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3 years ago

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A research report says that t(63) = 1.99; p = .03. From that information, can you reject the null hypothesis with 95% confidence

? No. Yes. It depends on the sample size. It depends on the size of the Type I error.

2 answers:

nika2105 [10]3 years ago

5 0

Answer:

Yes, we can reject the null hypothesis with 95% confidence.

Step-by-step explanation:

We are given that t(63) = 1.99; p = .03 and confidence level = 0.95

Confidence Level = 95% = 0.95

Significance level ( $\alpha$ ) = 1 - 0.95 = 0.05 or 5%

P- value = 0.03 or 3%

As we know that;

If p-value < significance level ⇒ Reject null hypothesis

If p-value > significance level ⇒ Accept null hypothesis

Since, P-value is less than significance level as 3% < 5%, so we can reject null hypothesis with 95% confidence.

Leona [35]3 years ago

4 0

Answer:

Yes, I can reject the null hypothesis with 95% confidence.

Step-by-step explanation:

Critical value t(63) = 1.99

p-value = 0.03

Confidence level (C) = 95% = 0.95

Significance level = 1 - C = 1 - 0.95 = 0.05

Conclusion:

Reject the null hypothesis because the p-value 0.03 is less than the significance level 0.05.

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Determine all prime numbers a, b and c for which the expression a ^ 2 + b ^ 2 + c ^ 2 - 1 is a perfect square .

kogti [31]

Answer:

The family of all prime numbers such that $a^{2} + b^{2} + c^{2} -1$ is a perfect square is represented by the following solution:

$a$ is an arbitrary prime number. (1)

$b = \sqrt{1 + 2\cdot a \cdot c}$ (2)

$c$ is another arbitrary prime number. (3)

Step-by-step explanation:

From Algebra we know that a second order polynomial is a perfect square if and only if $(x+y)^{2} = x^{2} + 2\cdot x\cdot y + y^{2}$ . From statement, we must fulfill the following identity:

$a^{2} + b^{2} + c^{2} - 1 = x^{2} + 2\cdot x\cdot y + y^{2}$

By Associative and Commutative properties, we can reorganize the expression as follows:

$a^{2} + (b^{2}-1) + c^{2} = x^{2} + 2\cdot x \cdot y + y^{2}$ (1)

Then, we have the following system of equations:

$x = a$ (2)

$(b^{2}-1) = 2\cdot x\cdot y$ (3)

$y = c$ (4)

By (2) and (4) in (3), we have the following expression:

$(b^{2} - 1) = 2\cdot a \cdot c$

$b^{2} = 1 + 2\cdot a \cdot c$

$b = \sqrt{1 + 2\cdot a\cdot c}$

From Number Theory, we remember that a number is prime if and only if is divisible both by 1 and by itself. Then, $a, b, c > 1$ . If $a$ , $b$ and $c$ are prime numbers, then $2\cdot a\cdot c$ must be an even composite number, which means that $a$ and $c$ can be either both odd numbers or a even number and a odd number. In the family of prime numbers, the only even number is 2.

In addition, $b$ must be a natural number, which means that:

$1 + 2\cdot a\cdot c \ge 4$

$2\cdot a \cdot c \ge 3$

$a\cdot c \ge \frac{3}{2}$

But the lowest possible product made by two prime numbers is $2^{2} = 4$ . Hence, $a\cdot c \ge 4$ .

The family of all prime numbers such that $a^{2} + b^{2} + c^{2} -1$ is a perfect square is represented by the following solution:

$a$ is an arbitrary prime number. (1)

$b = \sqrt{1 + 2\cdot a \cdot c}$ (2)

$c$ is another arbitrary prime number. (3)

Example: $a = 2$ , $c = 2$

$b = \sqrt{1 + 2\cdot (2)\cdot (2)}$

$b = 3$

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