Question

Question 1

An official of a large national union claims one-half of the union are women.

1. Using the sample information of 168 men and 232 women from the union, outline a six-step hypothesis test of the official’s claim.

2. Using the sample in part (a) as a pilot, how large of a sample would have to be taken to provide a margin of error of 0.03 or less? Show workings.

Answer 1

Answer)

A)

Ho : P= 0.5

Ha : p is not equal to 0.5

N = 168 + 232 = 400

P = 0.5

First we need to check the conditions of normality that is if n*p and n*(1-p) both are greater than 5 or not

N*p = 200

N*(1-p) = 200

Both the conditions are met so we can use standard normal z table to estimate the P-Value

Test statistics z = (oberved p - claimed p)/standard error

Standard error = √{claimed p*(1-claimed p)/√n

N = 400

Observed p = 232/400

Claimed p = 0.5

After substitution

Z = 3.2

From z table, P(z>3.2) = 0.0007

But our test is two tailed so, P-Value is = 2*0.0007 = 0.0014

As the P-Value is small, we reject the null hypothesis

For part b you have not provided the confidence level

Or alpha to estimate the sample size

But formula for margin of error is

MOE = z*√p*(1-p)/√n

MOE = 0.03 (given in the question)

Z = critical value for confidence level

(For 95%, critical value is 1.96

For 90%, critical value is 1.645

For 99%, critical value is 2.58)

P = point estimate

Substitute all the values to get the value of N