Question Paper Quantitative Methods-II (132): July 2006
	· Answer all questions. · Marks are indicated against each question.

1.

The following data relate to marketing expenditure in Rs.lac and the corresponding sales of a product in Rs.crores. Estimate the marketing expenditure to attain a sales target of Rs.40 crores.

Marketing expenditure	10	12	15	20	23
Product sales	14	17	23	21	25

(a) 43 lac (b) 33 lac (c) 37 lac (d) 27 lac (e) 53 lac.

(2 marks)

< Answer >

2.

The marks obtained by the 10 students in their graduation and the MBA entrance tests were found as given below. From these paired data, find the coefficient of correlation between two sets of marks.

Graduation	50	52	55	60	62	65	65	66	70	75
Entrance test	52	50	57	65	65	62	65	65	71	78

(a) 0.852 (b) 0.954 (c) 0.798 (d) 1.562 (e) 0.658.

(1 mark)

< Answer >

3.

From the given pair of observation (X, Y), we can estimate the value of Y in two ways as and , then the variation explained by the model is given by

(a) (b) (c)

(d) (e)

(1 mark)

< Answer >

4.

We construct a prediction interval around which would contain the actual Y, then would be the interval, where t is the appropriate t value with n–2 degrees of freedom, then the sample size n is?

(a) n > 30 (b) n < 30 (c) n 30 (d) n = 30 (e) n30.

(1 mark)

< Answer >

5.

Let b is an estimator of B. The random variable b is a normal distribution with mean B and the estimate of V(b) is given by. Therefore follows:

(a) t-distribution with n – 2 degrees of freedom

(b) F-distribution with n +1 degrees of freedom

(c) -distribution with n₁ + n₂ degrees of freedom

(d) F-distribution with n degrees of freedom

(e) t-distribution with n –1 degrees of freedom.

(1 mark)

< Answer >

6.

For which of the following distributions, the z-values and the observed values are same?

(a) Binomial distribution

(b) Standard normal distribution

(c) Normal distribution

(d) Lognormal distribution

(e) t-distribution.

(1 mark)

< Answer >

7.

The time that it takes to find a taxi when leaving a restaurant follows a left skewed distribution with a mean of 20 minutes and a variance of 100 minutes. If 64 restaurant patrons are randomly sampled and the average time that it takes for them to find a taxi is calculated, then what is the probability that the sample mean will be between 18 and 23 minutes?

(a) 0.4918 (b) 0.4452 (c) 0.9918 (d) 0.9152 (e) 0.9370.

(2 marks)

< Answer >

8.

An irregular six-faced die is thrown and the expectation that in 10 throws, it will give five even numbers is twice the expectation that it will give four even numbers. Of what probability the items in 10,000 sets of 10 throws each, would you expect it to give no even number?

(a) 1.000 (b) 0.549 (c) 0.257 (d) 0.159 (e) 0.413.

(1 mark)

< Answer >

9.

A manufacturer of cotter pins knows that 5% of his product is defective. If he sells cotter pins in boxes of 100 and guarantees that not more than 10 pins will be defective, what is the approximate probability that a box will fail to meet the guaranteed quality?

(a) (b) (c)

(d) (e) .

(1 mark)

< Answer >

10.

Which of the following finds the large applications in Statistical Quality Control in industry for setting control limits?

(a) F-distribution (b) t-distribution

(c) -distribution (d) Normal distribution (e) Both (a) and (c) above.

(1 mark)

< Answer >

11.

According to the factor reversal test if we interchange the price and quantity terms in an index number then the product of the original index number and the index number obtained by reversing the factors should be equal to

(a) One (b) One hundred (c) The value index

(d) The Fisher’s ideal index (e) The chain index number.

(1 mark)

< Answer >

12.

In a test of ANOVA the following are found:

Estimated population variance based on the variance among sample means = 36.25

and F-ratio is 2.56.

What is the estimated population variance based on the variance within the samples?

(a) 15.08 (b) 14.16 (c) 15.80 (d) 13.40 (e) 20.52.

(1 mark)

< Answer >

13.

A multiple regression relationship contains three independent variables. The standard error of estimate is 9.54 and = 625.48. The number of observations are

(a) 18 (b) 14 (c) 16 (d) 11 (e) 20.

(1 mark)

< Answer >

14.

The linear trend estimating equation for a time series is given below:

= 120 + 2.5x

where x = Year – 2001 and is the estimated value of the dependent variable Y.

The observed value of Y for the year 2005 is 146.

What is the relative cyclical residual for the year 2005?

(a) – 4.25 (b) 3.88 (c) 12.30 (d) 95.75 (e) 106.12.

(1 mark)

< Answer >

15.

Which of the following is not true about the statistical process control?

(a) Quality standards in manufacturing process and the service industry can be achieved through the application of statistical methods

(b) The variability present in the manufacturing process can be minimized to the extent possible

(c) The variations in the specifications of the product due to wearing of the machine is a systematic or non-random variation

(d) The variation arising out of measurement errors is a random variation

(e) If the process is out of control, then random variation will be observed.

(1 mark)

< Answer >

16.

In the test involving ANOVA the F-statistic is calculated on the basis of

(a) The variance of the largest sample

(b) The mean of the largest sample

(c) The standard deviation of the smallest sample

(d) The estimated population mean based on the mean among the sample variances and the estimated population variance based on the variance within the samples

(e) The estimated population variance based on the variance among the sample means and the estimated population variance based on the variance within the samples.

(1 mark)

< Answer >

17.

Which of the following is/are the example(s) of Poisson distribution?

I. The number of deaths in a city due to suicides.

II. The number of defective items in a box of 10 items.

III. The number of plane accidents per week.

(a) Only (I) above (b) Only (II) above

(c) Both (II)and (III) above (d) Both (I) and (III) above

(e) All (I), (II) and (III) above.

(1 mark)

< Answer >

18.

A Poisson distribution has a double mode at x =1 and x = 2. What is the probability that x will have one or the other of these two values?

(a) 0.542 (b) 0.569 (c) 0.4679 (d) 0.1326 (e) 0.4613.

(1 mark)

< Answer >

19.

It is known that the number of heavy trucks arriving at a railway station follows the Poisson distribution. If the average number of truck arrivals during a specified period of half an hour is 2, what is the probability that during a given half an hour no heavy truck will arrive?

(a) 0.1973 (b) 0.6482 (c) 0.1353 (d) 0.2528 (e) 0.3964.

(1 mark)

< Answer >

20.

The seasonal index sales of a company for four quarters of a year are 200, 180, 160 and 260 respectively. If the total sales in the first quarter are worth Rs.50,000, estimate the worth of sales expected during the second quarter of the year.

(a) Rs.25,000 (b) Rs.30,000 (c) Rs.45,000

(d) Rs.25,500 (e) Rs.55,000.

(1 mark)

< Answer >

21.

Which of the following methods is used to isolate the cyclical variation component from the time series?

