7 5 Histogra:
No. of wages No. of wages Histogra 00 8 50 05 00 50 9 7 0 4 6 6 8 8 0 0 4 4 6 Daily wages Frequency Polygon: Frequency Polygon 00 80 60 40 0 00 80 60 40 0 0 Mid 4 points 5 6 7 8
Lecture No 6 to 0. The value that occurs ost often in a set of data is called the: a) Mean b) Mode c) Geoetric ean d) Haronic ean Answer key: b). Find the ean of the following saple of distances of stars fro the earth: 8., 56.9, 4.6,.5 a) 8.0 b) 4.0 c) 8.0 d) 4.0 Answer Key= a). Which of the following is a true stateent about the edian? a) It is always one of the data values. b) It is influenced by extree values. c) Fifty percent of the observations are larger than the edian. d) It is the iddle value of the data values. Answer Key= d) 4. The G.M of, 4 and 8 is a).67 b) 4 c).4 d) 5 Answer Key= b)+4+8/=4 5. One of the ain disadvantages of the range is: a) It does not use all the observations in its calculations b) It cannot be influenced by an extree value c) It does not involve negative values d) It deals with open ended classes
Answer key = a) Question : Find the ode, for the distribution of exaination arks given below: Marks 0-9 40-49 50-59 60-69 70-79 80-89 90-99 Nuber of Students 8 87 90 04 85 0 The class that carries the highest frequency is 59.5-69.5 which the odal class is thus Also l = 59.5, f = 90, f =, f = 04 and h = 0 Hence Mode = l + f f f f f f h 04-90 Mode=59.5+ 0 04-90 + 04- =59.5+5.8=65.=65 Marks Question : Find the ean weight of 0 students at the Punjab University fro the frequency distribution: Weight (pounds) Nuber of students 0-9 0-9 4 0-9 7 40-49 8 50-59 5 60-69 8
70-79 80-89 6 90-99 5 00-09 0-9 Weight (pounds) Nuber of students(f) Class Mark () f 0-9 4.5 4.5 0-9 4 4.5 498.0 0-9 7 4.5 86.5 40-49 8 44.5 4046.0 50-59 5 54.5 86.5 60-69 8 64.5 96.0 70-79 74.5 68.5 80-89 6 84.5 07.0 90-99 5 94.5 97.5 00-09 04.5 409.0 0-9 4.5 4.5 n= f=0 f=8740 f f 8740 56.7 pounds 0 Question
If ode =5 and edian = 0, then find approxiate value of ean Mode= Median- Mean 5=(0)-(Mean) 5=90-(Mean) (Mean) =90-5 (Mean) =65 Mean=.5 Question 4: Find the haronic ean fro the following frequency distribution of weights: Weights (gras) 65-84 85-04 05-4 5-44 45-64 65-84 85-04 f 9 0 7 0 5 4 5 We calculate the haronic ean as below Weights (gras) f f(/) 65-84 9 74.5 0.08 85-04 0 94.5 0.058 05-4 7 4.5 0.4847 5-44 0 4.5 0.0745 45-64 5 54.5 0.06 65-84 4 74.5 0.09 85-04 5 94.5 0.057 60 0.5044
n H ( ) Hence H= f 60 H.gras 0.5044 Question 5 Find sei-inter quartile range, if Q= 4.6 and Q = 67.8 Q Q QD. 67.8 4.6 QD..74 Lecture No to 5. In a syetrical distribution Q=0, Median=0, then the value of Q is: a) 50 b) 5 c) 40*** d) 5. First two oents about 5 of a variable are and 4, then variance a) b) *** c) 4 d) 5. The lack of unifority is called. a) Skewness b) Dispersion*** c) Syetry d) Kurtosis 4. If variance of, 4, 6,8,0 is 8, then variance of 0,, 4, 6, 8 is: a) 6
b) 8*** c) 0 d) Zero 5. The dependent variable is also called: a) Regressand variable b) Predictand variable c) Explained variable d) All of these Q: Calculate M.D fro (i) ean (ii) edian fro the following frequency distribution. Also find the respective coefficients of M.D. Classes Frequency 0-4 5-9 4 0-4 8 5-9 0-4 5 5-9 9 40-44 Total 50 Classes F C.F Class Boundaries f f f (0.5-0)- (4+0.5) 9.5-4.5 7 7 7.8 7.8 5-9 4 5 4.5-9.5 7 68 48.8 5.
