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· Answer all questions. · Marks are indicated against each question.
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The
standard deviation of the number of successes in a binomial distribution is What is the probability of obtaining exactly eight successes? (a) 0.0197 (b) 0.1001 (c) 0.25 (d) 0.75 (e) 0.80. (1 mark) |
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In which of the following conditions will a binomial distribution be symmetrical? (a) Number of trials is 30 (b) Number of trials is 100 (c) Probability of success in any trial is less than 0.5 (d) Probability of failure in any trial is less than the probability of success (e) Probability of failure in any trial is equal to 0.5. (1 mark) |
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Variable T follows a normal distribution. The standardized value for T = 35 is 2 and the standardized value for T = 40 is 3. Which of the following are correct with regard to T? (a) Mean = 5 and variance = 5 (b) Mean = 25 and variance = 25 (c) Mean = 5 and variance = 25 (d) Mean = 25 and variance = 5 (e) Mean = 2.5 and variance = 2.5. (1 mark) |
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For a binomial probability distribution with parameters n and p (where n = number of trials and p = probability of success on any trial), the product of these two parameters will be equal to the (a) Mean (b) Standard deviation (c) Variance (d) Median (e) Range. (1 mark) |
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The standard normal distribution has a (a) Mean = 1 and standard deviation = 0 (b) Mean = 1 and standard deviation = 1 (c) Mean = 0 and standard deviation = 1 (d) Mean = 0 and standard deviation = 0 (e) Mean = –1 and standard deviation = 0. (1 mark) |
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The wind speed at a wind energy station is approximately normally distributed with a mean of 25 miles per hour and a standard deviation of 7 miles per hour. What percent of the time will the wind speed be below 15 miles per hour? (a) 7.64% (b) 42.36% (c) 57.64% (d) 92.36% (e) 50%. (1 mark) |
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Modern Books and Magazines sells a variety of books, magazines and newspapers. The manager of the shop is concerned about one particular newspaper, The City Times. This newspaper costs Rs.1.20 each to the shop and can be sold for Rs.1.50 each. Any copy of the newspaper not sold by the end of the day can be disposed off for Re.0.20 each to the paper packet manufacturers. The manager of the shop has recorded the daily sales of The City Times over the past 200 days and his observations are given below:
It is assumed that the number of copies of the newspaper sold in a day is a discrete random variable, which will assume only the numbers listed above and the shop will stock only the same number of copies. What is the optimal number of copies of the newspaper that should be stocked? (a) 200 (b) 300 (c) 400 (d) 500 (e) 1,400. (2 marks) |
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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 two of the absent students were girls? (a) 0.3768 (b) 0.6232 (c) 0.00104 (d) 0.00685 (e) 0.99315. (1 mark) |
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Which of the following is correct about the value for “Rapid development decision”? (a) It is the value of the outcome multiplied with the probabilities of outcomes (b) It is the value of the outcome added to the probabilities of outcomes (c) It is the value of the outcome multiplied with the total probability (d) It is the value of the outcome added to the possible outcomes (e) It is the value of the outcome multiplied with the probability of decision variables. (1 mark) |
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When a decision maker makes decision under conditions of risk (a) He cannot define all possible decision alternatives (b) He cannot quantitatively estimate the consequences of selecting any decision alternative (c) He cannot define the various states of nature that may occur (d) He can assign probabilities of occurrence to all the possible states of nature (e) He uses the regret criterion to make decision. (1 mark) |
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A sample has been taken from a population and the sample mean has been found to be 100. The upper limit of a 90 percent confidence interval is 112. What is the lower limit of this confidence interval? (a) 100 (b) 92 (c) 68.26 (d) 88 (e) 144.76. (1 mark) |
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The following details are available with regard to a hypothesis test on difference between means of two populations: H0: H1: n1 = 16 n2 = 9 The samples are independently collected. It is assumed that the two populations are normally distributed and have the same variance. If the significance level is 0.05 then (a) The critical value is –1.714 and the null hypothesis is rejected (b) The critical value is –1.714 and the null hypothesis is accepted (c) The critical
values are (d) The critical values is –2.069 and the null hypothesis is rejected (e) The critical values is –1.96 and the null hypothesis is accepted. (2 marks) |
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The following details are available with regard to a hypothesis test on a population mean:
The significance level is 0.05. For which of the following values of the sample mean will the null hypothesis be accepted? (a) 8.95 (b) 7.85 (c) 14.50 (d) 15.20 (e) 15.80. (1 mark) |
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A polling agency found that 387 people who planned to vote in the upcoming election said that they would vote for party A, 360 people planned to vote for party B and 253 are undecided. What is the 90% confidence interval to estimate the proportion of voters who plan to vote for party A in the next election? Assume that the voter population is large vis-à-vis the sample. (a) (0.341, 0.432) (b) (0.376, 0.398) (c) (0.357, 0.417) (d) (0.372, 0.402) (e) (0.362, 0.412). (1 mark) |
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A hypothesis test in which rejection of the null hypothesis occurs for values of the point estimator falling in either tails of the sampling distribution is called (a) The chi-square test (b) The t-test (c) A one tailed test (d) A two tailed test (e) Both (b) and (c) above. (1 mark) |
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The following details are available for a hypothesis test on population proportion: H0: p = 0.20 H1: p Sample size = 120 Sample proportion = 0.25 Significance level = 0.10 It is later known that the true population proportion is 0.20. Which of the following can be said with regard to the test? (a) There is insufficient information for doing the test (b) The chi-square 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) |
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The following details are available with regard to a hypothesis test on means of two populations: n1
= 24 n2
= 15 The samples collected from the two populations are independent. What is the estimated standard error of difference between means? (Mark your answer to the nearest integer). (a) 2 (b) 3 (c) 4 (d) 5 (e) 1. (1 mark) |
