Question

Suppose we know that 90% of people who seek treatment at the ER will list nausea as the presenting symptom. In a sample of 200 emergency admissions with a certain diagnosis at Kings County Hospital (Brooklyn, NY), 165 listed nausea as the presenting symptom. Does the data provide sufficient evidence to indicate that the percentage is less than 90%? Use the 6-step hypothesis testing procedure to answer this question.

STEP 1: STATE THE APPROPRIATE NULL AND ALTERNATIVE HYPOTHESES

STEP 2: DEFINE THE CRITICAL REGION

STEP 3: COMPUTE THE APPROPRIATE TEST STATISTIC

STEP 4: STATE YOUR STATISTICAL DECISION

STEP 5: STATE YOUR PRACTICAL CONCLUSION

STEP 6: REPORT THE P-VALUE

Answer 1

Solution :

This is the left tailed test .

The null and alternative hypothesis is

H₀ : p = 0.90

H_a : p < 0.90

n = 200

x = 165

= x / n = 165 / 200 = 0.825

P₀ = 0.90

1 - P₀ = 1 - 0.90 = 0.10

z = - P₀ / [P_{0 *} (1 - P₀ ) / n]

= 0.825 - 0.90 / [(0.90 * 0.10) / 200]

= -3.536

Test statistic = -3.536

P(z < -3.536) = 0.0002

P-value = 0.0002

= 0.05

P-value <

Reject the null hypothesis .

There is sufficient evidence to indicate that the percentage is less than 90% .