0-4 8 9.5-4.5 76 7 56 7.8 6.64 5-9 4 4.5-9.5 7 97.8. 0-4 5 9 9.5-4.5 480 45.7.55 5-9 9 48 4.5-9.5 7 8 7 7.7 64.5 40-44 50 9.5-44.5 4 84 6.7 4. Total 50 -- --- -- --- 86 ------ 84.4
f f 450 50 9 M.D(fro ean) 86 M.D 50 M.D 5.7 f f MD. Coefficient of M.D 5.7 Coefficient of M.D 9 Coefficient of M.D 0.97 Median n Median Group The class 50 th class th 5 class th h n Median l C f 5 Median 9.5 5 4 5 5 Median 9.5 5 Median 9.5 0. Median 9.8 M.D(fro edian) 84. M.D 50 M.D 5.6868 f f MD. Coefficient of M.D fro edian 5.6868 9.8 0.906
Q: The weights of 0 boxes of a certain brand of cereal have a ean content of 78 gra, with a S.D of 9.64 gra. If these boxes were purchased at 0 different stores and the average price per box is $.9 with a S.D $ 0.09. Can you conclude that the weights are relatively ore hoogeneous than the prices? CV S. 00 9.64 CV. 00 78 CV. 0.4700 CV..47 CV S. 00 0.09 CV. 00.9 CV. 0.69800 CV. 6.98 Weights are relatively ore hoogeneous than prices.
Q: The following frequency distribution gives the nuber of houses and the nuber of persons per house No. of Persons No. of houses 5 7 5 4 5 4 6 6 Calculate first four oents about =4. Convert these oents into oents about ean. f D 4 fd fd fd 4 fd 5 - -5 45-5 405 7 - -4 68-6 7 5 - -5 5-5 5 4 0 0 0 0 0 5 4 4 4 4 4 6 6 4 48 96 Total 00 - -48 76-4 8
' fd f ' ' 48 00 0.48 ' ' ' 76 00.76 fd f ' ' ' fd f 4 00.4 ' 4 ' 4 ' 4 8 00 8. fd f 4 ' ' 0.48 ( 0.48) 0.48 0.48 0 ' '.76 0.48.76 0.04.596
' ' ' '.4 ( 0.48)(.76) 0.48.4.544 0..544.56 0.068 ' ' ' ' ' ' 4 4 4 4 6 4 4 4 4 4 8. 4( 0.48)(.4) 6 0.48.76 0.48 8.00 4.498.40 0.59 0.550 4.65 5.9009 b 0.068 b.898 b 0.04 b b 4 5.9009 (.596) 5.9009 b.97 b.5
Q: Fit a regression line, Y a b fro the following data. Y 95 85 85 95 80 70 70 65 60 70 Student i Yi i Yi Y i Yi i Yi Y Y 95 85 7 8 89 64 6 85 95 7 8 49 4 6 80 70-7 4 49-4 4 70 65-8 - 64 44 96 5 60 70-8 -7 4 49 6 Su 90 85 70 60 470 Mean 78 77 Regression equation Ŷ a b
Y Y Y Y n 85 5 77 n 90 5 78 b b b i Yi Y i 470 70 0.644 a Y b a 77 (0.644)(78) a 77. a 6.768 Yˆ 6.768 0.644 Q: For a group of 0 values 45 470, Mode 4.7 Find the coefficient of Skewness?