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The extent to which the observed values of the dependent variable differ from the predicted values according to the regression line is measured by (a) The slope of the regression line (b) The intercept of the regression line (c) The standard error of estimate (d) The coefficient of correlation (e) The coefficient of determination. (1 mark) |
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A sample of size 96 has been taken from a population and the estimated standard error of proportion is found to be 0.05. What is the sample proportion? (a) 0.05 or 0.95 (b) 0.24 or 0.76 (c) 0.80 or 0.20 (d) 0.40 or 0.60 (e) 0.50. (1 mark) |
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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) |
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The movement of the dependent variable above and below the secular trend line over periods longer than one year is known as (a) Cyclical variation (b) Secular trend (c) Seasonal variation (d) Irregular variation (e) Medium term variation. (1 mark) |
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A sample of size 16 yielded a mean of 10. The population mean is known to be 12. What is the expected value of the sampling distribution of mean? (a) 10 (b) 12 (c) 2.5 (d) 3 (e) 1.732. (1 mark) |
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In which of the following conditions can the standard deviation of the sampling distribution of mean for a given sample size, be accurately known? (a) The population mean is known (b) The sample mean is known (c) The sample standard deviation is known (d) The population variance is known (e) The population median is known. (1 mark) |
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Other things remaining the same, if the confidence level is increased then the width of the confidence interval for the population mean will (a) Increase (b) Increase by twice the previous width (c) Increase by half the previous width (d) Decrease (e) Remain unchanged. (1 mark) |
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If the population is normally distributed then for any
given sample size, the sampling distribution of (a) A normal distribution (b) The standard normal distribution (c) A t-distribution (d) A t-distribution only when the sample size is 30 or less (e) A t-distribution only when the sample size is more than 30. (1 mark) |
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Which component of a time series the ‘Residual Method’ can isolate? (a) Secular variation (b) Cyclical variation (c) Seasonal variation (d) Irregular variation (e) Error Variation. (1 mark) |
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The following details are available with regard to a hypothesis test on means of two populations: H0: H1: n1 = 64
n2 = 36
The samples collected from the two populations are independent. What is the value of the test statistic? (a) 2.828 (b) 1.000 (c) 0.354 (d) 0.125 (e) 0. (1 mark) |
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The following details are available for a hypothesis test on population proportion: H0: p = 0.40 H1: p Sample proportion = 0.36 Sample size = 250 What is the value of the test statistic on the standardized scale? (a) 0.03098 (b) 0.096 (c) –1.291 (d) 1.291 (e) 11.62. (1 mark) |
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If (a) Degree of correlation between Y, X1 and X2 is high (b) Degree of correlation between Y and X2 is high (c) Degree of correlation between X1 and X2 is high (d) Value of a and b1 are higher compared to b2 (e) Value of a and b2 are higher compared to b1. (1 mark) |
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Which of the following is/are not required for the two-sample test of means for independent samples when both samples contain less than 30 observations? I. Both the populations are normally distributed. II. Both the
populations have equal variances. III. Both the samples
have equal means. IV. Both the samples are of equal sizes. (a) Only (I) above (b) Only (II) above (c) Only (III) above (d) Only (IV) above (e) Both (III) and (IV) above. (1 mark) |
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If the weighted aggregates price index for a set of prices was calculated as 117 using the Paasches method and as 121 using the Laspeyres method, what can be concluded from this? (a) The Paasches index is correct and Laspeyres index is incorrect (b) There is a trend towards more expensive goods (c) There is a trend towards less expensive goods (d) The Laspeyres index is correct and Paasches index is incorrect (e) The difference can be attributed to a poor estimation of consumer attitudes. (1 mark) |
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A fruit merchant in the fruit market of the city is worried about the rising prices of the fruits and its effect on the quantity consumed by the consumers. He wants to understand this effect by finding a suitable index. The following table produces the prices and quantities of some fruits in the year 1998 and 2004:
Find out Laspeyres price index number for the year 2004, taking 1998 as the base year. (a) 130.06 (b) 132.12 (c) 134.24 (d) 136.36 (e) 138.42. (1 mark) |
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The following details are available with regard to a basket of goods:
Value index calculated on the basis of the same basket of goods = 81.25 What is Paasches price index? (a) 123.10 (b) 64.38 (c) 126.21 (d) 155.34 (e) 81.25. (1 mark) |
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The following details are available with regard to a basket of goods: Weighted average price of the goods in current
year, using current year quantities as weights = Rs.60 Weighted average price of the goods in base
year, using current year quantities as weights = Rs.50 Weighted average price of the goods in current
year, using base year quantities as weights
= Rs.64 Weighted average price of the goods in the base
year, using base year quantities a weights
= Rs.51.20 What is Fisher’s ideal price index? (a) 120 (b) 122.47 (c) 125 (d) 117.2 (e) 128. (1 mark) |
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If the base year and current year quantities for a group of commodities are equal, though prices have changed, then the price index calculated for the group of commodities using the Marshall – Edgeworth method will be equal to which of the following? (a) Laspeyres price index (b) Unweighted aggregates price index (c) Unweighted average of relatives price index (d) Paasches price index (e) Both (a) and (d) above. (1 mark) |
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With regard to a basket of goods, we know the weighted average of the quantities consumed during the base year and the weighted average of the quantities consumed during the current year, where both the weighted averages have been computed on the basis of the current year prices as weights. Then which of the following can be calculated? (a) Laspeyres price index (b) Laspeyres quantity index (c) Paasches price index (d) Paasches quantity index (e) Unweighted aggregates price index. (1 mark) |
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The price of a commodity during 2000-2004 is given below. Fit a parabolic trend to this data and estimate the price of the commodity for the year 2005.
