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2 answers:
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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:
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:
Then, we can calculate the z-statistic as:
This test is a two-tailed test, so the P-value for this test is calculated as:
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.
Answer:
She can visit the cities in 192 ways
Step-by-step explanation:
If Sally starts from city A and after going to 4 cities she returns to home,
then that can be in 4! = 24 ways. (permutation of 4 distinct objects)
If between the time she visits cities, she comes home once,
then cities can be visited in,
= 72 ways. (permutation of 4 distinct objects and choosing 1 gap among the 3 )
If between the time she visits cities, she comes home twice,
then cities can be visited in,
=72 ways. (permutation of 4 distinct objects and finding 2 gaps among the 3)
If between the time she visits cities, she comes home thrice,
then cities can be visited in,
= 24 ways. (permutation of 4 distinct objects and choosing 3 gaps among the 3)
Now, she can't come home more than thrice in between the time she visits cities (since there are 4 cities to visit only)
So, the number of ways she can visit cities = (24+ 72 + 72 + 24)
= 192