Mean 45 0 45. n S n n 470 45 S 0 0 S 470 04.04 S 8.96 S 9.59 Now Skewness Mean M od e SK SD. 45. 4.7 SK 9.59.5 SK 9.59 SK 0.077 Lecture No 6 to 8 Multiple Choice Questions: (Saple questions). Two events say A and B are said to be utually exclusive events, when: a) Two events cannot occur sae tie.*** b) Two events can occur sae tie. c) All of the above d) None of the above
. A four-person coittee is to be fored fro a list of six persons. How any saple points are associated with the experient? a) 60 b) 5*** c) 50 d) 4. A club consists of four ebers. How any saple points are in the saple space when two officers: president and secretary are to be chosen? a) *** b) 6 c) 8 d) 0 4. In a throw ofcoin what is the probability of getting tails. a) b). c) /*** d) 0 5. If a card is drawn fro an ordinary deck of 5 playing cards, find the probability that the card is a red card. a) ½*** b) /4 c) / d) /6 Nuerical Questions (Saple questions) Q: Define Equally Likely Events. Two events A and B are said to be equally likely events, when one event is as likely to occur as the other event. Or when the chance of occurrence of both events are sae. Q: Define Exhaustive Events. Events are said to be exhaustive events when the union of utually exclusive events is the entire saple space. Q: Differentiate between Rule of Perutation and Rule of Cobination.
In rule of perutation, the order is iportant for the selection of r objects fro the set of n distinct objects. In rule of cobination, the order is not iportant for the selection of r objects fro the set of n distinct objects. Q4: In a box contains 4 red, 4 white and 5 green balls. In how any ways can red balls, white balls and green balls be chosen? Nubers of ways of choosing red balls fro 4 red balls 4 4! 4! 4! 4 C 6.!(4 )!!!!! Nubers of ways of choosing white balls fro 4 white balls 4 4! 4! 4! 4 C 6!(4 )!!!!!! Nubers of ways of choosing green balls fro 5 green balls 5 5! 5! 5 4! 5 4 C 0!(5 )!!!!!! By the counting principle black, white and red balls can be chosen 660 60. Q5:Two dice are rolled. a) Make a saple space b) Find the probability that the su of the outcoes is equal to at least 8. c) Find the probability of getting a sae nuber on both die. a) b) S = {(, ), (, ), (, ), (, 4) (, 5), (, 6), (, ), (, ), (, ), (, 4), (, 5), (, 6), (, ), (, ), (, ), (, 4), (, 5), (, 6), (4, ), (4, ), (4, ), (4, 4), (4, 5), (4, 6), (5, ), (5, ), (5, ), (5, 4), (5, 5), (5, 6), (6, ), (6, ), (6, ), (6, 4), (6, 5), (6, 6)} n(s)=6,
Let A be the event that su of at least 8 occurs. A {(,6),(,5),(,6),(4,4),(4,5),(4,6),(5,),(5,4),(5,5),(5,6),(6,),(6,),(6,4),(6,5),(6,6)} n(a)= 5 na ( ) 5 5 PA ( ). ns ( ) 6 b) Let B be the event that sae nuber appears on the die. B {(,),(,),(,),(4,4),(5,5),(6,6)} n(b)= 6 n(b) 6 P(B). ns ( ) 6 6 Lecture No to 7 Multiple Choice Questions: (Saple questions) Multiple Choice Questions: (Saple questions). When a coin is tossed 4 ties, the probability of getting the 4 heads is a) /6**** b) 5/6 c) /8 d) /. For a continuous rando variable, P(=) is: a) 0.5 b) 0**** c) d) Undefined. The second oent-ratio of a particular continuous probability distribution is.44. What does this value tell us about the shape of the distribution? a) Neither very peaked nor flatter. b) Higher peak than the noral distribution.****
c) Flatter peak than the noral distribution. d) Noral distribution. 4. In a bivariate probability distribution of and Y, if E()= and E(Y)=5,then E (4-Y) =? a) 0 b) 5 c) 0 d) **** 5. If the value of p is saller or lesser than 0.5 then the binoial distribution is classified as: a) Negatively skewed b) Positively skewed**** c) Syetric d) None of the above. Nuerical Questions (Saple questions) Q: Let be a rando variable with probability distribution x - 0 f(x) 0.5 0.50 0.0 0.05 0.5 Find E(). x f (x) xf(x) - 0.5-0.5 0 0.50 0 0.0 0.0 0.05 0.0 0.5 0.75 0.55 E( ) xf ( x) 0.55.
