Statistics for Economists Homework no. 1 Answer each of the questions below. Each question is worth 5 points. 1. Counting Rules

a) Suppose you plan to invest equal amounts of money in two of four ventures. How many ways can you arrange your investment portfolio assuming that you can invest in the same venture twice? Ans. 4 x 4=24 =16

b) Assume you can only invest once for each venture. How many different combinations of investments can you make if you select two from the four ventures?

Ans. 4! 6

2!(4 2)!=

−

c) Assume you can only invest once for each venture, and you have to rank your investments as top priority and low priority. If you select two from the four, how many different selections can you make where the differences in ranking is

considered a separate investment? Ans. 4! 12

(4 2)!=

−

2. Two methods were used to teach a high school algebra course. A sample of 75 scores was

selected for method 1, and a sample of 60 scores was selected for method 2, with the summary results given in the following: Method 1 Method 2 Sample mean 85 76 Sample Standard deviation 3 2 a) Compute for the standard deviation of the distribution of the differences between the

two means. b) Construct a 99 percent confidence interval for the difference in the mean scores for

the two methods. c) What can you conclude about these methods?

Answer: From the information given, we have nl = 75, n2 = 60, 1x = 85, 2x = 76, s1 = 3, s2=2,

2 21 2

1 2

s sn n+ =0.4320, α = 0.01, zα/2 =2.576. Thus, the 99 percent confidence interval estimate for

the difference of the means for the two methods (method 1 - method 2) is (85 -76) ± 2.576 x 0.3981 = 9 ± 1.1130. That is, we are 99 percent confident that the difference between the mean scores for the two teaching methods will lie between 10.1130 and 7.8870. Since both limits are positive, one may conclude that method 1 seems to be the better of the two methods.

3. Interval Estimation. a) In a random sample of 100 persons, 77 percent of them said that when they pray, they

pray for world peace. Determine a 90 percent confidence interval estimate for the true proportion of people who pray for world peace.

Answer

Since p̂ = 0.77, then p̂σ = ˆ ˆ(1 )p p− = 0 .0421. Since we need to find the 90 percent confidence interval, α= 10 percent = 0.1. From normal distribution table, zα/2 = 1.645. Thus, the 90 percent confidence interval estimate for the proportion of people who pray for world peace, using the formula, is 0.77 ± 1.645 x 0.0421 = 0.77 ±0.0692. That is, we are 90 percent confident that the true proportion of people who pray for world peace will lie between 0.7008 and 0.8392, or between 70.08 and 83.92 percent.

b) A researcher wants to determine the difference between the proportions of males and

females who believe in aliens. If a margin of error of ±0.02 is acceptable at the 95 percent confidence level, what is the minimum sample size that should be taken? Assume that equal sample sizes are selected for the two sample proportions.

Answer: In considering a large-sample confidence interval for the difference between two population proportions, that is, when nlpl > 5, nl(l -pl) > 5, n2p2 > 5, and n2(l -p2) > 5, we may need to estimate the sample sizes needed in order to collect data. The formula given here is for when the sample sizes are the same. In the absence of a variance, one has to make preliminary estimates of the true variance of the sample distribution. In this case, the often used estimate for the standard deviation = ( *(1 *))p p− is 0.5. [Your answer in this case depends on your assumed estimate of the standard deviation.] Also, this will be the equation for the minimum sample size. Based on the formula above, the solution for n is given below:

2/ 20.5* zn

Eα =

. Since we are given α=0.05 and E=0.02, then

21.960.5* 48020.02

n = =

This means that the sample should have at least 4802 males and 4802 females.

c) A random sample of size n1 = 36 selected from a normal distribution with standard deviation l = 4 has a mean 1x = 75. A second random sample of size n2 = 25 selected

from a different normal distribution with a standard deviation 2 = 6 has a mean 2x = 85. Find a 95 percent confidence interval for µ1 - µ2.

Answer: From the information given, we have nl = 36, n2 = 25, 1x = 75, 2x = 85, 1 = 4, 2 =

6,2 21 2

1 2n nσ σ

+ = 1.3728, = 0.05,zα/2 = 1.96. Thus, the 95 percent confidence interval estimate

for µ1 - µ2 is (75 - 85) ± 1.96 x 1.3728 = -10 ±2.6907. That is, we are 95 percent confident that the difference between the means will lie between -12.6907 and -7.3093. Since both limits are negative, one may conclude that the mean from population 2 is larger than the mean from population 1.