Question

Randi wants to know if at least 90% of the employees at her company are currently enrolled in a health insurance plan. She randomly samples 700 employees and finds that 651 of them are currently enrolled in a plan. Randi conducts a one-proportion hypothesis test at the 5% significance level, to test whether the true proportion of employees enrolled in a plan at this company is greater than 90%. For this test: H0:p=0.9; Ha:p>0.9, which is a right-tailed test. Use a TI-83, TI-83 Plus, or TI-84 calculator to test whether true proportion of employees enrolled in a health insurance plan at this company is greater than 90%. Identify the test statistic, z, and p-value from the calculator output, rounding to three decimal places.

Answer 1

Solution:

Given ,

p₀ = 90% = 0.90

Claim: p is greater than 90% i.e.p > 0.90

The hypothesis are

H₀ : p = 0.90 vs H₁ : p > 0.90

n = 700

x = 651

Use TI-83, TI-83 Plus, or TI-84 calculator to perform z test for proportion . 1 -PropZ Test

Press STAT and go to TESTS menu.

Go to 1 -PropZ TEST

p₀ : 0.90

x: 651

n : 700

Prop > p₀

calculate :

From output,

Test statistic z = 2.646

p value = 0.004

Since p value is less than given significance level 0.05 , we reject the null hypothesis.

There is sufficient evidence to support the claim that  true proportion of employees enrolled in a health insurance plan at this company is greater than 90%.