Q: Fro the following joint probability distribution, find g (). 4 h(y) 0.0 0.5 0.0 0.0 5 0.0 0.5 g(x) 5, g f y x 0.0 0.0 0.0 0.40 Q: Suppose has a p.d. given by - 0 f(x) c c 6c Deterine the value of c. f() is a probability distribution function, if f( x). f (x) c c 6c c c
Q4: Let be a continuous r. v. with p. d.f. f ( x) 6 x( x), 0 x. 0, otherwise. Check that f ( x) is a p. d. f. The f ( x) will be a p. d. f. if 0 f ( x) dx 6 x( x) dx 6xdx 6x dx 0 0 6x 6x 0 0 0 0 x x () () Hence f(x) is the p.d.f. Q5:
E( ) xg(x)dx x g(x) f (x, y) dy y 0 Let and Y be independent r. v ' s with joint p. d. f. f x y x xy x y 0, elsewhere. (, ), 0,0 Find the E(). x xy dy ( x ) 0 x dy 0 0 6x xy dy xy dy xy xy 0 6x x x x x, for 0 x. 0
E() xg( x) dx x x( x ) 0 x x x dx 0 0 dx dx 4 x x 4 0 0 4 9. 9 8 8 Practice Question Lecture -7 ) If saple of size give ean of 0 then population ean is: 0 0*** 0 40 ) If saples of size with replaceent give variance of.5 then population variance is: 6 7 8 5*** ) For α=0.05, the critical value of z0.05 is equal to.96***.58.645.
4) A rando saple of size n is taken fro the population with unknown ean and known variance. 95% confidence interval for population ean is :.96,.96 **** n n.58,.58 n n S.96,.96 n S n S S.58,.58 n n 5) If population proportion (p) is unknown, the standard error of the saple proportion (P) can be estiated by the forula: np p np P p p *** P P 6) Define an unbiased estiator. Solution An estiator is said to be unbiased if its expected values equals the corresponding population paraeter; otherwise it is said to be biased The statistic is said to be an unbiased estiator of,if E. The saple ean is an unbiased estiator of the population ean i.e. E 7) What is eant by confidence interval? Solution A range of values used to estiate a population paraeter is known as interval estiation or estiation by confidence interval and the interval (a, b) that will include the population paraeter with a high probability (e.g. 0.95 or 0.99) is known as confidence interval. 8) Students fro school A and B are copared on the basis of their scores on an aptitude test. Two rando saples of 90 and 00 students are selected fro school A and B respectively. The saple eans are 76.4 and 8., where as the saple standard deviation is 8. and 7.6 respectively. Establish a 98% confidence interval for the difference in population ean scores between students A and B. Solution
Let the population ean scores of students of Schools A and B be denoted by respectively. Fro the saple data we have and n 90, n 00, 76.4, 8., S 8., S 7.6. With - =0.98, we have /=0.0 and z.. 8. 7.6 4.8..5 or -4.8.68 or 7.48,. 0.0 Thus a 98% confidence interval for is S S z0.0 n n 76.4 8.. 90 00 9) If the standard deviation of the life ties of television tubes is estiated as 00 hours, how large a saple ust we take in order to be 95% confident that the error in the estiated ean life tie will not exceed 0 hours? Solution Here =00, the error, e=0 and z corresponding to the given probability is.96 Z / n = e.96 00.846 00 n 96.04 0 400 Hence the required saple size would be 97 0) Define one tailed and two tailed tests. Solution When the null hypothesis is tested against an alternative hypothesis of a greater than or a less than type the critical region is located on one end and the test is called a one tailed or one sided test. When the null hypothesis is tested against an alternative hypothesis of a not equal to type, the critical is located on both ends and test is called a two tailed or two sided test. Lecture No 8 to 9 Multiple Choice Questions: (Saple questions). If a significance level of 5% is used rather than %, the null hypothesis is: a) Less likely to be rejected
