Respuesta :
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:
[tex]H_0: \pi=0.3\\\\H_a:\pi\neq 0.3[/tex]
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:
[tex]\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.3*0.7}{9260}}\\\\\\ \sigma_p=\sqrt{0.000023}=0.005[/tex]
Then, we can calculate the z-statistic as:
[tex]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[/tex]
This test is a two-tailed test, so the P-value for this test is calculated as:
[tex]\text{P-value}=2\cdot P(z>2.089)=0.0367[/tex]
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.