(a) Residual method (b) Linear regression analysis

(c) Non-linear regression analysis (d) Ratio to Moving average method

(e) Multiple regression analysis.

(1 mark)

< Answer >

22.

The movement of the dependent variable above and below the secular trend line over periods longer than one year is known as

(a) Seasonal variation (b) Secular variation (c) Irregular variation

(d) Cyclical variation (e) Regular variation.

(1 mark)

< Answer >

23.

Which of the following is true, if the value of adjusting constant is greater than one, in the context of analyzing seasonal variations?

(a) Unadjusted means are more than seasonal indices

(b) Seasonal indices are less than zero

(c) Seasonal indices are less than 100

(d) Seasonal indices are greater than 100

(e) Seasonal indices are more than unadjusted means.

(1 mark)

< Answer >

24.

Variable Y increases as variable X increases. Variable X cannot explain 36 percent of the variations in Y. What is the coefficient of correlation between X and Y?

(a) 0.80 (b) 0.20 (c) 0.60 (d) – 0.64 (e) – 0.08.

(1 mark)

< Answer >

25.

If a deseasonalized value in a time series is 0.79 and the actual data corresponding to the deseasonalized value is 2, the seasonal index is

(a) 0.395 (b) 2.5316 (c) 253.16 (d) 2.1356 (e) 231.56.

(1 mark)

< Answer >

26.

Which of the following is a technique for studying seasonal variations?

(a) Regression analysis

(b) Residual method

(c) Hypothesis testing

(d) Ratio to moving average method

(e) Forecasting.

(1 mark)

< Answer >

27.

Which of the following is not required for making a point estimate of population mean?

(a) Observations in the sample

(b) Sum of the observations in the sample

(c) Sample size

(d) Sample standard deviation

(e) Sample mean.

(1 mark)

< Answer >

28.

In a test of hypothesis for a population mean it is found that the test statistic is equal to 1.60. The sample mean is 48 and the standard error of mean is 5. What is the value of the population mean according to the null hypothesis?

(a) 8 (b) 88 (c) 56 (d) 40 (e) 43.

(1 mark)

< Answer >

29.

The following details are available with regard to a random sample collected from a finite population:

Population size = 640

Finite population correction factor = 0.952

What is the sampling fraction?

(a) 0.0951 (b) 0.9063 (c) 0.0016 (d) 0.01 (e) 0.91.

(1 mark)

< Answer >

30.

Considering the process of collecting a simple random sample of a specific size from a population be an experiment, which of the following will not be a random variable?

(a) Sample mean (b) Sample variance

(c) Sample median (d) Sample range (e) Sample size.

(1 mark)

< Answer >

31.

A chi-square test of goodness of fit is used to

(a) Test the hypothesis about a population mean

(b) Test the hypothesis about a population proportion

(c) Test the hypothesis about difference between two population means

(d) Test the hypothesis about difference between two population proportions

(e) Test whether the data from a population follows a hypothesized probability distribution.

(1 mark)

< Answer >

32.

When estimating the population mean from a sample taken from a normal population with unknown variance, the t-distribution may be used with number of degrees of freedom equal to

(a) Sample size (b) Sample size + 1 (c) Sample size – 1

(d) 0 (e) 1.

(1 mark)

< Answer >

33.

The following regression relationship between two variables X and Y, has been obtained:

= 1236 – 104 X

Where X is the independent variable and Y is the dependent variable.

The following details are also available:

SY² = 19,00,400 ; SXY = 18,100 and = 612

Number of observations = 5

What is the approximate 90 percent prediction interval for Y if X = 7.50?

(a) 1,236 ± 456 (b) 456 ± 34.38 (c) 1,236 ± 104

(d) 456 ± 104 (e) 1,236 ± 34.38.

(2 marks)

< Answer >

34.

Which of the following statements is/are not true with respect to time series analysis?

I. Time-series analysis is used to detect patterns of change in statistical information over regular intervals of time.

II. Secular trends represent the long-term direction of a time series.

III. Of the four types of variations, cyclical variation is the most difficult to predict.

(a) Only (I) above (b) Only (II) above

(c) Only (III) above (d) Both (I) and (II) above

(e) Both (II) and (III) above.

(1 mark)

< Answer >

35.

For a paired data set of two variables X and Y, the covariance between X and Y is –242.5 and the standard deviation of the variable X is 12. The means of the observations of X and Y in the given data set are 6.2 and 4.1 respectively. If a regression line is plotted for estimating the value of Y for a given value of X, then what is the intercept of the regression line?

(a) –1.684 (b) 1.684 (c) –14.54 (d) – 6.34 (e) 14.54.

(1 mark)

< Answer >

36.

A company sells cornflakes in 1 kg boxes. It is aware that all the boxes do not have the same weight. The standard deviation of weights based on past data is 75 gms. A random sample of 50 boxes is drawn from a day’s production and the sample mean is 950 gms. Then which of the following would be the correct conclusion at 5% level of significance?

(a) It is in significantly less than 1 kg at 5% level of significance

(b) It is significantly less than 1 kg at 5% level of significance

(c) It is significantly equal to 1 kg at 5% level of significance

(d) It is in significantly equal to 1 kg at 5% level of significance

(e) It is significantly greater than 1 kg at 5% level of significance.

(2 marks)

< Answer >

37.

To estimate the accuracy of values predicted by a bivariate regression equation, it is useful to calculate the

(a) Mean of the values of the independent variable

(b) Mean of the observed values of the dependent variable

(c) Standard error of the estimate

(d) Mean of the estimated values of the dependent variable

(e) Standard deviation of the observed values of the dependent variable.

(1 mark)

< Answer >

38.

If the relative cyclical residual for an year is zero then the percent of trend for that year is

(a) 100 (b) 0 (c) 1 (d) 50 (e) 10.

(1 mark)

< Answer >

39.

A new shearing machine is set to cut off a piece of steel from a long bar. For various reasons the machine at some times cuts off a piece that is too long or too short. These unacceptable pieces are automatically dropped in a box and the operator of the shearing machine must count these defectives after every 100 pieces sheared off. The record after the first day of operation is:

Number of sheared off pieces	100	100	100	100	100
Number of defectives	5	6	7	4	8

Set out the lower control limit for the p chart.

(a) 0.012 (b) 0.011 (c) 0.001 (d) 0.000 (e) – 0.012.

(2 marks)

< Answer >

40.

Ten samples each of size 5 are drawn at regular intervals from a manufacturing process. The sample means and ranges are given below:

Sample No.	1	2	3	4	5	6	7	8	9	10
Sample Mean	49	45	48	53	39	47	46	39	51	45
Sample Range	7	5	7	9	5	8	8	6	7	6

Calculate the value of lower control limit for chart.

(a) 42.1825 (b) 42.2764 (c) 42.5820 (d) 42.6218 (e) 42.6416.

(1 mark)

< Answer >

41.

A sample of 15 people is taken at random from a group in which 40% favor a particular political stand. What is the probability that at least 4 individuals in the sample favor this political stand?