b) More likely to be rejected**** c) Just as likely to be rejected d) None of the above. In hypothesis testing, suppose the critical value is Z >.96. This value shows that the applied test is? a) One tailed test b) Two tailed test**** c) Right tailed test d) Left tailed test. The ter - β is called a) Level of the test b) Power of the test**** c) Size of the test d) Critical region 4. We can apply t-distribution to test the hypothesis, when a) Saple size is sall b) Saple size is large c) Population variance is known d) Population variance is unknown and saple size is sall**** 5. The t-distribution can never becoe narrower than: a) F-distribution b) Standard noral distribution**** c) Chi-square distribution d) Exponential distribution Nuerical Questions (Saple questions) Q: If n=0 and 0.05, then find the value of t. v Since ν = 0 = 9, and = 5%, therefore, the right-tail area is ½% and hence (using the t- table) we obtain t =.8 0.05(9)
Q: Calculate the pooled proportion Saple I: 0, n 00 Saple II: 0, n 5 P c for the given data. The pooled proportion P c is given by Pˆ c n n 0 0 50 0. 00 5 5 Q: If 50, n 60and p0 0.50, then find the z-test statistic for proportion. Test statistic: Z (where n p 0 p 0 0 0 0 ½ denotes the continuity correction) Here np = 600.50 =5. and = 50 n p Hence < np so use + ½ Z 50 5 60(0.50) 0.50 64.5 64.5 Z 5.9. 57.5.55 Q4: A saple of 00 electric light bulbs of type I showed a ean lifetie of 90 hours and a standard deviation of 90 hours. A saple of 75 bulbs of type II showed a ean of 0 hours with a standard deviation of 0 hours. Is there a difference between the ean lifeties of two types at a significant level of 5%? Step : H : 0or ( ) 0 H : 0or ( ) A
Step : Level of significance = 5%. Steps : ( ) ( ) z n n Step 4: 90 0 0 40 z.4. 90 0 6.5 00 75 Step 5: Critical Region: Reject z cal z H 0, z.4 z 0.05.4.96 we reject H if 0.05 0. Step 6: Conclusion: Since the calculated value of z =.4 is greater than the critical value of z =.96, so we reject H. 0 Or also we conclude that there is a difference between the two types of bulbs. Q5: Rando saples of 00 bolts anufactured by achine A and 00 bolts anufactured by achine B showed 9 and 5 defective bolts respectively. Test the hypothesis that achine B is perforing better than achine A. Use a 0.05 level of Significance.
Step : H : p p 0or (p p ) 0 H : p p 0or (p p ) c A Step : Level of significance = 5%. Steps : P P z pq c c n n where p n n Step 4: The proportions of non-defective bolts in the two saples are P 8 n 00 P 95 n 00 p c n n 0.905 0.905 8 95 76 0.9. 00 00 00 0.905 0.95 z 0.90.08 00 00 0.045.5. 0.0 Step 5: Critical Region: Reject H, if 0
z < z cal -.5< z 0.05.5.645 we don ' t reject H. 0 Step 6: Conclusion: Since the calculated value of z=-.5 is greater than the critical value of so we don t reject A. z =-.645, H 0.Or also we conclude that achine B is not perforing better than achine Lecture 4 & 4 The F-distribution always ranges fro: 0 to 0 to - - to + ***** 0 to + When we want to test the equality of two variances we usually use: F-test**** Chi-square test ANOVA Z-test If s =. and s=.5 then the value of F-Test statistic is:.5****.45..67
Which is the basic assuption is required for F-Distribution? The two rando saples are independently and randoly selected**** The two rando saples are dependent. The two saples are not necessary to follow noral distribution. None of these. If F has an F-distribution with v and v degree of freedo, then /F: Has F distribution with v and v degree of freedo. Has F distribution with v and v degree of freedo***** Has t-distribution with v degree of freedo. Has z distribution. Analysis of Variance (ANOVA) is a test for equality of: Variances***** Means Proportions Only two paraeters When testing for independence in a contingency table with rows and 4 coluns, there are error degrees of freedo. 5 6*** 7 In one way ANOVA, the degree of freedo of Mean square between is the Nuber of groups inus *****