(a) Rs.310.50 (b) Rs.202.02 (c) Rs.391.48 (d) Rs.219.20 (e) Rs.302.52. (2 marks) |
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The standard deviation of the sampling distribution of the mean is also called the (a) Standard error of the mean (b) Sum of squares (c) Sum of squared deviations (d) Variance (e) Sample standard deviation. (1 mark) |
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Pradeep Khanna, a second year MBA student, is doing a study of companies going public for the first time. He is curious to see whether or not there is a significant relationship between the sizes of the offering (in crores of rupees) and the price per share after the issue. The data are given below:
Calculate the coefficient of determination for the above data set. (a) 0.4523 (b) 0.5842 (c) 0.6753 (d) 0.7664 (e) 0.8575. (2 marks) |
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The manager of Pioneer Enterprises wants to formulate a relationship between total cost and sales revenues, which may be used to estimate the total cost for a given level of sales. He collects the following information from the past records.
Formulate a relationship between sales and total cost with sales as the independent variable. Find the slope of the regression line. (a) 0.533 (b) 0.622 (c) 0.711 (d) 0.899 (e) 0.988. (1 mark) |
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The coefficient of correlation between variables X and Y is 0.80. The coefficient of variation of X is 40% and the coefficient of variation of Y is 20%. The mean of X is 20 and the mean of Y is 50. What is the covariance between 5X and 4Y? (a) 8 (b) 10 (c) 800 (d) 640 (e) 1280. (1 mark) |
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The following data shows the number of house loan applications processed per day by four workers of Oxpro Finance Pvt.Ltd. observed over the number of days:
Find the variance among the sample means. (a) 5.50 (b) 6.50 (c) 7.50 (d) 8.50 (e) 9.50. (2 marks) |
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The security supervisor at a big department store wants to know if the store apprehends relatively more shoplifters during the Diwali festival season than in the weeks before or after the season. He gathered data on the number of shoplifters apprehended in the store during the months of September, October and November over the past 6 years. The information are as follows:
It is to be tested at a significance level of 5 percent whether the average number of the shoplifters is same in each of the three months. Which of the following statements is false? (a) The value of the test statistic is 6.67 (b) The appropriate probability distribution for the test is the F-distribution (c) The test statistic falls in the rejection region. (d) The critical value for the test is 3.68 (e) At a 5% significance level it can be concluded that the average number of shoplifters is same in each of the three months. (3 marks) |
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In response to a number of complaints about slow mail delivery, the Postmaster General initiated a preliminary investigation. An investigation follows nine letters from Mumbai to Pune, to estimate the variance in delivery time. The staff collects a sample of 9 mail deliveries and finds the sample standard deviation to be 23 hours. Postmaster General wants a 95% confidence interval for the variance in the mail delivery. Find the upper limit of the confidence interval for the estimated population variance. (a) 1721.07 (b) 1831.14 (c) 1941.28 (d) 2051.42 (e) 2161.56. (1 mark) |
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The standard deviation of a normal population is hypothesized to be 50. An observed sample of 30 yields a sample standard deviation of 57. What is the value of the test statistic for the appropriate statistical test? (a) 33.25 (b) 34.36 (c) 35.47 (d) 36.58 (e) 37.69. (1 mark) |
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Two random samples were drawn from two normal populations and their values are:
We want to test whether the population B has larger variance than the population A. What conclusions can be most appropriately drawn from the above test at a significance level of 5%? (a) The variances of the two populations are equal (b) The difference between the variances of the two populations is 38 (c) The difference between the variances of the two populations is 76 (d) The variance of the population A is larger than the variance of the population B (e) The variance of the population B is larger than the variance of the population A. (2 marks) |
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A survey of 320 families, each with 5 children, revealed the following distribution:
A statistician wants to test the hypothesis that male and female births are equally probable. What is the value of the correct test statistic for the above test? (a) 4.13 (b) 5.14 (c) 6.15 (d) 7.16 (e) 8.17. (2 marks) |
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A random sample of 400 persons was selected from each of three age groups and each person was asked to specify which of three types of TV programmes, A, B and C, is preferred. The results are shown in the following table.
We want to test whether there is an association between age group and preference for TV programmes. What is the value of the correct test statistic for the above test? (a) 172.12 (b) 174.23 (c) 176.46 (d) 178.82 (e) 180.50. (1 mark) |
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A new shearing machine is set to cut off a piece of steel from a long bar. For various reasons the machine at 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 are sheared off. The record after the first day of operation is:
Set out the upper control limit for the p chart. (a) 0.128 (b) 0.130 (c) 0.132 (d) 0.134 (e) 0.136 (1 mark) |
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A drilling machine bores holes with a mean diameter of 0.5230 cm and a standard deviation of 0.0032 cm. Calculate the lower control limit for means with samples of size 4 each. (a) 0.5174 cm (b) 0.5176 cm (c) 0.5178 cm (d) 0.5180 cm (e) 0.5182 cm. (1 mark) |
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The expected value of a binomial probability distribution is (a) Always equal to the variance (b) Always more than the variance (c) Always less than the variance (d) Always equal to zero (e) Always equal to 1. (1 mark) |
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The following data provides the values of sample range R, for samples of size 5 each.