(a) 0.0047 (b) 0.0219 (c) 0.0634 (d) 0.0905 (e) 0.9095.

(2 marks)

< Answer >

42.

A quality control assistant was asked to obtain the control limits for R-chart. He randomly picked up 10 samples of size 10 each from the assembly line and generated the following range values of a given product quality characteristic.

Range:

2.5

3.7

4.6

3.8

4.0

3.9

4.2

3.8

3.9

4.3

Compute the upper control limit for the R-chart.

(a) 5.02 (b) 6.88 (c) 4.36 (d) 9.21 (e) 7.12.

(1 mark)

< Answer >

43.

A small sample has been taken from a population and the sample mean has been found to be 62. The upper limit of a 95 percent confidence interval for population mean is 81.60. The population variance is known to be 2,400. What is the sample size?

(a) 30 (b) 24 (c) 36 (d) 64 (e) 16.

(2 marks)

< Answer >

44.

From the following data obtain the regression equation of X on Y with Y as the independent variable.

X	6	2	10	4	8
Y	9	11	5	8	7

(a) = 16.4 – 1.3Y (b) = 16.4 + 1.3Y (c) = 1.3 – 16.4 Y

(d) = 1.3 + 16.4 Y (e) = – 1.3 +16.4 Y.

(1 mark)

< Answer >

45.

Let the sample of a size is considered as a lot. When the lot size varies, then which of the following charts is better to construct?

I. R charts. II. P charts. III. charts.

(a) Only (I) above (b) Only (II) above

(c) Only (III) above (d) Both (I) and (II) above

(e) Both (I) and (III) above.

(1 mark)

< Answer >

46.

When the lower control limit for R chart is zero, it means

(a) The control chart has no limit

(b) The control chart is under control

(c) The control chart is out of control

(d) The control chart limits to zero

(e) The R chart is not suitable.

(1 mark)

< Answer >

47.

A sample analysis of examination results of 200 MSc’s is made. It was found that 65 students got first division, 61 secured a second division, 42 secured a third division and the rest were failed. An analyst wants to test whether these figures are commensurate with the general examination results which is in the ratio of 3 : 2 : 4 : 1, for the various categories (in the same order as above) respectively. Which of the following conclusions can be drawn from the given data, at a significance level of 5%?

(a) There is no difference between the MSc results and general examination results

(b) There is a significant difference between the MSc results and general examination results

(c) The number of first divisions in MSc results are more than the general examinations results

(d) The number of failed students are more in the MSc results than the general examinations results

(e) No conclusion can be drawn, as the data is insufficient.

(2 marks)

< Answer >

48.

The given information shows the distribution of digits in numbers chosen at random from a telephone directory.

Number	0	1	2	3	4
Frequency	1107	853	997	966	933
Number	5	6	7	8	9
Frequency	1075	964	972	1026	1107

An eager student wants to test whether the numbers may be taken to occur equally frequently in the directory. Which of the following conclusions can be drawn on the basis of the appropriate statistical test at a significance level of 5%?

(a) The digits are uniformly distributed in the directory

(b) The digits are not uniformly distributed in the directory

(c) The digits 1 and 7 occur more than any other digits in the directory

(d) The digit 9 occurs least number of times in the directory

(e) The digits 0, 1, 4 and 6 occur more than other digits in the directory.

(2 marks)

< Answer >

49.

The regional manager of a retail chain believes that the number of customers entering each of the five stores in his division is same in any week. In a given week the managers of the five stores report the following number of customers in their stores:

2568, 1698, 3695, 1596, 5874.

The regional manager wants to test his belief on the basis of above data. What is the critical value for the above test at a significance level of 1%?

(a) 7.779 (b) 9.210 (c) 11.345 (d) 13.277 (e) 15.086.

(1 mark)

< Answer >

50.

A welfare society claiming to be promoters of sex education sought the views of parents from the three Indian cities of Srinagar, Ahmedabad and Kolkatta on introducing sex education at the school level. The views of 80 parents selected at random from each of the three states were found as under:

	Srinagar	Ahmedabad	Kolkatta
Support	50	20	45
Reject	30	60	35

Do the sample data provide enough evidence to support the view that the proportion of parents in favour of introducing sex education in schools is the same in all the three states? Use level of significance.

(a) H₁ is accepted, thus the proportion of parents in favour of introducing sex education in schools is the same in all the three states

(b) H₀ is rejected, thus proportion of parents in favour of introducing sex education in schools is not same in all the three states

(c) H₁ is rejected, thus the proportion of parents in favour of introducing sex education in schools is not the same in all the three states

(d) H₀ is accepted, thus proportion of parents in favour of introducing sex education in schools is not same in all the three states

(e) The result leads to either type I error or type II error.

(2 marks)

< Answer >

51.

Which of the following is not a disadvantage of simulation?

(a) These are very expensive and take a long time to develop

(b) It does not generate optimal solutions to the problems

(c) Each application of simulation is ad hoc to a great extent

(d) Simulation model does not produce answers by itself, whereas the user has to provide

(e) It is suitable to analyze large and complex real-life problems, which cannot be solved by usual quantitative methods.

(1 mark)

< Answer >

52.

A sample of 64 farm laborers engaged in paddy harvesting shows an average monthly wage rate of Rs.200 with a standard deviation of Rs.9. Using 5% level of significance verify if the sample result indicates that their current average monthly wage rate is higher than Rs.198.

(a) Current wage is higher than Rs.198

(b) Current wage is lower than Rs.198

(c) Current wage is not equal to Rs.198

(d) Current wage is equal to Rs.198

(e) Data insufficient to decide.

(1 mark)

< Answer >

53.

A proportion of 20% in a sample of 100 persons is found to exhibit a particular characteristic. The estimated standard error of proportion is

(a) 0.04 (b) 0.08 (c) 0.16 (d) 0.20 (e) 0.80.

(1 mark)

< Answer >

54.

A multiple regression relationship contains two independent variables. The standard error of estimate is 4.8. Error sum of squares = 576. Then find the number of data points?

(a) 24 (b) 25 (c) 26 (d) 27 (e) 28.

(2 marks)

< Answer >

55.

From an association consisting of 540 individuals, a sample of 60 individuals is taken. From this sample, the average age of the individuals is found to be 31 years and the standard deviation is found to be 6.84 years.

A 95 percent confidence interval for the mean age of the individuals in the association has to be constructed. The lower and upper confidence limits of the confidence interval are

(a) 24.16 years and 37.84 years respectively

(b) 28 years and 33 years respectively

(c) 29.37 years and 32.63 years respectively

(d) 30.17 years and 31.83 years respectively

(e) 17.6 years and 44.41 years respectively.

(2 marks)

< Answer >

56.

As the sample size increases, the interval estimate for the population mean at a given confidence level will

(a) Increase in width

(b) Remain the same

(c) Increase initially and then decrease in width

(d) Decrease in width

(e) Double in width for increase in sample size by one observation.

