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Anon25 [30]
3 years ago
13

A researcher collated data on Americans' leisure time activities. She found the mean number of hours spent watching television e

ach weekday to be 2.7 hours with a standard deviation of 0.2 hours. Jonathan believes that his football team buddies watch less television than the average American. He gathered data from 40 football teammates and found the mean to be 2.3. Which of the following are the correct null and alternate hypotheses?
A. H0: u=2.7; Ha: u < 2.7
B. H0: u ≠ 2.7; Ha: u=2.3
C. H0: u=2.7; Ha: u≠ 2.7
D. H0: u=2.7; Ha: u≥ 2.3
Mathematics
2 answers:
Nataly [62]3 years ago
8 0

Answer:

option a

Step-by-step explanation:

did on edge

Semmy [17]3 years ago
5 0

Answer: A. H_0:\mu=2.7\ ;\ H_a:\mu

Step-by-step explanation:

Let \mu be the population mean.

Given : A researcher collated data on Americans' leisure time activities. She found the mean number of hours spent watching television each weekday to be 2.7 hours with a standard deviation of 0.2 hours.

i.e. \mu=2.7

Jonathan believes that his football team buddies watch less television than the average American.

i.e.\mu

Since, the null hypotheses contains equals to sign.

Then, the correct null and alternate hypotheses will be :-

H_0:\mu=2.7\ ;\ H_a:\mu

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defon
Answer for problem 46 is choice A
Answer for problem 47 is choice B
Answer for problem 48 is choice E

-----------------------------------------------------------------------------
-----------------------------------------------------------------------------

Work Shown

Problem 46
Equation 1: 3x+y = 17
Equation 2: x+3y = -1
Add equation 1 to equation 2 to get 4x+4y = 16. Divide every term by 4 to get x+y = 4. Then finally multiply both sides by 3 to get 3x+3y = 12
That shows why the answer is choice A

---------------------------------
Problem 47)

If y hours pass by, then y-(2/3)y=y/3 is the time value (2/3)y hours ago

So,
Distance = rate*time
d = r*t
d = x*(y/3)
d = (xy)/3
That's why the answer is choice B

---------------------------------
Problem 48)
Let L1,L2,L3 be the three lists where
L1 = {a1,a2,a3,...,ak} there are k scores here
L2 = {a1,a2,...,a10} there are 10 scores here
L3 = {a11,a12,...,ak} the remaining k-10 scores
S(L1) = sum of the scores in list L1
M(L1) = mean of L1 = 20 = S(L1)/k
M(L2) = mean of L2 = 15 = S(L2)/10
S(L1) = 20k
S(L2) = 150
S(L1) = S(L2)+S(L3)
M(L1) = [S(L2)+S(L3)]/k
20 = [150+S(L3)]/k
20k = 150+S(L3)
S(L3) = 20k-150
M(L3) = [S(L3)]/(k-10)
M(L3) = (20k-150)/(k-10)
So that shows why the answer is choice E
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4 years ago
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