Question

A billing company that collects bills for​ doctors' offices in the area is concerned that the percentage of bills being paid by medical insurance has risen.​ Historically, that percentage has been 30​%. An examination of 9 comma 260 ​recent, randomly selected bills reveals that 31​% of these bills are being paid by medical insurance. Is there evidence of a change in the percent of bills being paid by medical​ insurance?
1) Write the appropriate hypotheses. Let p be the proportion of bills being paid by medical insurance.
A. H0: p = 0.30
HA: p not equal to 0.30
B. H0: p = 0.29
HA: p note qual to 0.29
C. H0: p = 0.30
HA: p > 0.30
D. H0: p = 0.30
HA: p < 0.30
E. H0: p = 0.29
HA: p < 0.29
F. H0: p = 0.29
HA: p > 0.29
2) Check the assumptions and conditions.
A. The Independence Assumption is met.
B. The 10% Condition is met.
C. The Randomization Condition is met.
D. The Success/Failure Condition is met.
3) Perform the test and find the P-value. What is the test statistic? What is the P-value?
4) State your conclusion.
A. Fail to reject H0. There is not sufficient evidence to suggest that the percentage of bills paid by medical insurance has changed.
B. Fail to reject H0. There is sufficient evidence to suggest that the percentage of bills paid by medical insurance has changed.
C. Reject H0. There is sufficient evidence to suggest that the percentage of bills paid by medical insurance has changed.
D. Reject H0. There is not sufficient evidence to suggest that the percentage of bills paid by medical insurance has changed.
E) Do you think this difference is meaningful?
A. Yes, the conclusion of the hypothesis test shows that the difference has financial significance to the company.
B. No, the conclusion of the hypothesis test shows that the difference has no financial significance to the company.
C. Not necessarily. The difference is statistically significant, but the actual change in the proportion is small. We cannot tell from the P-value alone if the difference has financial significance to the company.Get more help from Chegg

Answer 1

Answer:

1) A. H0: p = 0.30

HA: p not equal to 0.30

2) A. The Independence Assumption is met.

C. The Randomization Condition is met.

D. The Success/Failure Condition is met.

3) Test statistic z = 2.089

P-value = 0.0367

4) C. Reject H0. There is sufficient evidence to suggest that the percentage of bills paid by medical insurance has changed.

Step-by-step explanation:

1) This is a hypothesis test for a proportion.

The claim is that there is a significant change in the percent of bills being paid by medical​ insurance.

As we are looking for evidence of a difference, no matter if it is higher or lower than the null hypothesis proportion, the alternative hypothesis is defined by a unequal sign.

Then, the null and alternative hypothesis are:

$H_0: \pi=0.3\\\\H_a:\pi\neq 0.3$

2) Cheking the conditions:

The independence assumption and the randomization condition are met as the bills were selected randomly from the population.

The 10% condition can not be checked, as we do not know the size of the population.

The success/failure condition is met as the products np and n(1-p) are bigger than 10 (the number of successes and failures are both bigger than 10).

3) The significance level is assumed to be 0.05.

The sample has a size n=9260.

The sample proportion is p=0.31.

 

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.3*0.7}{9260}}\\\\\\ \sigma_p=\sqrt{0.000023}=0.005$

Then, we can calculate the z-statistic as:

$z=\dfrac{p-\pi-0.5/n}{\sigma_p}=\dfrac{0.31-0.3-0.5/9260}{0.005}=\dfrac{0.01}{0.005}=2.089$

This test is a two-tailed test, so the P-value for this test is calculated as:

$\text{P-value}=2\cdot P(z>2.089)=0.0367$

As the P-value (0.0184) is smaller than the significance level (0.05), the effect is  significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that there is a significant change in the percent of bills being paid by medical​ insurance.