An emergency fund is defined as a savings account that has a balance equal to at least two months' living expenses. An article in a financial magazine claims that 80% of American adults do not have an emergency fund. To investigate this claim, a financial advisor selects a random sample of 150 Americans and finds that 112 do not have an emergency fund. The financial advisor would like to know if the data provide convincing evidence that the true proportion of American adults who do not have an emergency fund is less than 80%.

Are the conditions for inference met for conducting a z-test for one proportion?

A. Yes, the random, 10%, and large counts conditions are all met.
B. No, the random condition is not met.
C. No, the 10% condition is not met.
D. No, the large counts condition is not met.

The conditions for inference needed to conduct a z-test for one proportion are met, including the random condition, the 10% condition, and the large counts condition, based on the information given.

Therefore, the correct answer is: option "Yes, the random, 10%, and large counts conditions are all met.

Explanation:

To determine if the conditions for inference are met for conducting a z-test for one proportion, we must check the following conditions:

Random: The sample of Americans must be chosen randomly.
10% Condition: The sample size must be no more than 10% of the entire population.
Large Counts Condition: Both np and n(1-p) must be at least 10, where n is the sample size and p is the proportion in question.

Assuming the financial advisor has selected a random sample, the random condition is met.

As the population of American adults is large, well over 1500, the 10% condition is met.

For the large counts condition, we can calculate np = 150 x 0.8 = 120 and n(1-p) = 150 x 0.2 = 30.

Since both of these are greater than 10, the large counts condition is also met. Therefore, all conditions for conducting a z-test for one proportion are satisfied.