In a sample of 49 adolescents who served as the subjects in an immunologic study, one variable of interest was the diameter of a skin test reaction to an allergen. The sample mean and standard deviation were 21 and 11 mm erythema, respectively. a. Use a z- or t-distribution? Why?

b. One-sided or two-sided test? Why?

c. Can it be concluded from these data that the population mean is less than 24 mm erythema?

d. What is the range on the p-value?

Answers

Answer 1

Answer:

Step-by-step explanation:

a) We would use a t distribution because the population standard deviation is unknown.

b) it is a one sided test because we are trying to determine if the population mean is less than 24 mm erythema. The lesser than means that it is a left tailed test.

c) To determine the p value, we would apply the formula,

t = (x - µ)/(s/√n)

Where

x = sample mean = 21

µ = population mean = 24

s = samples standard deviation = 11

n = number of samples = 49

t = (21 - 24)/(11/√49) = - 1.91

Since n = 49

Degrees of freedom, df = n - 1 = 49 - 1 = 48

We would determine the p value using the t test calculator. It becomes

p = 0.031

Assuming alpha = 0.05, then

Since alpha, 0.05 > than the p value, 0.031, then we would reject the null hypothesis. Therefore, At a 5% level of significance, it can be concluded from these data that the population mean is less than 24 mm erythema.

d) p value = 0.031

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Suppose that an experiment has five possible outcomes, which are denoted {1,2,3,4,5}. Let A be the event {1,3,4} and let B be the event {2,4,5}. (Notice that we did not say that the five outcomes are equally likely: the probability distributions could be anything.) For each of the following relations, tell whether it could possibly hold. If it could, give a numerical example using a probability distribution of your own choice: if it could not, explain why not (what rule is violated)a. P(A) = P(B)
b. P(A) = 2P(B)
c. P(A) = 1 - P(B)
d. P(A) + P(B) > 1
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f. P(A) - P(B) > 1

Answers

Answer:

a. P(A) = P(B)

c. P(A) = 1 - P(B)

a and c are true . The rest are false.

Step-by-step explanation:

Two events A and B are said to be equally likely when one event is as likely to occur as the other. In other words each event should occur in equal number in repeated trials. For example when a fair coin is tossed the head is likely to appear as the tail, and the proportion of times each side is expected to appear is 1/2.

So when the events A= {1,3,4} B = {2,4,5} are equally likely then suppose their probability is 1/2.

a. P(A) = P(B) True

1/2= 1/2

b. P(A) = 2P(B) False

1/2 is not equal to 1

c. P(A) = 1 - P(B) True

1/2= 1-1/2= 1/2

d. P(A) + P(B) > 1 False

1/2 + 1/2 is not greater than 1

e. P(A) - P(B) < 0 False

1/2-1/2= 0 is not less than 0

f. P(A) - P(B) > 1 False

1/2-1/2= 0 is not greater than 1

Final answer:

The relationships between the probabilities are evaluated and explained.

Explanation:

a. P(A) = P(B) could possibly hold if P(A) = 1/3 and P(B) = 1/3.

b. P(A) = 2P(B) could not hold, as probabilities cannot exceed 1.

c. P(A) = 1 - P(B) could possibly hold if P(A) = 2/3 and P(B) = 1/3.

d. P(A) + P(B) > 1 could possibly hold if P(A) = 1/3 and P(B) = 1/2.

e. P(A) - P(B) < 0 could not hold, as the difference between probabilities cannot be negative.

f. P(A) - P(B) > 1 could not hold, as the difference between probabilities cannot exceed 1.

Learn more about Probability here:

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HELPPPPPP ASAPPPPPPPPP!!!!!!!!!!

Answers

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Answer in three complete sentences to explain how much of your income is in each category.

Answers

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