Calculate the value for the center line of the R chart. (a) 6.1 (b) 6.2 (c) 6.3 (d) 6.5 (e) 6.6. (1 mark) |
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Which of the following is not a characteristic of normal probability distribution? (a) The mean of the normally distributed population lies at the center of its normal curve (b) It is multi-modal (c) The mean, median and mode are equal (d) Both the tails extend indefinitely and never meet the horizontal axis (e) It is symmetric curve. (1 mark) |
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A certain drug is claimed to be effective in curing cold. In an experiment on 500 persons with cold, half of them were given the drug and half of them were given sugar pills. The patients’ reactions to the treatment are recorded in the following table:
Which of the following conclusions can be drawn on the basis of above data, at 5% level of significance using the appropriate statistical test? (a) The use of the
drug and getting cured from cold are associated (b) The use of the
drug and getting cured from cold are not associated (c) The mean number
of patients cured from cold is equal to the mean number of patients not cured (d) The mean number of
patients cured from cold is not equal to the mean number of patients not
cured (e) No conclusion can be drawn from the given information. (1 mark) |
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A statistician is trying to explore the relationship between per unit cost (Y) and volume of production (X). He finds that the coefficient of determination is 0.7744 and the estimated regression line is Y = 0.8 – 0.01X. The coefficient of correlation X is (a) 0.80 (b) –0.88 (c) 0.88 and –0.88 (d) 0.80 and –0.80 (e) 0.88 and –0.80. (1 mark) |
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The manager of a food chain outlet wants to ensure that the variability in the service time by their bearers is in control. He does a sampling of the average time taken by the bearers in serving the customers in the outlet. He finds the mean of the sample range to be 1.5 minutes. If he has collected five samples, each containing ten observations to conduct the test then what is the upper control limit (in minutes) of the R chart? (Use the relevant D4 factor). (a) 0.335 min (b) 1.005 min (c) 1.500 min (d) 2.005 min (e) 2.666 min. (1 mark) |
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Given a probability distribution for a random variable, the covariance of the variable with itself is equal to (a) 0 (b) 1 (c) The variance of the variable (d) The standard deviation of the variable (e) The expected value of the variable. (1 mark) |
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In the linear regression model, the residuals are assumed to (a) Have a binomial distribution with a probability of success of 0.50 (b) Have a normal
distribution with a mean of 0 (c) Have a chi-square
distribution with a mean of 0 (d) Have a t-distribution
with a mean of 1.00 (e) Have a normal distribution with unknown mean. (1 mark) |
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The slope and the intercept on y-axis for a simple regression line are 3 and 5 respectively. If the independent variable has a value of 5 then the estimated value of the dependent variable is (a) 3 (b) 5 (c) 20 (d) 27 (e) 28. (1 mark) |
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Which of the following probability distributions is a distribution derived from the ratio of two random variables each of which follows a chi-square distribution? (a) Normal distribution (b) Hypergeometric distribution (c) t-distribution (d) F-distribution (e) Binomial distribution. (1 mark) |
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The estimated regression equation of the variable Y
on X is: (a) 10 (b) 8 (c) 12 (d) 20 (e) 5. (1 mark) |
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A test of ANOVA is applied to data obtained from four samples, where each sample contains six observations. At a 5 percent significance level what is the critical value for the test? (a) 3.13 (b) 3.10 (c) 3.49 (d) 2.87 (e) 3.52. (1 mark) |
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If the probability of success in a Bernoulli experiment is 0.60 and the test is repeated 67 times, then the standard deviation of the results of this experiment is (a) 3.37 (b) 3.61 (c) 3.76 (d) 3.92 (e) 4.01. (1 mark) |
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The covariance between two variables X and Y is 256. If the variance of variable X is 625 and the coefficient of correlation between X and Y is 0.64, the variance of the variable Y is approximately (a) 4 (b) 8 (c) 16 (d) 64 (e) 256. (1 mark) |
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Which of the following can indicate the magnitude of association between the variables in an estimated simple regression relationship? (a) Coefficient of variation (b) Coefficient of determination (c) Slope of the regression relationship (d) Intercept of the regression relationship (e) Mean of independent variable. (1 mark) |
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Which of the following is false about index numbers? (a) Index numbers are expressed as a percentage (b) Index number for the base year is always 100 (c) Index number has the unit of the variable that is being compared (d) Index number is calculated as a ratio of the current value of the variable to its value in the base year (e) Index number for a group of commodities indicates the overall change in the variable being compared. (1 mark) |
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In a simple regression relationship, the minimum possible value of standard error of estimate is (a) 0 (b) 1 (c) 0.50 (d) – (1 mark) |
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The following information are available with regard to a sampling distribution of mean: Sample size = 64 Probability that the sample mean is less than 65 = 0.0228 Population standard deviation = 20 It is assumed that the Central Limit Theorem will be applicable and the population mean is greater than 65.What is the population mean? (a) 60 (b) 65 (c) 70 (d) 75 (e) 5. (1 mark) |
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Which of the following statements is false regarding Fisher’s ideal price index? (a) It is the geometric mean of Laspeyres and Paasches price indices (b) It takes into account both current year and base year prices and quantities (c) It satisfies both time reversal and factor reversal tests (d) It is free from bias (e) For the same basket of goods Fisher’s ideal price index will be more than both Laspeyres and Paasches price indices. (1 mark) |
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The mean and the variance of a simple random sample are found to be 12 and 9.6 respectively. The sample size is 6 and it is less than 5 percent of the population size. It is known that the population is normally distributed. What is the 95 percent confidence interval for the population mean, on the basis of the given information? (a) (9.432, 14.571) (b) (8.748, 15.252) (c) (8.905, 15.095) (d) (7.40, 21.60) (e) (8.252, 15.748). (1 mark) |
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The following is the matrix of correlation coefficients between three variables Y, X1 and X2.