(1 mark)

< Answer >

57.

The following information are provided with regard to a sample:

n = 8

and

What is the point estimate of population standard deviation?

(a) 3.209 (b) 4.367 (c) 2.828 (d) 8.268 (e) 9.625.

(1 mark)

< Answer >

58.

A sample has been taken from a population and the sample proportion has been found to be 0.65. The estimated standard error of proportion is found to be 0.17. What is the sample size?

(a) 4.58 (b) 7.87 (c) 4.00 (d) 6.20 (e) 25.85.

(1 mark)

< Answer >

59.

A computer store purchases untested microchips for computers, from a wholesaler. It has to estimate the proportion of faulty microchips in the consignments supplied by the wholesaler. From a large consignment sent by the wholesaler a sample of 150 microchips were tested and 39 microchips were found to be faulty.

A 98 percent confidence interval has to be constructed for the true population proportion of faulty microchips. The lower and upper confidence limits of the confidence interval are

(a) 0.231 percent and 1.029 percent respectively

(b) 0.1766 percent and 0.3433 percent respectively

(c) 0.788 percent and 12.012 percent respectively

(d) 0.8788 percent and 0.9142 percent respectively

(e) 0.8508 percent and 9.492 percent respectively.

(1 mark)

< Answer >

60.

The following details are available with regard to a hypothesis test on population mean:

H₀:

H₁:

n = 36 ; ;

Significance level = 0.05

It is later known that the true population mean is 10. Which of the following can be said with regard to the test?

(a) There is insufficient information for doing the test

(b) The t-distribution should be used

(c) The test does not lead to either type I or type II error

(d) The test leads to a type I error

(e) The test leads to a type II error.

(2 marks)

< Answer >

61.

Gopal Fabricators purchased a used drilling machine from Rohan Automobiles Ltd. It was informed to Gopal Fabricators that there is a probability of 5% of the output from the machine will be defective. It is expected that this rate of defective output will continue in future.

Find out the probability that maximum three out of every ten jobs done on that machine will be defective.

(a) 0.5949 (b) 0.6959 (c) 0.7969 (d) 0.8979 (e) 0.9989.

(2 marks)

< Answer >

62.

Which of the following is not correct with regard to a value index?

(a) It measures changes in net monetary worth

(b) It combines price changes and quantity changes

(c) It is not as useful as price indices or quantity indices

(d) It cannot distinguish between the price changes and quantity changes

(e) An increase in the value index indicates with certainty that the prices have increased.

(1 mark)

< Answer >

63.

The following data pertain to three commodities:

Commodity	Price in 2005 (Rs./kg)	Price in 2003 (Rs./kg)
Rice	11.50	9.50
Wheat	13.50	8.50
Pulses	26	20

The base year is 2003. The Unweighted aggregates price index for the year 2005 is approximately

(a) 115.62 (b) 125.45 (c) 134.21 (d) 250.55 (e) 410.00.

(1 mark)

< Answer >

64.

Which of the following is/are the use(s) of index numbers?

I. Establishing trends.

II. Policy making regarding wage payment in an organization.

III. Determining the purchasing power of the currency.

IV. Deflating the time series data.

(a) Only (I) above (b) Only (II) above

(c) Both (II) and (IV) above (d) Both (III) and (IV) above

(e) All (I), (II), (III) and (IV) above.

(1 mark)

< Answer >

65.

The prices and quantities of some commodities consumed by a family in the years 2002 and 2005 are given below:

Commodity	Year
	2002		2005
	Price (Rs./kg)	Quantity (kg)	Price (Rs./kg)	Quantity (kg)
Coffee	600	10	800	8
Amul	14.50	150	15.00	120
Pulses	90	15	120	12
Grams	13.50	500	15.50	480
Wheat	12.00	300	13.50	250

What is the price index by Laspeyre’s method for the basket of commodities consumed by the family, for the year 2005, considering 2002 as the base year?

(a) 183.33 (b) 118.50 (c) 102.92 (d) 112.07 (e) 120.00.

(2 marks)

< Answer >

66.

The following data are collected from a fruit merchant:

Commodity	Year
	2000		2005
	Price	Quantity	Price	Quantity
Mangoes	Rs. 10 per kg.	500 kgs.	Rs. 15 per kg.	400 kgs.
Bananas	Rs. 8 per doz.	600 doz.	Rs. 12 per doz.	500 doz.
Oranges	Rs. 30 per kg.	350 kgs.	Rs. 50 per kg.	300 kgs.
Pineapple	Rs. 8 per piece	450 pieces	Rs. 12 per piece	400 pieces
Apple	Rs. 40 per kg.	300 kgs.	Rs. 70 per kg.	200 kgs.

What is the price index by Marshall-Edgeworth method for the fruits traded by the merchant, for the year 2005, considering 2000 as the base year?

(a) 160.38 (b) 181.82 (c) 162.87 (d) 165.63 (e) 122.22.

(2 marks)

< Answer >

67.

The weighted average of price relatives using base values as weights is same as the

(a) Unweighted aggregates price index

(b) Unweighted aggregates quantity index

(c) Laspeyres price index

(d) Laspeyres quantity index

(e) Paasche’s price index.

(1 mark)

< Answer >

68.

The following index values pertain to a year:

Fisher’s ideal price index = 112.5

Laspeyres price index = 101.9

Paasche’s price index for the year is approximately

(a) 172.45 (b) 167.37 (c) 142.50 (d) 124.20 (e) 153.74.

(1 mark)

< Answer >

69.

Which of the following is true with regard to Fisher’s ideal price index?

(a) It does not consider the base year prices

(b) It does not consider the base year quantities

(c) It does not consider the current year prices

(d) It does not consider the current quantities

(e) It is the geometric mean of the Laspeyre’s and Paasche’s price indices.

(1 mark)

< Answer >

70.

Calculate the value of the test statistic for the hypothesis test on population standard deviation, from the following data:

Sample size = 32

Hypothesized value of the population standard deviation = 6

Sample standard deviation = 0.19

(a) 0.0310 (b) 0.1650 (c) 0.1765 (d) 0.1860 (e) 0.1900.

(1 mark)

< Answer >

71.

If the sum of squares due to error is 86 and the total sum of squares is 186, then what percentage of the changes in the dependent variable is explained by the changes in the independent variable?

(a) 72.17% (b) 69.22% (c) 53.76% (d) 47.92% (e) 27.12% .

(1 mark)

< Answer >

72.

On conducting a test of ANOVA on 3 samples of different sizes, as 11, 9 and 10. The number of degrees of freedom for the denominator of the F-ratio is

(a) 14 (b) 15 (c) 27 (d) 17 (e) 18.

(1 mark)

< Answer >

73.

Which of the following distributions can be identified by a pair of degrees of freedom?

(a) t-distribution (b) Binomial distribution

(c) Normal distribution (d) F-distribution

(e) Chi square distribution.

(1 mark)

< Answer >

74.

