Foundations for Econometrics Matriculation

Name: EC2303 Foundations for Econometrics Matriculation #: PS4 Tutorial Section (W#):

Week 9: Problem Set 4

DUE: Monday 21 October, 12pm

• Please hand in to the economics office (AS2 – L6)

• Late assignments will not be accepted.

• Show your work to get full credit.

• Carry through fractions or exact decimals when possible, or round to four decimal places. Please round final answers to four decimal places.

1. NCT 7.7 Suppose that x1 and x2 are random samples of observations from a population with mean µ and variance s2. Consider the following three point estimators, X, Y , and Z of µ:

X = 1

2 x1 +

1

2 x2

Y = 1

4 x1 +

3

4 x2

Z = 1

3 x1 +

2

3 x2

(a) Show that all three estimators are unbiased

(b) Which of these estimators is the most efficient?

(c) Find the relative efficiency of X with respect to each of the two other estimators

2. NCT 7.13 A personnel manager has found that historically, the scores on aptitude tests given to applicants for entry-level positions follow a normal distribution with a standard deviation of 32.4 points. A random sample of nine test scores from the current group of applicants had a mean score of 187.9 points. Find an 80% confidence interval for the population mean score of the current group of applicants.

3. NCT 7.47 The quality control manager of a chemical company randomly sampled twenty 100-pound bags of fertilizer to estimate the variance in the pounds of impurities. The sample variance was found to be 6.62. Find a 95% confidence interval for the population variance in the pounds of impurities. What assumption, if any, did we make about the underlying population distribution?

4. NCT 7.69 A politician wants to estimate the proportion of constituents favoring a contro- versial piece of proposed legislation. Suppose that a 99% confidence interval that extends at most 0.05 on each side of the sample proportion is required. How many sample obser- vations are needed?

1

 

 

EC2303 – Foundations for Econometrics Version: October 14, 2013

5. NCT 7.77 An instructor in a class of 417 students is considering the possibility of a take- home final examination. She wants to take a random sample of class members to estimate the proportion who prefer this form of examination. If a 90% confidence interval for the population must extend at most 0.04 on each side of the sample proportion, how large a sample is needed?

6. NCT 7.79 Suppose that the owner of a recently opened convenience store in Kuala Lumpur, Malaysia, wants to estimate how many pounds of bananas are sold during a typical day. The owner checks his sales records for a random sample of 16 days and establishes that the mean number of pounds sold per day is 75 pounds, and that the sample standard deviation is 6 pounds. Estimate the mean number of pounds the owner should stock each day to a 95% confidence level.

7. NCT 7.81 The following data represent the number of audience members per week at a theater in Paris during during the last year (The theater was closed for 2 weeks for refur- bishment). Estimate overall average weekly attendance with a 95% interval estimate.

163 165 94 137 123 95 170 96 117 129

152 138 147 119 166 125 148 180 152 149

167 120 129 159 150 119 113 147 169 151

116 150 110 110 143 90 134 145 156 165

174 133 128 100 86 148 139 150 145 100

8. NCT 7.97 A corporation employes 148 sales representatives. A random sample of 60 of them was taken, and it was found that, for 36 of the sample members, the volume of orders taken this month was higher than for the same month last year. Find a 95% confidence interval for the population proportion of sales representatives with a higher volumes of orders.

9. NCT 7.99 The president’s policy on domestic affairs received a 45% approval rating in a recent poll. The margin of error was given as 0.035. What sample size was used for this poll if we assume a 95% confidence level?

10. For a finite sample taken without replacement that is large relative to the population,

σ2 X̄

= σ2

n

N − n N − 1

.

(a) Show that n = Nσ2

(N − 1)σ2 X̄ + σ2

(b) Show that n = n0N

n0 + (N − 1) , where n0 =

z2 α/2

σ2

ME2

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