How much variation in Y is explained by X1 if X2 is kept constant? (a) 33 % (b) 58 % (c) 61 % (d) 72 % (e) 83 %. (1 mark) |
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Which pattern in the (a) Increasing or decreasing trend (b) Cycles (c) Hugging the center line (d) Hugging the control limits (e) Randomly scattered means between the control limits. (1 mark) |
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The following details are available with regard to a hypothesis test on a population mean:
The significance level is 0.05. For which of the following values of the sample mean will the null hypothesis be rejected? (a) 13.53 (b) 14.25 (c) 14.50 (d) 15.15 (e) 15.50. (1 mark) |
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Assume z is a standard normal random variable. P(-2.0 < z < -1.0) = (a) 0.8185 (b) 0.1469 (c) 1.0000 (d) 0.1359 (e) 0.4772. (1 mark) |
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The sample mean is an unbiased estimator of the population mean because (a) The expected value of sample mean is greater than the population mean (b) The expected value of sample mean is less than the population mean (c) The expected value of sample mean is equal to the population mean (d) The expected value and variance of sample mean are equal (e) The expected value and standard deviation of sample mean are equal. (1 mark) |
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Which of the following is true with regard to multicollinearity in multiple regression analysis? (a) It increases the influence of the individual variables in the model (b) It reduces the predictive power of the model (c) It increases the reliability of the regression coefficients (d) It reduces the effectiveness of sensitivity analysis of the model (e) It arises if there is no correlation among the independent variables. (1 mark) |
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A test of independence of two factors is to be conducted on the basis of a contingency table. The contingency table has 2 rows and 4 columns. Which of the following distributions should be used? (a) Chi-square distribution with 6 degrees of freedom (b) Chi-square distribution with 8 degrees of freedom (c) Chi-square distribution with 3 degrees of freedom (d) F-distribution with 2 degrees of freedom in the numerator and 4 degrees of freedom in the denominator (e) F-distribution with 1 degree of freedom in the numerator and 3 degrees of freedom in the denominator. (1 mark) |
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Other things remaining the same, if the sample size increases then the width of the confidence interval for the population mean (a) Decreases (b) Decreases by exactly the square root of the increase in the sample size (c) Decreases by exactly the reciprocal of the square root of the increase in the sample size (d) Decreases by exactly the increase in the sample size (e) Increases. (1 mark) |
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The following details are available for a hypothesis test on population proportion: H0: p = 0.40 H1: p Value of the test statistic = 1.5492 Sample size = 360 What is the sample proportion? (a) 0.44 (b) 0.56 (c) 0.40 (d) 0.60 (e) 0.04. (1 mark) |
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The manager of the quality department for a tyre manufacturing company wants to study the average tensile strength of rubber used in making a certain brand of radial tyre. The population is normally distributed and the population standard deviation is known. She wants to test the null hypothesis that the mean tensile strength is equal to 800 pounds per square inch against the alternative hypothesis that the mean tensile strength is more than 800 pounds per square inch. She has taken a sample of 25 observations and the test statistic is calculated to be 1.695. If the significance level is 0.05, then (a) The critical value is 1.645 and the null hypothesis is accepted (b) The critical value is 1.645 and the null hypothesis is rejected (c) The critical value is 1.711 and the null hypothesis is accepted (d) The critical value is 1.711 and the null hypothesis is rejected (e) The critical value is 1.282 and the null hypothesis is rejected. (1 mark) |
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The following data provides the values of sample mean
Calculate the value of the lower control limit for the (a) 4.458 (b) 4.347 (c) 4.236 (d) 4.807 (e) 4.944. (1 mark) |
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Ocimum Labs is a renowned company manufacturing biochemistry based products. It has produced clones of popular genes such as DNA, RNA and Oops. The number of pairs sold for each of the genes on different days is as follows:
An appropriate test is to be performed, for testing the equality of the mean number of pairs sold for each of the three genes. What is the estimated population variance on the basis of the variance within the samples? (a) 3.0833 (b) 5.4853 (c) 7.6453 (d) 9.0833 (e) 12.4533. (1 mark) |
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Which of the following is true with regard to the method of least squares used in simple regression analysis? (a) It minimizes the sum of the squared values of the independent variable (b) It minimizes the sum of the squared differences between the values of the independent variable and their mean (c) It minimizes the sum of the squared differences between the actual values of the dependent variable and the estimated values of the dependent variable as per the regression equation (d) It minimizes the sum of the squared differences between the values of the dependent variable and their mean (e) It minimizes the sum of the squared values of the dependent variable. (1 mark) |
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Which of the following is false with regard to the maximin criterion of decision-making? (a) It is assumed that the decision maker is pessimistic (b) It is assumed that whatever strategy is selected nature will respond in such a way as to produce the least possible gain (c) It is used to make decisions under conditions of uncertainty (d) It is used to make decisions under conditions of risk (e) It is used when probabilities of the various states of nature are not known. (1 mark) |
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Which of the following is/are the parameter(s) for describing a normal probability distribution? (a) Mean only (b) Standard deviation only (c) Degrees of freedom (d) Maximum and minimum values of the normal random variable (e) Both mean and standard deviation. (1 mark) |
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The expected value of a continuous uniform random variable is (a) Independent of the upper and lower limits of the range of values of the random variable (b) Equal to the arithmetic mean of the upper and lower limits of the range of values of the random variable (c) Equal to the product of the upper and lower limits of the range of values of the random variable (d) Equal to the geometric mean of the upper and lower limits of the range of values of the random variable (e) Equal to the difference between the upper and lower limits of the range of values of the random variable. (1 mark) |
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The confidence level being same and the population variance not being known, a confidence interval for the population mean which is narrower than another for the same population, is certainly (a) Based on a sample with lower mean (b) Based on a sample of lower size (c) Based on a sample with higher variance (d) Based on a sample with lower estimated standard error of mean (e) Based on a sample with higher range. (1 mark) |
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A fitted regression equation is given by (a) 5 (b) 10 (c) 0 (d) 15 (e) -5. (1 mark) |
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Other things remaining the same, if there are two populations with known variances then, a random sample collected from the population with the lower variance will produce a confidence interval for the population mean which is _____________ that produced by a random sample collected from the population with the higher variance. (a) Wider than (b) Narrower than (c) As wide as (d) A constant multiple determined by the confidence level, of the width (e) Half as wide as. (1 mark) |
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. The probability of success in a single trial is 1/4. 
and the null hypothesis is accepted
0.20
0.40
= a + b1X1 + b2X2,
where X1 and X2 are two independent variables, then the
problem of multicollinearity will arise when the
. Let Z =Y + 2. What is the intercept of the
estimated regression equation of Z on X (X is the
independent variable)?
