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Leno4ka [110]
2 years ago
10

A null hypothesis is being tested. How does the confidence level affect the testing process?

Mathematics
1 answer:
never [62]2 years ago
4 0
The confidence level C and the significance level alpha are linked through the equation

alpha = 1-C

So for instance, if the confidence level is C = 95% = 0.95 then alpha is 

alpha = 1-C
alpha = 1-0.95
alpha = 0.05

meaning we have a 5% significance level. The larger C gets, the smaller alpha gets and vice versa. It turns out that 0 < C < 1 and also 0 < alpha < 1. 

The closer C gets to 1, the alpha value gets closer to 0. The smaller alpha gets, the harder it is to reject the null. Why is that? If we have a fixed p value, say p = 0.02 then we reject the null if alpha > pvalue. But we fail to reject the null when alpha < pvalue. For very small alpha values, we're going to fail to reject H0 no matter how small the pvalue is. The pvalue would have to be really small for H0 to be rejected.

In short, I'm saying that if the confidence level is high, then the chance of rejecting the null hypothesis is low (or rare)

This is why the answer is choice A
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Answer:

1. Steve's age is 18 and Anne's age is 8.

2. Max's age is 17 and Bert's age is 11.

3. Sury's age is 19 and Billy's age is 9.

4. The man's age is 30 and his son's age is 10.

Step-by-step explanation:

1. Let us assume that:

S = Steve's age now

A = Anne's age now

Therefore, in four years, we have:

S + 4 = (A + 4)2 - 2

S + 4 = 2A + 8 - 2

S + 4 = 2A + 6 .................. (1)

Three years ago, we have:

S - 3 = (A - 3)3

S - 3 = 3A - 9 ................................ (2)

From equation (2), we have:

S = 3A - 9 + 3

S = 3A – 6 …………. (3)

Substitute S from equation (3) into equation (1) and solve for A, we have:

3A – 6 + 4 = 2A + 6

3A – 2A = 6 + 6 – 4

A = 8

Substitute A = 8 into equation (3), we have:

S = (3 * 8) – 6

S = 24 – 6

S = 18

Therefore, Steve's age is 18 while Anne's age is 8.

2. Let us assume that:

M = Max's age now

B = Bert's age now

Therefore, five years ago, we have:

M - 5 = (B - 5)2

M - 5 = 2B - 10 .......................... (4)

A year from now, we have:

(M + 1) + (B + 1) = 30

M + 1 + B + 1 = 30

M + B + 2 = 30 .......................... (5)

From equation (5), we have:

M = 30 – 2 – B

M = 28 – B …………………… (6)

Substitute M from equation (6) into equation (4) and solve for B, we have:

28 – B – 5 = 2B – 10

28 – 5 + 10 = 2B + B

33 = 3B

B = 33 / 3

B = 11

Substituting B = 11 into equation (6), we have:

M = 28 – 11

M = 17

Therefore, Max's age is 17 while Bert's age is 11.

3. Let us assume that:

S = Sury's age now

B = Billy's age now

Therefore, now, we have:

S = B + 10 ................................ (7)

Next year, we have:

S + 1 = (B + 1)2

S + 1 = 2B + 2 .......................... (8)

Substituting S from equation (7) into equation (8) and solve for B, we have:

B + 10 + 1 = 2B + 2

10 + 1 – 2 = 2B – B

B = 9

Substituting B = 9 into equation (7), we have:

S = 9 + 10

S = 19

Therefore, Sury's age is 19 while Billy's age is 9.

4. Let us assume that:

M = The man's age now

S = His son's age now

Therefore, now, we have:

M = 3S ................................... (9)

Five years ago, we have:

M - 5 = (S - 5)5

M - 5 = 5S - 25 ................ (10)

Substituting M from equation (9) into equation (10) and solve for S, we have:

3S - 5 = 5S – 25

3S – 5S = - 25 + 5

-2S = - 20

S = -20 / -2

S = 10

Substituting S = 10 into equation (9), we have:

M = 3 * 10

M = 30

Therefore, the man's age is 30 and his son's age is 10.

5 0
3 years ago
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