Answer:
a) Null and alternative hypothesis
b) Point estimate d = -$78
c) Test statistic t = -2.438
P-value = 0.0113
Reject H0. This indicates that the average automobile premium in Pennsylvania is lower than in the nation.
Step-by-step breakdown:
This is a statistical test for the average population mean.
The hypothesis posits that car insurance in Pennsylvania is notably less expensive compared to the national average.
Accordingly, the null and alternative hypotheses are:
The significance level is set at 0.05.
The sample size is n=25.
The sample mean equates to M=1425.
A point estimate of the difference between the Pennsylvania mean premium and the national average can be computed using the sample mean:
Given that the standard deviation of the population is unknown, we approximate it using the sample standard deviation, which is s=160.
The estimated standard error of the mean is determined with the formula:
Then, we can calculate the t-statistic as:
The degrees of freedom for this sample size stand at:
This constitutes a left-tailed test, with 24 degrees of freedom and t=-2.438, rendering the P-value as (per a t-table):
As the P-value (0.0113) falls below the significance level (0.05), the results prove significant.
Thus, the null hypothesis obtains dismissal.
At a 0.05 significance level, there's sufficient evidence to assert that car insurance in Pennsylvania costs notably less than the national average.
Answer: Ana should produce more fudge, and Leo should produce more toffee.
Step-by-step explanation:
Comparative advantage defined: A person or country has a comparative advantage producing a good if their opportunity cost for it is lower than someone else's.
In this case, Ana and Leo will both benefit if they focus on the product for which they have the lower opportunity cost.
(a) Ana’s production possibilities:
If she spends all her time making fudge: 3 pounds; or toffee: 2 pounds.
Opportunity cost of 1 pound fudge = 2/3 = 0.66 pounds toffee
Opportunity cost of 1 pound toffee = 3/2 = 1.5 pounds fudge
(b) Leo’s production opportunities:
All time making fudge: 4 pounds; or toffee: 5 pounds.
Opportunity cost of 1 pound fudge = 5/4 = 1.25 pounds toffee
Opportunity cost of 1 pound toffee = 4/5 = 0.8 pounds fudge
Since Ana’s opportunity cost for fudge is lower than Leo’s, she should specialize in fudge production.
Conversely, Leo has a lower opportunity cost for toffee, so he should focus on producing toffee.
Let Jacob, Carol, Geraldo, Meg, Earvin, Dora, Adam, and Sally be denoted as J, C, G, M, E, D, A, and S respectively. In part IV, we need to identify the pairs of potential clients that could potentially be selected. The sample space consists of all possible outcomes, therefore we create a set of all valid pairs, listed as follows: {(J, C), (J, G), (J, M), (J, E), (J, D), (J, A), (J, S), (C, G), (C, M), (C, E), (C, D), (C, A), (C, S), (G, M), (G, E), (G, D), (G, A), (G, S), (M, E), (M, D), (M, A), (M, S), (E, D), (E, A), (E, S), (D, A), (D, S), (A, S)}. We can verify the number of elements in the sample space, n(S) is 1+2+3+4+5+6+7=28. This gives us the answer to the first question: What is the count of pairs of potential clients that can be randomly selected from the pool of eight candidates? (Answer: 28.) II) What is the chance of a certain pair being chosen? The chance of picking a specific pair is 1/28, as there’s just one way to select a particular pair out of the 28 possible options. III) What is the probability that the selected pair consists of Jacob and Meg or Geraldo and Sally? The probability of selecting (J, M) or (G, S) is 2 out of 28, which equates to 1/14. Answers: I) 28 II) 1/28 ≈ 0.0357 III) 1/14 ≈ 0.0714 IV) {(J, C), (J, G), (J, M), (J, E), (J, D), (J, A), (J, S), (C, G), (C, M), (C, E), (C, D), (C, A), (C, S), (G, M), (G, E), (G, D), (G, A), (G, S), (M, E), (M, D), (M, A), (M, S), (E, D), (E, A), (E, S), (D, A), (D, S), (A, S).}
Response: 7
Detailed explanation:
A Venn diagram can help visualize this problem.
There are a total of 5 students interested in both French and Latin.
Out of these, 3 students also want to learn Spanish, meaning only 2 students want solely French and Latin.
Moreover, there are 5 students who wish to study only Latin.
This results in 1 student wanting both Latin and Spanish, calculated by 11 - 5 - 3 - 2.
There are 8 students who desire only Spanish, and 4 students who want both Spanish and French.
In the same manner, those wishing to study only French amount to 16 - 4 - 3 - 2 = 7.
