Question

1. Complete the problems below.

a. Find the critical value z a/2 that corresponds to 93% confidence level.

Z a/2 = _____

Q1 (ROUND TO 2 DECIMAL PLACES)

b.You plan to conduct a survey to estimate the percentage of people who eat breakfast. Find the number of people who must be surveyed if you want to be 93% confident that the sample percentage is within two percentage points of the true population percentage.

(USE YOUR ANSWER FROM PROBLEM #1a for critical values)

Assume that nothing is known about the percentage to be estimated. n=__________Q2

Assume that studies have shown that 34% of the people eat breakfast. n=___________Q3

c.

In a survey of 1002 people, 701 said they voted in a recent presidential election. Construct a 95% confidence interval for the proportion of people who said they voted in a recent presidential election.

(ROUND TO 3 DECIMAL PLACES)

Confidence Interval ( ___________________Q4 , ___________________Q5)

Answer 1

(a) The 2 tailed critical value for = (100 - 93) / 100 = 7% = 0.07 is 1.81

(0.07/2 = 0.035, and 1 - 0.035 = 0.965. using this p value, from the standard normal tables the closest value is 1.81)

_____________________

(b) We have that the ME = Zcritical * Sqrt(p * (1-p)/n).

Therefore n = (Zcritical/ME)² * p * q

Zcritical at = 0.03 (2 tail) = 1.81 and ME is given as 2% = 0.02

(i) When nothing is known, p = 1 - p = 0.5

Therefore n = (2.81 / 0.02)² * 0.5 * 0.5 = 2047.56.

Rounding to the next Integer, n = 2048

(ii) p = 0.34, 1 - p = 1 - 0.34 = 0.66

Therefore n = (2.81 / 0.02)² * 0.34 * 0.66 = 1837.89.

Rounding to the next Integer, n = 1838

____________________

(c) = 701 / 1002 = 0.7

The Zcritical (2 tail) for = 0.05, is 1.96

The Confidence Interval is given by ME, where

The Lower Limit = 0.7 - 0.028 = 0.672

The Upper Limit = 0.7 + 0.028 = 0.728

The 95% Confidence Interval is (0.672 , 0.728)

________________________________