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Inessa [10]
3 years ago
6

Four different leadership styles used by Big-Six accountants were investigated. As part of a designed study, 15 accountants were

randomly selected from each of the four leadership style groups (a total of 60 accountants). Each accountant was asked to rate the degree to which their subordinates performed substandard fieldwork on a 10-point scale—called the "substandard work scale". The objective is to compare the mean substandard work scales of the four leadership styles. The data on substandard work scales for all 60 observations were subjected to an analysis of variance.
Interpret the results of the ANOVA F test shown on the printout for a = 0.05.
a. At a = 05, there is insufficient evidence of differences among the substandard work scale means for the four leadership styles.
b. At a = .05, three is no evidence of differences among the substandard work scale means for the four leadership styles.
c. At a = .05, there is sufficient evidence of difference among at least two of the substandard work scale means for the four leadership styles.
d. None of the above.
Mathematics
1 answer:
Flauer [41]3 years ago
5 0

Answer:

c. At a = .05, there is sufficient evidence of difference among at least two of the substandard work scale means for the four leadership styles.

Step-by-step explanation:

From the given output, df(Between)=3

df (Between) =k-1, where, k denotes number of groups. Therefore, k-1=3, k=4.

Thus the null hypothesis is as follows:

H0:mu1=mu2=mu3=mu4 (there is no difference in substandard work scale means for four leadership styles)

H1: At least one mean scale is different from another.

Per rule, reject H0, if p value corresponding to F test statistic is less than given level of significance, alpha=0.05.

Here, F(3, 56)=6.056, p<0.05 (the p value is 0.0012). Therefore, reject H0, and conclude that at least one substandard work scale mean of a leadership is significantly different from a substandard work scale mean of another leadership.

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Can we obtain a diagonal matrix by multiplying two non-diagonal matrices? give an example
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Yes, we can obtain a diagonal matrix by multiplying two non diagonal matrix.

Consider the matrix multiplication below

\left[\begin{array}{cc}a&b\\c&d\end{array}\right]   \left[\begin{array}{cc}e&f\\g&h\end{array}\right] =  \left[\begin{array}{cc}a e+b g&a f+b h\\c e+d g&c f+d h\end{array}\right]

For the product to be a diagonal matrix,

a f + b h = 0 ⇒ a f = -b h
and c e + d g = 0 ⇒ c e = -d g

Consider the following sets of values

a=1, \ \ b=2, \ \ c=3, \ \ d = 4, \ \ e=\frac{1}{3}, \ \ f=-1, \ \ g=-\frac{1}{4}, \ \ h=\frac{1}{2}

The the matrix product becomes:

\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}\frac{1}{3}&-1\\-\frac{1}{4}&\frac{1}{2}\end{array}\right] = \left[\begin{array}{cc}\frac{1}{3}-\frac{1}{2}&-1+1\\1-1&-3+2\end{array}\right]= \left[\begin{array}{cc}-\frac{1}{6}&0\\0&-1\end{array}\right]

Thus, as can be seen we can obtain a diagonal matrix that is a product of non diagonal matrices.
8 0
3 years ago
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Corey bought 2 1/2 liters of paint for $ 60.<br> What was the cost per liter of paint?
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It costs $24 per liter.
4 0
3 years ago
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The figure shows two parallel lines KL and NO cut by the transversals KO and LN.
Sever21 [200]

Answer:

(A)  

Step-by-step explanation:

From the given figure, we have to prove whether the two given triangles are congruent or similar.

Thus, From the figure, ∠3=∠4 (Vertically opposite angles)

Since, KL and NO are parallel lines and KO and LN are transversals, then

measure angle 1= measure angle 5 that is ∠1=∠5(Alternate angles).

Thus, by AA similarity rule, ΔKLM is similar to ΔONM.

Thus, Option A that is Triangle KLM is similar to triangle ONM because measure of angle 3 equals measure of angle 4 and measure of angle 1 equals measure of angle 5 is correct.

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3 years ago
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Answer:

35 homeruns

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3 years ago
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Part 5:
miss Akunina [59]

4

One in 1200 are not particularly good odds. On the other hand, winning the lotto is 1 chance in 13,000,000 which if you've ever played the lotto you know that those odds are good enough to insure that if you played for the rest of your life and you are 18 not expect to live to 80 and you have 104 [given 2 draw a week] chances of winning per year, it likely won't happen. One in 1200 is better but still not good, especially with only 1 draw.


3

As a fraction her probability of winning is 1/2000 which is 0.000833333 as a decimal. You can put that in as


1

÷

1200

=

if you are not sure how your calculator works.


2

Sample Space = {1,2,3,4 .... 1198,1199,1200}

The outcome depends on sophies number. Either 1 number can be chosen or all of them can.


1

The sample space is the integers from 1 to 1200 inclusive.

8 0
3 years ago
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