0.40
chart. (Use the relevant A2 factor)
Suggested Answers
Quantitative Methods-II (132): January 2006
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Answer : (a) Reason : p = \ q = 1– p = 1 – Variance = npq \ or 3 = or n = 16 \ P(r = 8) = |
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Answer : (e) Reason : A binomial distribution is symmetrical if the probability of success and the probability of failure are equal. Since p + q = 1, this means that the probability of failure is equal to 0.5. |
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Answer : (b) Reason :
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Answer : (a) Reason : The product of the two parameters n and p will be equal to the mean. |
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Answer : (c) Reason : Self-explanatory |
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Answer : (a) Reason : P(x <
15) = |
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Answer : (b) Reason : Profit per newspaper sold = 1.50 – 1.20 = Re.0.30 Loss per unsold newspaper = 1.20 – 0.20 = Re.1.00 Deriving the probabilities:
From the above table we can see that the maximum expected daily profit (Rs.64) is associated with the stock level of 300 newspapers. Hence the optimal number of the newspaper, The City Times, that should be stocked is 300. Working notes: Conditional profit for any stock level = Number of newspapers demanded and sold x 0.30 – Number of unsold newspapers x 1.00. |
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Answer : (a) Reason : The number of absentees who are girls 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 girls = 11 n = No. of trials = No. of absent students = 5 x = Desired number of successes = 2 P(x = 2) = |
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Answer : (a) Reason : The value for Rapid development decision will be multiplying the value of the outcome with the probabilities of outcomes. |
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Answer : (d) Reason : When a decision maker makes decisions under conditions of risk he can assign probabilities of occurrence to the various states of nature. |
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Answer : (d) Reason : UCL = Or \LCL = |
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Answer : (a) Reason :
Since both the samples are small and the population variances are unknown the appropriate distribution is t distribution with 16 + 9 – 2 = 23 d.o.f. This is a left-tailed test. At a significance level is 0.05 the critical value is –1.714. The observed test statistic falls in the rejection region. Hence we reject the null hypothesis. |
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Answer : (c) Reason : Under the null hypothesis, the non-standardized critical values for this test are = Hence the null hypothesis will be rejected for sample means less than 9.06 and more than 14.94 and accepted for the values in between. |
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Answer : (e) Reason : Assuming the population of the voters to be large vis-à-vis the sample. Point estimate of population
proportion of voters willing to vote for party A = \90% confidence interval for proportion in favour of party A = = = (0.362, 0.412) |
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Answer : (d) Reason : A hypothesis test in which rejection of the null hypothesis occurs for values of the point estimator in either tails of the sampling distribution is called a two-tailed test. |
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Answer : (c) Reason : H0: p = 0.20 H1: p This is a large sample test of proportion. So we can use the normal
approximation to the binomial.
\
z = At a = 0.10, the critical values are ±1.645. The test statistic is less than the right tail critical value. So it falls in the acceptance region. \ We accept H0. The true proportion is 0.40. So H0 is true. Hence the test does not lead to either type I or type II error. |
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Answer : (e) Reason : Estimated standard error of difference between means:
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Answer : (c) Reason : The standard error of estimate for any
regression equation is given by se = That represents the extent to which the observed values of the dependent variable differ from the predicted values according to the regression line. The factors mentioned in the other alternatives do not represent that fact. So the alternative (c) is correct |
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Answer : (d) Reason : Or \ or or \ |
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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 = |
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Answer : (a) Reason : The movement of the dependent variable above and below the secular trend line over periods longer than one year is known as cyclical variation. |
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Answer : (b) Reason : The expected value of sampling distribution of mean = Population mean = 12. |
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Answer : (d) Reason : The standard deviation of the sampling distribution of mean for a given sample size, can be accurately known if the population variance is known. |
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Answer : (a) Reason : Other things remaining the same, if the confidence level is increased then the width of the confidence interval for the population mean will increase. |
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Answer : (c) Reason : If the population is normally distributed
then for any given sample size, the sampling distribution of |
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Answer : (b) Reason : Residual method facilitates in isolating cyclical variation from a time series. |
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Answer : (c) Reason : We use t-test of difference between two means i.e. Value of the test statistic =
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Answer : (c) Reason : Test
statistic =
\ Test statistic = |
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Answer : (c) Reason : As the degree of correlation between the two independent variables increases, the problem multicollinearity arises. Other options are wrong. |
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Answer : (e) Reason : The populations must follow the normal distribution and have equal standard deviations, but the sample sizes do not have to be the same or the sample means need not be equal. |
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Answer : (c) Reason : The difference between Paasche Index and Laspeyres Index reflects the change in consumption patterns of the commodities. Generally, Laspeyres and Paasche methods tend to produce opposite extremes in index values computed from the same data. The use of Paasche Index requires the use of new quantity weights for each period considered. By contrast to Laspeyres Index, Paasche Index generally tends to underestimate the prices or has a downward bias. As people tend to spend less on goods when their prices are rising, the use of Paasche method which is based on current weighting, produces an index which does not estimate the rise in prices showing a downward bias. Hence from above discussion, we can infer that option (c) is correct. |
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Answer : (c) Reason : We calculate the following to find the Laspeyres Index: Now the values of We know that Laspeyres index = = |
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Answer : (c) Reason : Paasches
price index = = Value
index = 126.21. |
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Answer : (b) Reason : Fisher
ideal price index = = = = = 122.47. |
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Answer : (e) Reason : In the given situation the Marshall-Edgewroth price index is equal to the Laspeyres price index
= Laspeyres price index Note: By the same logic it also be said to be equal to Paasches price index. |
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Answer : (d) Reason : Paasches
quantity index = =
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Answer : (d) Reason : Parabolic
trend equation : To determine the value of a, b and c, we solve the following normal equations: åY =
na + cåx2
åxY = b å x2 åx2Y = aåx2 + cåx4.