Of the following which are the instances, of where the Poisson distribution may be successfully applicable:

I. Number of suicides reported in a particular city.

II. The number of defective material in packing, manufactured by a good concern.

III. Number of faulty blades in a packet of 100.

IV. The emission of radioactive particles.

(a) Only (I) above (b) Only (IV) above

(c) Both (I) and (II) above (d) Both (III) and (IV) above

(e) All (I), (II), (III) and (IV) above.

(1 mark)

< Answer >

75.

When the result of taking that the decision is uncertain, then which of the following figure is to be drawn so that it represents the above situation?

(a) Circle (b) Rectangle (c) Oval (d) Square (e) Crossed oval.

(1 mark)

< Answer >

76.

There are 25 students in a class, which consists of 14 boys and 11 girls. 5 students of the class were absent on a particular day. What is the probability that at least 4 boys were absent?

(a) 0.20725 (b) 0.03768 (c) 0.24493 (d) 0.75507 (e) 0.79275.

(2 marks)

< Answer >

77.

An automobile manufacturer is planning to launch a new model of scooter. In order to find out the opinion of the prospective customers about the proposed model, the firm has taken a random sample from the audience, which attended a preview of the proposed model. The results obtained are given below:

Opinion of the persons	Age group (in years)
	Below 20	20 - 30	30 - 40	40 - 50	Total
Persons who liked the model	140	80	50	30	300
Persons who disliked the model	60	50	30	60	200
Total	200	130	80	90	500

A chi-square test has to be performed to find out whether the opinion of the persons and the age groups are independent. At a 5 percent significance level what is the conclusion?

I. The chi-square value for the sample data is greater than the critical value.

II. The opinion of the persons and the age groups are independent.

III. The opinion of the persons and the age groups are dependent.
IV. The data is insufficient to perform a chi-square test.

(a) Only (I) above (b) Only (II) above (c) Only (IV) above
(d) Both (I) and (II) above (e) Both (I) and (III) above.

(3 marks)

< Answer >

78.

Phizer Tech. Ltd. used three different periods to train its new employees. The number of units of output produced per hour by different employees trained by one of the three training periods are given below:

Period A	-	-	55	44	45	50
Period B	64	-	52	56	44	48
Period C	46	42	48	45	57	42

You are required to test, whether the three different training periods lead to different levels of productivity. The significance level of the test is 5 percent.

Then the estimate of the population variance on the basis of the variance among the sample means is

(a) 45.25 (b) 48.36 (c) 52.58 (d) 55.65 (e) 57.25.

(2 marks)

< Answer >

79.

In a Bernoulli process, the probability of success is 0.50. If the number of trials is 13 then the probability of getting 6 successes is equal to the probability of getting

(a) 4 successes (b) 6 successes (c) 7 successes

(d) 10 successes (e) 13 successes.

(2 marks)

< Answer >

Suggested Answers
Quantitative Methods-II (132): July 2006

:1.

Answer : (c)

Reason : Let the marketing expenditure be denoted by X and product sales by Y, with regression equation of X on Y as

X = a + bY

The necessary computations for obtaining the regression equation are as follows:

X	Y	XY	X^2	Y^2
10	14	140	100	196
12	17	204	144	289
15	23	345	225	529
20	21	420	400	441
23	25	575	529	625
= 80	= 100	= 1684	=1398	2080

Substituting for the regression coefficient of X on Y,

b = =

Using the required values a can be obtained as

a = – b= 80/5 – (1.05)(100/5) = –5.0

Now substituting a and b values in the regression equation, we have

X = a + bY X = (–5.0) + (1.05)Y

When Y = 40(Rs crores), the corresponding X value is

X = (–5) + (1.05)40 = 37 (Rs lac)

That is, to achieve a sales target of Rs 40 crore, there is a need to spend Rs 37 lac on marketing.

< TOP >

2.

Answer : (b)

Reason : Let the marks in graduation be denoted as X and those in the entrance test as Y. Then the necessary computations are as under:

X	Y	XY	X^2	Y^2
50	52	2600	2500	2704
52	50	2600	2704	2500
55	57	3135	3025	3249
60	65	3900	3600	4225
62	65	4030	3844	4225
65	62	4030	4225	3844
65	65	4225	4225	4225
66	65	4290	4356	4225
70	71	4970	4900	5041
75	78	5850	5625	6084
= 620	= 630	= 39630	=39004	40332

Substituting for the coefficient of correlation between X and Y,

=

< TOP >

3.

Answer : (b)

Reason : The variation explained by the model is given by . So the option (b) is true.

< TOP >

4.

Answer : (b)

Reason : The option (b) is correct, since the appropriate t-value is applicable only when the sample size is less than 30.

< TOP >

5.

Answer : (a)

Reason : Since b is an estimator of B. The random variable b is a normal distribution with mean B and the estimate of V(b) is given by. Therefore follows t-distribution with n-2 degrees of freedom.

< TOP >

6.

Answer : (b)

Reason : For standard normal distribution mean = 0 and standard deviation = 1. If the observed value is ‘x’ then the corresponding z–value would be (x – 0)/1 = x.

< TOP >

7.

Answer : (e)

Reason : By Central Limit Theorem for large samples the sample mean is approximately normally distributed with mean = population mean and standard deviation

\ = m = 20

= = = = = 1.25

\ = = P(–1.60 < z < 2.40)

= P(z < 2.40) – P(z < –1.60) = P(z < 2.40) – P(z > 1.60)

= P(z < 2.40) – [1 – P(z < 1.60)]

= (0.50 + 0.4918) – 1 + (0.50 + 0.4452) = 0.937.

< TOP >

8.

Answer : (b)

Reason : Let p be the probability of getting an even number in a throw of a die.

Then the probability of getting x even numbers in ten throws of a die is given by:

P(X =x) =

We are given that P(x=5) = 2P(x=4)

Hence the required number of times that in 10,000 sets of 10 throws each, we get no even number

=

< TOP >

9.

Answer : (a)

Reason : Here n= 100,

P = Prob. of defective pin = 5% = 0.05 (since p is small, we may use Poisson distribution.

= Mean number of defective pins = np =1000.05 =5.

Let the random number X denote the number of defective pins in a box of 100. then by Poisson probability law, the probability of x defective pins in a box is:

< TOP >

10.

Answer : (d)

Reason : The large applications in Statistical Quality Control in industry for setting control limits is found by Normal distribution.

< TOP >

11.

Answer : (c)

Reason : (a) This is the wrong answer. According to the factor reversal test the product of the original index and the index obtained through reversing the factors is not equal to one.

(b) This is the wrong answer. According to the factor reversal test the product of the original index and the index obtained through reversing the factors is not equal to one hundred.

(c) This is the right answer. According to the factor reversal test if we interchange the price and quantity terms in an index number then the product of the original index number and the index number obtained by reversing the factors should be equal to the value index.

(d) This is the wrong answer. According to the factor reversal test the product of the original index and the index obtained through reversing the factors is not equal to the Fisher’s ideal index.

