Question

Suppose you are a researcher studying the study habits of college students. The parameter you wish to measure is the number of minutes students study each week. From previous research you know that this parameter is distributed Normally and has a standard deviation of 50 minutes. (Please only answer one part at a time to give other students a chance to answer as well! Start with the first one!)

1. If you wish to construct a 90% confidence interval that has a margin of error of 5 minutes, how many students do you need in your sample?

2. Suppose you wish to construct a 90% confidence interval that has a margin of error of only 2 minutes. How will this affect the number of students you need in your sample? How many students will you need?

3. Repeat the calculation in Part (1) above for an 80% confidence interval.

4. Repeat the calculation in Part (2) above for an 80% confidence interval.

Answer 1

The standard deviation is known from the previous study =   = 50 minutes.

1. If you wish to construct a 90% confidence interval that has a margin of error of 5 minutes, how many students do you need in your sample?

Here we need to find sample size ( n ).

The formula of sample size is given by:

............. ( 1 )

Where is a critical value for given level of significance.

E = margin of erro = 5

Let's find the value of Z_c

c = confidence level = 0.90

Therefore level of significance = = 1 - c = 1 - 0.90 = 0.10

/2 = 0.10 / 2 = 0.05

We want to find Z_c such that P( Z > Z_c ) = /2 = 0.05

This implies that  P( Z < Z_c ) = 1 -  0.05 = 0.95

Let's use excel:

Z_c = "=NORMSINV(0.95)" = 1.645

E = 5 and = 50

Plug these values in equation 1), so we get

So answer of first part is 271

Note that if we get n in decimal form then we round it to the next integer. Because sample size is not in decimal form.