A catering company’s recipe for salad uses a ratio of 2 c of dressing to 4 lb of vegetables. (Hint: c = cups; lb = pounds) (a) H
ow many cups of dressing will be needed if 30 pounds of vegetables are used? (b) Draw a model for the ratio. The butterfly method works well for this.
1 answer:
Answer:
A. 15 cups
B. Diagram below
We are given that,
The ratio of dressing to the vegetables is 2 cups : 4 pounds.
A. It is given that the vegetables are of 30 pounds.
Let the number of cups be x.
Since, the ratio of dressings and vegetables will remain same.
We have,
i.e.
i.e. x= 15
Thus, we will need 15 cups of dressings for 30 pounds of vegetables.
B. As we have obtained the fractions,
2 cups : 4 pounds and 15 cups : 30 pounds
Using the butterfly method, we see from the diagram that these fractions are equivalent because the product of the numbers is equal to 60.
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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.