656 = 5a + 10c ....….(i) 195 = 10b ……(ii) 1371 = 10a + 34c …….(iii) Solving the above three equations we get the value of a = 122.772 , b = 19.5 , c = 4.214 . Therefore
the estimating equation is Y2005 = 122.772 + 19.5 ´ 3 + 4.214 ´ 32 = 219.198 or Rs.219.20 (approx.) |
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Answer : (a) Reason : The standard deviation of the sampling distribution of the mean is also called the standard error of the mean. |
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Answer : (a) Reason : To compute the coefficient of correlation between the two variables we tabulate them as below:
The mean values
r = \ r = The coefficient of determination = r2 = (0.6725)2 = 0.4523. |
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Answer : (d) Reason : The equation
of the regression line is where, X = Sales (Rs. in lakhs) Y = Total costs (Rs. in lakhs) b =
b = |
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Answer : (e) Reason : Coefficient of variation = Or
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Answer : (d) Reason : This problem requires the application of ANOVA. Employee 1: Mean Employee 2: Mean Employee 3: Mean Employee 4: Mean Grand mean Variance among the sample means: First estimate of the population
variance using variance among the sample means = = |
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Answer : (e) Reason : Required: ANOVA H0 : mSept = mOct = mNov H1 : mSept., mOct and mNov are not all equal
The means of the samples are:
The grand mean, We calculate Between – column variance as below:
We calculate Within-column variance as below:
s12 = = 81.06668. s22 = = 47.76668. s32 = = 21.50. So, = 50.1111. Therefore, F statistic = Between-column variance / Within-column variance = 334.2279 / 50.1111 = 6.67. Degrees of freedom for numerator = Number of samples – 1 = 3 – 1 = 2. Degrees of freedom for denominator = Total number of elements present in all the samples – Number of samples = 18 – 3 = 15. From F distribution table, at 0.05 level of significance, and above degrees of freedom, the critical value of F statistic comes out to be 3.68. The value of calculated F statistic of 6.67 is outside acceptance region (in rejection region) Therefore, we can conclude that the average number of shoplifters differs significantly (not same) during these months. Hence from above discussion, we can infer that option (e) is correct. |
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Answer : (c) Reason : The upper confidence limit for the
estimated population variance is given by The value of Chi square statistic from the table with 0.025 area in the left tail and (9 – 1) = 8 degrees of freedom is 2.180. \ The upper confidence limit
for the estimated population variance is |
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Answer : (e) Reason : It is given that sample standard deviation = 57: Hypothesized value of the population standard deviation = 50, Or s2 = 502 Sample Size = 30. The value of the Chi-square
statistic is |
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Answer : (a) Reason : We set the null hypothesis for this test as: ‘The variances of the two populations are same’ We set the alternative hypothesis for this test as: ‘The variances of the population B is larger than the variance of the population A’. We calculate the variances of the
two samples as follows:
The test statistic F The critical value can be found by looking at the F table with: The critical value from the table is F(10, 8, 0.05) = 3.35. As we find that the test statistic is less than the right tail critical value; so it falls in the acceptance region, and we accept the null hypothesis. Therefore we conclude that at 5% degree of significance that the variances of the two populations are equal. |
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Answer : (d) Reason : Let us take the null hypothesis on the assumption that male and female births are equally probable, the probability of male birth is p = 1/2. If H0 is true then the
number of male children in each family will follow a binomial distribution
with number of trials (n) = 5, and probability of success (p) = This is a chi-square test of
goodness of fit. It is assumed that the number of male births follows a
binomial distribution with n = 5 and p = f (x) = To get the expected frequencies, multiply f (x) by the total number n = 320. The calculations are shown below:
The value of Chi square
statistic, c2
= |
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Answer : (e) Reason : This is a chi-square test of independence The
following contingency table is obtained for the given data:
The
value of Chi square statistic, c2 = |
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Answer : (c) Reason : Given
that
The mean of proportions = 0.30 / 5 = 0.06 Upper
Control Limit for p chart = |
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Answer : (e) Reason : Lower
Control Limit for |
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Answer : (b) Reason : The expected value of a binomial probability distribution is always more than the variance. |
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Answer : (c) Reason : The
central line for the R chart = Mean of the sample range (
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Answer : (b) Reason : There is only one mode in case of a normal probability distribution |
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Answer : (a) Reason : This is a chi-square test of independence. Let us take the null hypothesis that there is no difference in the drug and sugar pills as far as their effect on curing cold is concerned (i.e., getting cured and the usage of the drug are not associated) Since it is 2 ´ 3 table, the degrees of freedom would be (2 – 1)(3 – 1) = 2, the critical value at 5% level of significance is 5.991. Expected frequency = Expected frequencies are computed and arranged in
the following table:
The value of the test statistic (chi square
statistic), c2
= We find that the chi square statistic is less than the critical value. Therefore, we accept the null hypothesis. Hence we conclude that the use of the drug and getting cured are not associated. |
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Answer : (b) Reason
: |
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Answer : (e) Reason : The Upper Control Limit for the R chart is
given as, = 1.5 ´ 1.777 = 2.666 minutes. |
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Answer : (c) Reason : The covariance of any random variable with itself is equal to the variance of the variable. |
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Answer : (b) Reason : In the linear model, the residuals are assumed to have a normal distribution with a mean of zero. |
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Answer : (c) Reason : If y is dependent on variable x then we
can write that the estimated value of y, |
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Answer : (d) Reason : The F distribution is a distribution of the ratio of two random variables each of which follows a chi-square distribution. |
