Question

1. Suppose we wish to find the required sample size to find a 90% confidence interval for the population proportion with the desired margin of error. If there is no rough estimate   of the population proportion, what value should be assumed for  ?

		0.90
		0.10
		0.50
		0.05

2. An analyst takes a random sample of 25 firms in the telecommunications industry and constructs a confidence interval for the mean return for the prior year. Holding all else constant, if he increased the sample size to 30 firms, how are the standard error of the mean and the width of the confidence interval affected?

	Standard error of the mean	Width of confidence interval
A	Increases	Becomes wider
B	Increases	Becomes narrower
C	Decreases	Becomes wider
D	Decreases	Becomes narrower

		1.96(10.24/6)
		1.96(3.20/6)
		1.645(3.20/6)
		1.645(10.24/6)

3. For a given confidence level and population standard deviation, which of the following is true in the interval estimation of the population mean?

		If the sample size is bigger, the interval is narrower.
		If the population size is smaller, the interval is narrower.
		If the population size is bigger, the interval is narrower.
		If the sample size is smaller, the interval is narrower.

Answer 1

1. The margin of error- ME = (critical value)* where is the sample proportion, n is the sample size i.e number of samples under consideration.

If the value of is not given, we assume it to be 0.5

2. The confidence interval for the mean of a population, when sample size n, the sample mean , and standard deviation s, is given is [(critical value)*s/√n], where the critical value, z* depends on the % of confidence we want to calculate. i.e. for 95% confidence interval, consider =(100-95)/100=0.05.

Find the z-score for P(-z*≤ Z ≤ z*)= 0.95. such that the areas to the left of -z* and to the right of z* is 0.025= /2

Now given n = 25, standard error of the mean =s/√n= s/5

Now keep everything else constant and change n=30

Thus new standard error of the mean = s/√30=s/5.48 which implies the standard error for mean decreased.

Since standard error has decreased, the margin of error ME=(Critical value)*(Standard Error) decreases, thus the width of confidence interval also decreases.

Therefore, as sample size increases, Standard error decreases and the width of confidence interval also decreases

Option D

3. As explained in question.2 for a given confidence level and population standard deviation, if the sample size is bigger, the width of the confidence interval for the population mean becomes narrower