(e) This is the wrong answer. According to the factor reversal test the product of the original index and the index obtained through reversing the factors is not equal to the chain index number.

< TOP >

12.

Answer : (b)

Reason : The required estimate of population variance based on the variance within the sample will be = 14.16.

< TOP >

13.

Answer : (d)

Reason : The number of data points can be obtained as

=

< TOP >

14.

Answer : (c)

Reason : Year = 2001

\ x = 2005 – 2001 = 4

Relative cyclical residual =

Y = 146

\ Relative cyclical residual for the year 2005 = = 12.30

< TOP >

15.

Answer : (e)

Reason : (a) This is true about statistical process control. Quality standards in manufacturing process and the service industry can be achieved through the application of statistical methods

(b) This is true about statistical process control. The variability present in the manufacturing process can be minimized to the extent possible

(c) This is true about statistical process control. The variations in the specifications of the product due to wearing of the machine is a systematic or non-random variation

(d) This is true about statistical process control. The variation due to an error made by the personnel in the measurement of the product specifications is a random variation.

(e) This is not true about statistical process control. If the process is ‘out of control’, it indicates the presence of non-random pattern in the system. The management should first identify the cause that variation and eliminate it. This elimination or the reduction of the systematic variation results in process being brought ‘in control’. Once this is done, the whole can be redesigned to improve or reduce the incidence of random or inherent variability.

< TOP >

16.

Answer : (e)

Reason : The F statistic is calculated on the basis of two estimates of the population variance viz.,

The estimated population variance based on the variance among the sample means and the estimated population variance based on the variance within the samples.

< TOP >

17.

Answer : (d)

Reason : The constant probability p of success for each trial is very small and the number of trials must be indefinitely large. The instances (I) and (III) are applicable to Poisson distribution. (I) and (III) are the examples of the Poisson distribution. So (d) is the correct answer.

< TOP >

18.

Answer : (a)

Reason : We know that the Poisson distribution is bimodal, then the two modes are at points

x =and x =.

Where is the parameter of the Poisson distribution. Therefore, since we given that the two modes are at the points x = 1 and x = 2, we find that = 2.

P(X = x) =x 0, 1, 2, 3, …

P(X=1)=e^–2 2 and P(X=1)+P(X=2) = 2 e^–2 + 2 e^–2 = 0.542

< TOP >

19.

Answer : (c)

Reason : Given = 2, the desired probabilities using the Poisson distribution are as under:

P(X = 0;2)== 0.1353

< TOP >

20.

Answer : (c)

Reason : Estimated sales for the 2^nd quarter

=

= = Rs. 45000.

< TOP >

21.

Answer : (a)

Reason : (a) This is the right answer. A time series consisting of annual data for longer periods is depicted by trend lines. This facilitates us to isolate the component of secular trend variation from the series and examine it for cyclical, seasonal and irregular component. We use “Residual Method” to isolate the cyclical variation component. This method uses one of the two measures namely Percent of trend and relative cyclical residual measure. Both these measures are expressed in terms of percentage.

(b) This is the wrong answer. Cyclical component of the time series is not isolated by Linear regression analysis.

(c) This is the wrong answer. Cyclical component of the time series is not isolated by non-linear regression analysis.

(d) This is the wrong answer. Seasonal component of the time series is isolated by ratio to moving average method.

(e) This is the wrong answer. Cyclical component of the time series is not isolated by multiple regression analysis.

< TOP >

22.

Answer : (d)

Reason : (a) This is a wrong answer. The movement of dependent variable above and below the secular trend line over periods longer than one year is not known as Seasonal variation.

(b) This is a wrong answer. The movement of dependent variable above and below the secular trend line over periods longer than one year is not known as Secular variation.

(c) This is a wrong answer. The movement of dependent variable above and below the secular trend line over periods longer than one year is not known as Irregular variation.

(d) This is the right answer. The movement of dependent variable above and below the secular trend line over periods longer than one year is known as Cyclical variation.

(e) This is a wrong answer. The movement of dependent variable above and below the secular trend line over periods longer than one year is not known as Regular variation.

< TOP >

23.

Answer : (e)

Reason : Seasonal index = Unadjusted means ´ Adjusting constant

From the above equation, if the value of adjusting constant is greater than one, it can be understood that seasonal index is more than unadjusted mean.

< TOP >

24.

Answer : (a)

Reason : Since Y increases as X increases, they are positvely correlated. Therefore the coefficient of correlation between them is the positive square root of the coefficient of determination.

< TOP >

25.

Answer : (c)

Reason : Deseasonalized value = ´ 100

\Seasonal Index = ´ 100

=´ 100 = 253.16

< TOP >

26.

Answer : (d)

Reason : The technique used for studying seasonal variations is known as ratio to moving average method

< TOP >

27.

Answer : (d)

Reason : Sample standard deviation is not required for making a point estimate of population mean

< TOP >

28.

Answer : (d)

Reason : Test statistic =

or 1.60 =

or m = 48 – (1.6 ´ 5) = 40

< TOP >

29.

Answer : (a)

Reason : The finite population correction factor is given by

< TOP >

30.

Answer : (e)

Reason : Considering the process of collecting a simple random sample of a specific size from a population to be an experiment the sample size is not a random variable because it is specified and constant.

< TOP >

31.

Answer : (e)

Reason : A chi-square test of goodness of fit is used to test whether the data from a population follows a hypothesized probability distribution.

< TOP >

32.

Answer : (c)

Reason : When estimating population mean from a sample taken from a normal population with unknown variance the t-distribution may be used with number of degrees of freedom equal to sample size – 1.

a, b, d & e. represent incorrect degrees of freedom for using the t distribution

< TOP >

33.

Answer : (b)

Reason : Standard error of estimate, s_e =

SY² = 19,00,400 (given)

SY = x n = 612 x 5 = 3060

SXY = 18,100 (given)

= 1,236 – 104X

\ a = 1,236 and b = -104

\ s_e = = 14.61.

The sample size is less than 30 and the standard deviation of the population is not known. Hence the appropriate distribution to be used for the prediction interval is the t-distribution.

Degrees of freedom = n – 2 = 5 – 2 = 3.

The appropriate t–value is 2.353 (from the t–table)

The limits of the prediction interval are given below:

For, x = 7.50, = 1236 – 104 (7.50) = 456

The prediction interval is: t (s_e) = 456 ± 2.353 (14.61)

= 456 ± 34.38

< TOP >

34.

Answer : (c)

Reason : Time-series analysis is used to detect patterns of change in statistical information over regular intervals of time.

Secular trend represent the long-term direction of a time series. Of the four types of variation, irregular variation is the most difficult to predict. Hence, Statement III is false. Alternative (c) is the answer.

< TOP >

35.

Answer : (e)

Reason : We know that the regression coefficient b =

\ b =

Given and

Now, The Y intercept, a =

= 4.1 – (- 1.684 ´ 6.2) = 14.54.

