Question

1. A university Dean is interested in determining the proportion of students who receive
some sort of financial aid. Rather than examine the records for all students, the dean randomly selects
200 students and finds that 118 of them are receiving financial aid. By hand, and using formulas for the
“Margin of Error” (i.e. not calculator confidence interval), find a 98% confidence interval to estimate
the true proportion of students on financial aid. (Use 3 decimal places for proportions.)

2. Using the t-Distribution Tables, find the critical t-value that corresponds to a 99%
confidence, with n =20 .

3. A researcher wishes to estimate the current proportion of the households with two
or more computers. How large a sample is needed in order to be 95% confident that the sample
proportion will not differ from the true population proportion by more than 4%? (A previous older
study indicated that 20% of households had two or more computers in their household.)

4. The mean replacement time for a random sample of 12 cd players is 8.6 years , with a
standard deviation of 3.9 years . Construct a 90% confidence interval for the population standard
deviation,  . (Assume the data are normally distributed.) (Use 2 decimal places.)

Answer 1

We would be looking at question 1 here :

From standard normal tables, we get:
P(Z < 2.326) = 0.99

Therefore, due to symmetry, we get here:
P( -2.326 < Z < 2.326) = 0.98

The sample proportion here is computed as:
P = x/n = 118/200 = 0.59

Therefore the confidence interval here for the true proportion is computed as:

This is the required 98% confidence interval for proportion here.