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Answer : (c) Reason : Slope of the estimated regression equation of Z on X = Intercept of the estimated regression equation of Z on X =
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Answer : (b) Reason : The distribution is F distribution with the following dof: No. of degrees of freedom for the numerator = k – 1 = 4 – 1 = 3 No. of degrees of freedom for the denominator = nT – k = (4 ´ 6) – 4 = 20 \ Critical value at 5% significance level = 3.10. |
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Answer : (e) Reason : Standard
deviation of binomial distribution = |
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Answer : (e) Reason : or \ |
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Answer : (b) Reason : Coefficient of determination = r2 (= proportion of explained variations) \
Magnitude of coefficient of correlation, r = |
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Answer : (c) Reason : This is not true for index numbers. Index number is a ratio of the price, quantity or value of the commodities under study and therefore does not have any unit. |
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Answer : (a) Reason : The minimum possible value of Se is 0 (i.e. when the regression equation is a perfect estimator of dependent variable). |
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Answer : (c) Reason : By Central Limit Theorem for large samples
the sample mean is approximately normally distributed with mean = population
mean and standard deviation
From tables are find that k = –2.0 \z = –2 = or – 2 ´ 2.50 = 65 – m or m = 65 + 5 = 70. |
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Answer : (e) Reason : Fisher’s Ideal Index is the geometric mean of Laspeyres Index and Paasche’s Index. So its value falls between the values of Laspeyres and Paasches indices. Hence, from above discussion, we can infer that option (e) is false regarding Fisher’s Ideal Index. Options (a), (b), (c) and (d) are all true regarding Fisher’s Ideal Index. |
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Answer : (b) Reason : The sample size is small and the population variance is not known; so the appropriate distribution to apply for the confidence interval is t distribution with 6 – 1 = 5 degrees of freedom.
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Answer : (a) Reason : We know that the variation in Y explained
by X1 if X2 is kept constant is given by the partial
coefficient of determination denoted by Therefore = = » 0.33. i.e. 33%. |
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Answer : (a) Reason : In |
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Answer : (a) Reason : Under the null hypothesis, the non-standardized critical values for this test are =
Hence the null hypothesis will
be rejected for sample means less than |
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Answer : (d) Reason : P(-2.0 < z < -1.0) = 0.4772 – 0.3413 = 0.1359. |
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Answer : (c) Reason : The sample mean is an unbiased estimator of the population mean because the expected value of sample mean is equal to the population mean |
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Answer : (d) Reason : Multicollinearity reduces the effect of individual variables on the model. So It reduces the effect of any sensitivity analysis. |
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Answer : (c) Reason : Chi-square distribution with (4 – 1)(2 – 1) = 3 degrees of freedom. |
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Answer : (a) Reason : Other things remaining the same, if the sample size increases then the width of the confidence interval for the population mean decreases, because the standard error of mean decreases. |
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Answer : (a) Reason : Standardized test statistic = p0 = 0.40
\ or |
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Answer : (b) Reason : The population is normally distributed and the population standard deviation is known. So the standard normal distribution should be used. The test is right tailed and the significance level is 0.05. So the critical value for the test is 1.645. Since the test statistic is more than the critical value it falls in the critical region and we reject the null hypothesis. |
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Answer : (e) Reason : The central line of the Mean of the sample means or
grand mean, Mean of sample range = Lower Control Limit of = 9.56 - 0.577 ´ 8 = 4.944. (The value of A2 for sample of size 5 is 0.577 from control chart factors table) |
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Answer : (d) Reason : The means
of the samples are We calculate Within column variance as below:
s12
= s22
= s32
= So, |
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Answer : (c) Reason : The method of least squares in simple regression analysis minimizes the sum of the squared differences between the actual values and estimated values of the dependent variable. |
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Answer : (d) Reason : The maximin criterion of decision making is useful to a pessimistic decision maker under conditions of uncertainty, not under conditions of risk. |
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Answer : (e) Reason : The parameters that describe a normal random variable are mean and standard deviation. |
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Answer : (b) Reason : The mean of a continuous uniform random variable is equal to the arithmetic mean of the upper and lower limits of the range of values of the random variable. |
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Answer : (d) Reason : The confidence level being same and the population variance not being known, a confidence interval for the population mean which is narrower than another for the same population, is certainly based on a sample with lower estimated standard error of mean. |
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Answer : (e) Reason : At
the point X=100, Y=90, |
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Answer : (b) Reason : Other things remaining the same, if there are two populations with known variances then, a random sample collected from the population with the lower variance will produce a confidence interval for the population mean which is narrower than that produced by a random sample collected from the population with the higher variance. |
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= n 
= P(z < –1.43) = 1 - (0.4236 + 0.5) = 0.0764
= 200
= 0.20
= 0.30
= 0.40
= 0.10
= 1.00
.
0.20
=
=
= 0.0365.
5164
6932
= 2002
= a + bX
.
=
= 28.5.
=
=
=
=
=
= 14.5.
= 
(mean for October); 
(mean for November)
;
= [(43 + 37 + 59 + 55 + 38 + 48) + (54 + 41 + 48 + 35 + 50
+ 49) + (36 + 28 + 34 + 41+ 30 + 32 )] / 18
= 
= 334.2279.
.
= 
x (81.06668 + 47.76668 + 21.50)
.
.
)2
)2
= 80
= 83
.
= 7.16
= 180.50.
)
= 3.522
= 0.7744
and accepted for the values above 
.
chart is = Mean of the sample means (
)
= 
chart = 
(mean sale of DNA); 
(mean sales of RNA); 
(Mean sales of Oops)
.
= 
x 
= 9.083.
