Question

.           Matwirth corporation has instituted a new disciplinary policy for employees who fail a random drug test. The personnel office wishes to test to see if there is a difference in employee behavior based on the new policy. The records of 200 employee drug tests from before the new policy were randomly selected and 200 drug test results from after the new policy were also randomly selected. In the sample from before the new policy, 3.50% tested positive, while 1.00% in the sample from after the new policy tested positive.

            For the hypotheses,

            H₀: p₁ = p₂   and   H_A: p₁ > p₂,

            calculate the test statistic.

Answer 1

Solution :

Given that,

n₁ = 200

Point estimate = sample proportion = ₁ = x₁ / n₁ = 0.0350

n₂ = 200

Point estimate = sample proportion = ₂ = x₂ / n₂ = 0.001

The value of the pooled proportion is computed as,

= (1 n1 + ₂ n2 ) / ( n₁ + n₂ )

= (7 + 2 ) / ( 200 + 200)

= 0.0225

1 - = 0.9775

Test statistics

z = (₁ - ₂ ) / *(1-) ( 1/n₁ + 1/n₂ )

= (0.0350 - 0.01 ) / ( 0.0225 * 0.9775) (1/200 + 1/200 )

= 1.686