Late payment of medical claims can add to the cost of
health care. An article (M. Freudenheim, “The Check Is Not
in the Mail,” The New York Times, May 25, 2006, pp. C1,
C6) reported that for one insurance company, 85.1% of the
claims were paid in full when first submitted. Suppose that
the insurance company developed a new payment system in
an effort to increase this percentage. A sample of 200 claims
processed under this system revealed that 180 of the claims
were paid in full when first submitted.
a. At the 0.05 level of significance, is there evidence that
the population proportion of claims processed under this
new system is higher than the article reported for the previous
system?
b. Compute the p-value and interpret its meaning.
To test if there is evidence that the population proportion of claims processed under the new system is higher than what was reported in the article, we can perform a hypothesis test.
a. The null and alternative hypotheses for this test are:
Null hypothesis (H0): The population proportion of claims paid in full under the new system is the same as what was reported in the article (p = 0.851).
Alternative hypothesis (Ha): The population proportion of claims paid in full under the new system is higher than what was reported in the article (p > 0.851).
We will conduct a one-sample proportion z-test at the 0.05 level of significance.
b. To compute the p-value, we can use the normal distribution table or a statistical software. However, since we have the sample size of 200 and the sample proportion of 180/200 = 0.9, we can use the standard normal distribution to find the p-value.
The test statistic (z) can be calculated using the formula:
z = (p̂ - p) / √(p(1-p)/n)
Where p̂ is the sample proportion, p is the hypothesized population proportion, and n is the sample size.
Using the given values, we have:
p̂ = 0.9
p = 0.851
n = 200
Calculating the test statistic:
z = (0.9 - 0.851) / √(0.851(1-0.851)/200)
z = 0.049 / √(0.7257/200)
z ≈ 0.049 / 0.03122
z ≈ 1.569
Now, we can find the p-value associated with this test statistic. The p-value is the probability of observing a test statistic as extreme as or more extreme than the one calculated, assuming the null hypothesis is true.
Referring to the standard normal distribution table, we find that the probability associated with a z-value of 1.569 is approximately 0.0583.
Interpreting the meaning of the p-value:
The p-value of 0.0583 means that there is approximately a 5.83% chance of observing a sample proportion as extreme as, or more extreme than, 0.9, assuming the null hypothesis is true.
Since the p-value (0.0583) is greater than the significance level (0.05), we fail to reject the null hypothesis. This means that there is not enough evidence to conclude that the population proportion of claims paid in full under the new system is higher than what was reported in the article.
To answer these questions, we need to conduct a hypothesis test for a proportion. Let's go through the step-by-step process.
Step 1: State the hypotheses:
The null hypothesis (H0): The population proportion of claims processed under the new system is equal to the proportion reported for the previous system (p = 0.851).
The alternative hypothesis (Ha): The population proportion of claims processed under the new system is higher than the proportion reported for the previous system (p > 0.851).
Step 2: Set the significance level:
The significance level (α) is given as 0.05.
Step 3: Compute the test statistic:
We will use the z-test formula for testing proportions. The formula is:
z = (p̂ - p) / √(p * (1 - p) / n)
Where:
p̂ is the sample proportion (180/200 = 0.9)
p is the proportion reported for the previous system (0.851)
n is the sample size (200)
Substituting the values, we get:
z = (0.9 - 0.851) / √(0.851 * (1 - 0.851) / 200)
Step 4: Determine the critical value:
Since the alternative hypothesis is one-tailed (we are testing if the proportion is higher), and the significance level is 0.05, we need to calculate the critical value for a one-tailed test using a z-table.
Looking up the critical value at α = 0.05 (one-tailed), we find it to be approximately 1.645.
Step 5: Compute the test statistic and make a decision:
Calculating the value of z from the formula, we find:
z = 3.084
Since z (3.084) is greater than the critical value (1.645), we can reject the null hypothesis.
Step 6: Compute the p-value:
The p-value is the probability of observing a test statistic as extreme as the one calculated (or more extreme) under the null hypothesis.
To find the p-value, we use the z-score to find the corresponding area under the standard normal curve using a z-table. The p-value can be calculated as P(Z > z), where Z represents the standard normal distribution.
From the z-table, the area to the right of z = 3.084 is approximately 0.0005.
Since this is a one-tailed test, the p-value is 0.0005.
Step 7: Interpret the p-value:
The p-value obtained is 0.0005. This means that if the null hypothesis is true (the proportion of claims processed under the new system is equal to the proportion reported for the previous system), the probability of observing a sample with a sample proportion of 0.9 or higher would only be 0.0005.
Since the p-value is less than the significance level (0.05), we can reject the null hypothesis. There is evidence that the population proportion of claims processed under the new system is higher than the proportion reported for the previous system.
In summary:
a. Yes, there is evidence at the 0.05 level of significance to suggest that the population proportion of claims processed under the new system is higher than the proportion reported for the previous system.
b. The p-value is 0.0005, which represents a very small probability of observing a sample proportion as extreme as 0.9 or higher under the null hypothesis.