< TOP >

36.

Answer : (b)

Reason : Here denotes the mean weight of boxes, the null and alternative hypothesis are

H₀ = ( = 1000) and H₁: (<1000) respectively.

Z = [

= (–50) (7.07)/75= – 4.71

the critical value of z at 5% level of significance is z_0.05 = –1.64

since the calculated value of z is smaller than its critical value at 5% level of significance, the alternative hypothesis H₁: (<1000) is accepted.

Therefore the sample mean is significantly less than 1 kg.

< TOP >

37.

Answer : (c)

Reason : To estimate the accuracy of values predicted by a regression equation, it is useful to calculate the standard error of the estimate (s_e). The larger the s_e, the worse the fit of the regression line.

< TOP >

38.

Answer : (a)

Reason : Relative cyclical residual =

= Percent of trend – 100

\ 0 = Percent of trend – 100 Þ Percent of trend = 100.

< TOP >

39.

Answer : (d)

Reason : Given that

Number of sheared off pieces	100	100	100	100	100
Number of defectives	5	6	7	4	8
Proportion defective	0.05	0.06	0.07	0.04	0.08

The mean of proportions = 0.30 / 5 = 0.06

Lower Control Limit for p chart =

Since the lower control limit cannot be negative, we set the lower control limit as zero.

< TOP >

40.

Answer : (b)

Reason : The central line of the chart is = Mean of the sample means ()

Mean of the sample means or grand mean, = 462 /10 = 46.2

Mean of sample range = = åR / n = 68 /10 = 6.8

Lower Control Limit of chart = - A₂

= 46.2 - 0.577 ´ 6.8 = 42.2764

(The value of A₂ for sample of size 5 is 0.577 from control chart factors table)

< TOP >

41.

Answer : (e)

Reason : n = 15

p = 0.40

q = 1– 0.40 = 0.60

P (r ³ 4) = 1– P (r £ 3)

= 1–

=

= = 0.9095

< TOP >

42.

Answer : (b)

Reason : Given 10 range values the mean range then the upper limit of R control chart is

given that

for the sample size n= 10 and D₄ = 1.777 according to table.

Thus, the upper control limit is

< TOP >

43.

Answer: (b)
Reason: Since the population variance is known, the distribution to be used is the N(0, 1) distribution not with standing that the sample size is small.

\ UCL =

or = UCL – = 81.60 – 62 = 19.60

or =

or n = = = = 24

Since the population variance (s²) is known the standard error of mean will be based on s

< TOP >

44.

Answer : (a)

Reason : We obtain the regression equation using the following table:

X	Y	XY	Y^2
6	9	54	81
2	11	22	121
10	5	50	25
4	8	32	64
8	7	56	49
åX = 30	åY = 40	åXY = 214	åY^2 = 340

The regression equation of X on Y is X_c = a + bY

To determine the values of a and b the following two normal equations are to be solved:

å X = na + b å Y

å XY = a å Y + b å Y²

Substituting the values, we get:

30 = 5 a + 40 b

214 = 40 a + 340 b

The values of a = 16.4 and b = -1.3

= 16.4 – 1.3Y.

< TOP >

45.

Answer : (b)

Reason : When the lot(sample) size varies , there is a definite case for constructing p charts.

< TOP >

46.

Answer : (a)

Reason : When the lower control limit for R chart is zero, it means that the control chart has no limit.

< TOP >

47.

Answer : (b)

Reason : This is a chi-square test of goodness of fit.

Let us take the null hypothesis that there is no difference in the MSc results and general examinations results. On the basis of ratio 3 : 2 : 4 : 1, the expected number of students, getting first division, second division, third division and failed should be 60, 40, 80 and 20 respectively.

The following table is generated for observed and expected frequencies:

Category	Observed frequency (O)	Expected frequency (E)	(O–E)^2 / E
First Division	65	60	0.416
Second Division	61	40	11.025
Third Division	42	80	18.05
Failed	32	20	7.2

The value of Chi square statistic, c² = = 36.691

The table value of chi-square for 4 – 1 = 3 degrees of freedom at 5% level of significance is 7.815. The calculated value of chi-square is greater than the table value. The null hypothesis is rejected. Hence it can be included that the MSc results are not commensurate with the general examination results (i.e., the MSc results are different from general examination results)

< TOP >

48.

Answer : (b)

Reason : This is a test of goodness of fit. The null hypothesis for the given test would be, “The digits are uniformly distributed in the directory”.

The expected frequency (E) for each digit =

= = 1,000

The value of Chi square statistic, c² = = 58.542

The critical value at 5% level of significance for 9 degrees of freedom for chi square distribution is 16.919.

Since the chi square statistic exceeds the critical value it falls in the rejection region. So we reject the null hypothesis. Therefore we conclude at 5% level of significance that the digits are not uniformly distributed in the directory.

< TOP >

49.

Answer : (d)

Reason : This is a test of chi-square goodness of fit.

There are 5 elements therefore the number of degrees of freedom is (n – 1) = 4.

The chi-square value for the 4 degrees of freedom and 1% level of significance is 13.277.

< TOP >

50.

Answer : (b)

Reason : H₀: the proportion of parents in favour of introducing sex education in schools is the same in the three states .

H₁: the proportion is not the same

we have the test statistic is

as v = 2df, H_o is rejected in favour of H₁ if , where

				Total
	50	20	45	115
	30	60	35	125
Total	80	80	80	240

Expected value of 50 = , similarly we get all the expected values of the observations.

oi	50	20	45	30	60	35
ei	38.3	38.3	38.3	41.6	41.6	41.6

Since the calculated value of =25.9 is greater than , H_o is rejected. This means that the proportion of parents in favour of introducing sex education at the school level is not the same in the three states.

< TOP >

51.

Answer : (e)

Reason : It is suitable to analyze large and complex real-life problems which cannot be solved by usual quantitative methods. This is the advantage of simulation.

< TOP >

52.

Answer : (a)

Reason :

Since z = 1.78 is greater than =1.645, H₀ is rejected and we conclude that the current wage rate is higher than Rs 198.

< TOP >

53.

Answer : (a)

Reason : Estimated standard error of proportion = =

= = 0.04

< TOP >

54.

Answer : (e)

Reason : In multiple regression relationship:

Standard error of estimate,

Given : ESS = 576

n = ?

k = 2

s_e = 4.8

\4.8 =

\

or

or n =

< TOP >

55.

Answer : (c)

Reason :

s = 6.84

n = 60 N = 540

Sampling fraction,

\ Estimated standard error of mean,

=

=

= 0.833

Since the sample size is greater than 30 the normal distribution will be used.

Z-values for the upper and lower confidence limit are

\ Upper confidence limit =

Lower confidence limit =

< TOP >

56.

Answer : (d)

Reason : As the sample size increases, the interval estimate for the population mean at a given confidence level will decrease in width

< TOP >

57.

Answer : (b)

Reason :

< TOP >

58.

