Answer with A B C D or #. Correct answer gets brainlest.
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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:
a) H0: There is no association between level of education and TV station preference (Independence)
H1: There is association between level of education and TV station preference (No independence)
b)
c)
d) Since the p value is lower than the significance level we enough evidence to reject the null hypothesis at 5% of significance, and we can conclude that we have dependence between the two variables analyzed.
Step-by-step explanation:
A chi-square goodness of fit test "determines if a sample data matches a population".
A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".
Assume the following dataset:
High school Some College Bachelor or higher Total
Public Broadcasting 15 15 10 40
Commercial stations 5 25 10 40
Total 20 40 20 80
We need to conduct a chi square test in order to check the following hypothesis:
Part a
H0: There is no association between level of education and TV station preference (Independence)
H1: There is association between level of education and TV station preference (No independence)
The level os significance assumed for this case is
The statistic to check the hypothesis is given by:
Part b
The table given represent the observed values, we just need to calculate the expected values with the following formula
And the calculations are given by:
And the expected values are given by:
High school Some College Bachelor or higher Total
Public Broadcasting 10 20 10 40
Commercial stations 10 10 20 40
Total 20 30 30 80
Part b
And now we can calculate the statistic:
Now we can calculate the degrees of freedom for the statistic given by:
Part c
In order to find the critical value we need to look on the right tail of the chi square distribution with 2 degrees of freedom a value that accumulates 0.05 of the area. And this value is
Part d
And we can calculate the p value given by:
And we can find the p value using the following excel code:
"=1-CHISQ.DIST(33.75,2,TRUE)"
Since the p value is lower than the significance level we enough evidence to reject the null hypothesis at 5% of significance, and we can conclude that we have dependence between the two variables analyzed.