Nuber of groups ultiplied by, where n is the saple size Nuber o f groups Total nuber of observations inus The basic entity or unit on which the experient is perfored, is called Experiental unit **** Treatent unit Factor Levels of factor A ter referring to the aount of balancing, blocking and grouping of the experiental units is Local control **** Spurious effect Systeatic error Extraneous factor Q: Find Coefficient of association AB 55, 75, A 45, B 5
AB 55, 75, A 45, B 5 Q Q AB A B AB A B 5575 455 5575 455 45 565 Q 45 565 500 Q 9750 Q 0.5 This indicates negative association between A and B. Q: Test the independence by a siple approach between intelligences of fathers and sons. Sons Intelligent Not intelligent Total Intelligent 00 00 00 Not intelligent 00 400 500 Total 400 600 000 Let A denotes intelligent fathers and B denotes Intelligent Sons Then
AB 00, ( A)( B) (400)(500) 00 n 000 ( A)( B) ( AB) n 00 00 Thus there is positive association between intelligent fathers and intelligent sons. It eans that intelligent fathers have intelligent sons. Q:Fro the following table, test the hypothesis that the flower color is independent of flatness of leaf. Use α=0.05 Flat Leaves Lean Leaves Total White Flowers 99 6 5 Red Flower 0 5 5 Total 9 4 60 : Null and alternative Hypothesis: : the flower color is independent of faltness of leaves. H o H : Level of significance: : the flower color is not independent of faltness of leaves. 0.05 : Test Statistic 4: Critical region: f f o e f e 0.05().84 5. Calculations:
f o f e f o f e f o f e f f o f e e 99 00.4 -.4.988 0.00 0 8.59.4.988 0.07 6 4.59.4.988 0.057 5 6.4 -.4.988 0.0 60 60 0 =0.494 6: Conclusion: The calculated value of 0.494 lies in acceptance region. Thus Ho: The flower color is independent of flatness of leaves is accepted. It eans that the color of flowers have no relation with flatness of leaves. Q4: A rando saple of 00 people each fro cities gave weekly incoe figures as suarized below, with ij the incoe of the ith person fro the jth city. City Saple Size ij ij A 00 0000 05600 B 00 00 596000 C 00 9800 00000 Set up an Analysis of Variance table. State and test the standard hypothesis fro the table: : Null and alternative Hypothesis:
: H o H : Level of significance: : atleast two of the eans are not equal. 0.05 : Test Statistic 4: Critical region: Treatent Mean square F Error Mean Square v k, v n k 00 97 F.0 5. Calculations: A B C Total N 00 00 00 00 ij 0000 00 9800 000 05600 596000 00000 7500 ij ij n j 000000 54400 960400 4800
CF. n 000 CF. 00 CF. 0. j ij Total Su of Squares ij. T SS 7500 0. T SS 548966.67 Treatent SS TrSS 4800 0. TrSS 466.67 n j ij C. F C F ESS TSS TrSS ESS 548966.67 466.67 ESS 57500 Source of variation ANOVA TABLE d.f Su of Squares Mean squares F TrSS k-=-= 466.67 57.4 F=.7 Error SS n-k=00-=97 57500 809.76 Total n-=00-=99 5487966.67 6: Conclusion: The calculated value of F=.7 fallsin rejection region. So we reject our null hypothesis, and conclude that at least two treatent eans are not equal.
Q5: Coplete an ANOVA table. Source of Variation D.f Su of Square Mean Square Treatent SS ---- 0 Block SS ---- 9 ----- Error SS ---- ---- Total 9 ----- Source of Variation D.f Su of Square Mean Square Treatent SS k-=4-= 0 0 Block SS r-=5-=4 9 98 Error SS (r-)(k-)= 0 Total 9 94