Answer : (b)

Reason :

= or 0.17 =

or n = = = 7.87 n = 7.87.

< TOP >

59.

Answer : (b)

Reason :

Estimated standard error or proportion,

= 0.0358

Z – Values for the upper and lower confidence limit are

Upper confidence limit =

Lower confidence limit =

< TOP >

60.

Answer : (e)

Reason : H₀ : m = 15

H₁ : m < 15

The sample is large. So we shall use the normal distribution.

= = = 7.08

z = = = –1.19.

At a = 0.05, the left tail critical value is –1.645.

The test statistic is more than the left tail critical value. So it falls in the acceptance region.

\ We accept H₀.

But the true mean is 10. So H₀ is false.

\ This leads to type II error.

< TOP >

61.

Answer : (e)

Reason : The binomial distribution should be used.

According to the binomial distribution formula, probability of ‘r’ success in ‘n’ trials, P(r) =

p = Probability of being defective = 5% = 0.05

q = Probability of not being defective = 1– p = 1 – 0.05 = 0.95

n = Number of trials = 10

r = Number of successes i.e., defective jobs in this case = 3

Probability of maximum 3 out of 10 jobs being defective is

P(£ 3) = P(0) + P(1) + P(2) + P(3)

= ¹⁰C₀ (0.05)⁰ (0.95)¹⁰ + ¹⁰C₁ (0.05)¹ (0.95)⁹ + ¹⁰C₂ (0.05)² (0.95)⁸ + ¹⁰C₃ (0.05)³ (0.95)⁷

= 1 ´ 1 ´ 0.5987 + 10 ´ 0.05 ´ 0.6302 +

45 ´ 0.0025 ´ 0.6634 + 120 ´ 0.000125 ´ 0.6983

= 0.5987 + 0.3151 + 0.0746 + 0.0105

= 0.9989.

< TOP >

62.

Answer : (e)

Reason : A value index can not distinguish between the price changes and quantity changes. Hence an increase in the value index cannot indicate with certainty that the prices have increased

< TOP >

63.

Answer : (c)

Reason : Unweighted aggregates price index =

=

= 134.21

< TOP >

64.

Answer : (e)

Reason : All the alternatives are stating the different uses of an index number.

< TOP >

65.

Answer : (e)

Reason : Laspeyres price index =

P₀ = Base year (2002) price

Q₀ = Base year (2002) quantity

P₁ = Current year (2005) price

Q₁ = Current year (2005) quantity

\ Laspeyres price index for the year 2005 =

=

< TOP >

66.

Answer : (c)

Reason : Marshall-Edgeworth price index =

P₁ = Current year (2005) price

P₀ = Base year (2000) price

Q₁ = Current year (2005) quantity

Q₀= Base year (2000) quantity

\ Marshall-Edgeworth price index for 2005 =

=

=

= 162.87

< TOP >

67.

Answer : (c)

Reason : (a) This is the wrong answer. The weighted average of price relatives using base values as weights is not same as the unweighted aggregates price index.

(b) This is the wrong answer. The weighted average of price relatives using base values as weights is not same as the unweighted aggregates quantity index.

(c) This is the right answer. The weighted average of price relatives using base values as weights is same as the Laspeyres price index.

(d) This is the wrong answer. The weighted average of price relatives using base values as weights is not same as the Laspeyres quantity index.

(e) This is the wrong answer. The weighted average of price relatives using base values as weights is not same as the Paasche’s price index.

< TOP >

68.

Answer : (d)

Reason : Fisher’s ideal price index =

Paasches price index =

= = 124.20.

< TOP >

69.

Answer : (e)

Reason : a. Fisher’s ideal price index considers base year prices.

b. Fisher’s ideal price index considers base year quantities.

c. Fisher’s ideal price index considers current year prices.

d. Fisher’s ideal price index considers current year prices.

e. Fisher’s ideal price index is the geometric mean of the Laspeyres and Paasche’s price indices

< TOP >

70.

Answer: (a)

Reason: The appropriate test statistic is the c² statistic, is given by

c² =

< TOP >

71.

Answer : (c)

Reason : Regression sum of squares = 186 – 86 = 100. So, the percentage of changes in the dependent variable, which is explained by the changes in the independent variable, is 100/186 = 0.5376 or 53.76%.

< TOP >

72.

Answer : (c)

Reason : The number of degrees of freedom for the denominator of the F-ratio

= n_T – k where (n_T = Total number of elements = 11 + 9 + 10 = 30)

(k = Number of samples = 3)

= 30 – 3

= 27.

< TOP >

73.

Answer : (d)

Reason : F distribution is identified by a pair of degrees of freedom.

< TOP >

74.

Answer : (e)

Reason : The constant probability p of success for each trial is very small and the number of trials is indefinitely large.

So, all the given examples follows Poisson distribution.

< TOP >

75.

Answer : (a)

Reason : When the result of taking that decision is uncertain, then we draw a small circle to represent the result.

< TOP >

76.

Answer : (c)

Reason : The number of absentees who are boys follows a hypergeometric distribution. The following details are available:

N = No. of elements in the population = 25

r = No. of elements in the population labelled success = No. of boys = 14

n = No. of trials = No. of absent students = 5

x = Desired number of successes

In a hypergeometric distribution P(x) =

< TOP >

77.

Answer : (e)

Reason : H₀ : Opinion of the persons is independent of the age group

H₁ : Opinion of the persons is dependent on the age group.

Expected frequency for any cell =

Row	Column	fo
1	1	140		20	400	3.333
1	2	80		2	4	0.051
1	3	50		2	4	0.083
1	4	30		–24	576	10.667
2	1	60		–20	400	5.000
2	2	50		–2	4	0.077
2	3	30		–2	4	0.125
2	4	60		24	576	16
						35.336

Number of degree of freedom = (2 – 1) (4 – 1) = 3

Critical value :

The value for the sample data is much greater than the critical value. Hence we reject the null hypothesis and conclude that the opinion of the persons and age groups are dependent.

< TOP >

78.

Answer : (c)

Reason : The observations are stated in the following table:

Methods	Observations						Sum	Mean
A			55	44	45	50	194	48.5
B	64		52	56	44	48	264	52.80
C	46	42	48	45	57	42	280	46.67

Grand mean =

The estimate of the population variance on the basis of the variance among the sample means

=

< TOP >

79.

Answer : (c)

Reason : Probability of ‘r’ successes in ‘n’ trials, in a Bernoulli process, is given by,

P (r) =

Where, q = 1 – p since p = 0.50 then q = 0.50.

Thus, the above equation can be modified and written as:

P(r) = =

Now, = \ = (n = 13, r = 6)

P(r = 6) = (0.50)¹⁰

P(r = 7) = (0.50)¹⁰.

Since, = we can say that P(r = 6) = P(r = 7)

\ In the given case, probability of getting 6 successes is equal to the probability of getting 7 successes.

< TOP >

< TOP OF THE DOCUMENT >