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Lelu [443]
3 years ago
14

A new pet food product for small dogs has been developed to help with tooth decay. Fifty-two dogs participate in the study. Each

dog owner picks a card from a standard deck out of a hat. If the card is red, their dog will be in the treatment group. If the card is black, their dog will be in the placebo group. Which experimental design was used?
a. Randomized Block Design
b. Completely Randomized Design
c. Case-Control Design
d. Matched-Pair Design
Mathematics
1 answer:
Kitty [74]3 years ago
6 0

Answer: B. Completely Randomized Design

Step-by-step explanation:

A completely randomized design is an experimental design such that treatments are assigned at random completely in order for each experimental unit to have same chance of receiving treatment.

In this scenario, since all the dogs are randomly assigned treatment by picking a card without taking into consideration other factors, then we can infer that thus is a completely randomized design.

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The symbol ⊕ is used to denote exclusive or, so p ⊕ q ≡ (p v q) ^ ~ (p ^ q).
Elan Coil [88]

I'll use text forms of the logical symbols because copying/pasting on mobile is annoying.

⊕ = XOR

∨ = OR

∧ = AND

~ = NOT

↔ = IFF (meaning "if and only if")

→ = IMP (short for "implies")

and for logical equivalence I'll use a regular equal sign.

Then

p XOR q = (p OR q) AND NOT (p AND q)

(a) By definition of XOR, we have

p XOR p = (p OR p) AND NOT (p AND p)

= p AND NOT p

and this is tautologically FALSE.

(b) To rewrite XOR, first distribute the NOT over the AND using DeMorgan's law:

NOT (p AND q) = NOT p OR NOT q

Then we can interchange the AND and ORs using their respective distributive properties. That is,

p XOR q = (p OR q) AND (NOT p OR NOT q)

= (p AND (NOT p OR NOT q)) OR (q AND (NOT p OR NOT q))

= ((p AND NOT p) OR (p AND NOT q)) OR ((q AND NOT p) OR (q AND NOT q))

Both (p AND NOT p) and (q AND NOT q) are tautologically FALSE, so the truth value of either OR depends only on the other statements, so we have

p XOR q = (p AND NOT q) OR (q AND NOT p)

Recall the definition of IFF:

p IFF q = (p IMP q) AND (q IMP p)

Also recall that IMP is logically equivalent to

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p IFF q = (p OR NOT q) AND (q OR NOT p)

so that its negation is

NOT (p IFF q) = NOT ((p OR NOT q) AND (q OR NOT p))

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= (NOT p AND q) OR (NOT q AND p)

The statements in bold match, so p XOR q is indeed equivalent to NOT (p IFF q).

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