Students with D's / total students : 3 / (7 + 9 + 11 + 3 + 2) =
3 / 32 = 0.0937...to turn to a percent, multiply by 100 = 9.375....rounded =
9.4% <== ur probability

3 0

2 years ago

A billing company that collects bills for doctors' offices in the area is concerned that the percentage of bills being paid by

omeli [17]

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.

5 0

2 years ago

A study was done to determine the average number of homes that a homeowner owns in his or her lifetime. For the 60 homeowners su

const2013 [10]

Answer:

D) (3.67, 4.73)

Step-by-step explanation:

Confidence Interval for the true average number of homes that a person owns in his or her lifetimecan be calculated using M±ME where

M is the average number of home owned (4.2)

ME is the margin of error from the mean

And margin of error (ME) can be calculated as

ME= $\frac{z*s}{\sqrt{N} }$ where

z is the corresponding statistic in the given confidence level(1.96)

s is the standard deviation of the sample(2.1)

N is the sample size (60)

Putting the numbers we get ME= $\frac{1.96*2.1}{\sqrt{60} }$ ≈0.53

Then the 95% confidence interval is 4.2±0.53 or (3.67, 4.73)

6 0

2 years ago