Find the p-value using Excel (not Appendix D): (Round your answers to 4 decimal places.) p-value (a) Right-tailed test t = 1.465
, d.f. = 11 (b) Two-tailed test t = 2.522, d.f. = 12 (c) Left-tailed test t = –1.952, d.f. = 22
1 answer:
Answer:
(a) 0.0855
(b) 0.0268
(c) 0.0319
Step-by-step explanation:
The <em>p</em>-value is well defined as per the probability, [under the null hypothesis (H₀)], of attaining a result equivalent to or greater than what was truly observed.
A small <em>p</em>-value (typically <em>p</em> ≤ 0.05) specifies sturdy proof against the null hypothesis (H₀), so we discard H₀. A large <em>p</em>-value (<em>p</em> > 0.05) specifies fragile proof against the H₀, so we fail to discard H₀.
(a)
Use the Excel function "=T.DIST.RT(1.465,11)" to compute the right-tailed <em>p</em>-value for a test statistic of, <em>t</em> = 1.465 and s degrees of freedom of, <em>df</em> = 11.
<em>p</em>-value = 0.0855
(b)
Use the Excel function "=T.DIST.2T(2.522,12)" to compute the two-tailed <em>p</em>-value for a test statistic of, <em>t</em> = 2.522 and s degrees of freedom of, <em>df</em> = 12.
<em>p</em>-value = 0.0268
(c)
Use the Excel function "=T.DIST(-1.952,22,TRUE)" to compute the left-tailed <em>p</em>-value for a test statistic of, <em>t</em> = -1.952 and s degrees of freedom of, <em>df</em> = 22.
<em>p</em>-value = 0.0319
You might be interested in
Answer:
1) increasing on (-∞,-1] ∪ [1,∞), decreasing on [-1,0) ∪ (0,1]
is local maximum,
is local minimum
2) increasing on [1,∞), decreasing on (-∞,0) ∪ (0,1]
is absolute minimum
3) increasing on (-∞,0] ∪ [8,∞), decreasing on [0,4) ∪ (4,8]
is local maximum,
is local minimum
4) increasing on [2,∞), decreasing on (-∞,2]
is absolute minimum
5) increasing on the interval (0,4/9], decreasing on the interval [4/9,∞)
is local minimum,
is absolute maximum
Step-by-step explanation:
To find minima and maxima the of the function, we must take the derivative and equalize it to zero to find the roots.
1)
and
So, the roots are and
The function is increasing on the interval (-∞,-1] ∪ [1,∞)
The function is decreasing on the interval [-1,0) ∪ (0,1]
is local maximum,
is local minimum.
2)
and
So the root is
The function is increasing on the interval [1,∞)
The function is decreasing on the interval (-∞,0) ∪ (0,1]
is absolute minimum.
3)
and
So the roots are and
The function is increasing on the interval (-∞,0] ∪ [8,∞)
The function is decreasing on the interval [0,4) ∪ (4,8]
is local maximum,
is local minimum.
4)
has no solution and
is crtitical point.
The function is increasing on the interval [2,∞)
The function is decreasing on the interval (-∞,2]
is absolute minimum.
5) for
So the root is
The function is increasing on the interval (0,4/9]
The function is decreasing on the interval [4/9,∞)
is local minimum,
is absolute maximum.
Answer:
The correct answer is 0.486.
Step-by-step explanation:
Total number of employees at the home office of Gibraltar Insurance Company is 270 + 340 = 610
Number of employees on flex time are 350, out of which 170 are men and 180 are women.
We need to find the probability that out of the given employees, a randomly chosen one is on flex time and is a man.
Favorable outcomes are 170 and total outcome is 350.
Thus the probability is given by = 0.4857≈ 0.486