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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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Step-by-step explanation:

The Canada Urban Transit Association has reported that the average revenue per passenger trip during a given year was $1.55. If we assume a normal distribution and a standard deviation of 5 $0.20, what proportion of passenger trips produced a revenue of Source: American Public Transit Association, APTA 2009 Transit Fact Book, p. 35.

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