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Anna35 [415]
4 years ago
12

2. The variation within the sample means is

Mathematics
1 answer:
sergey [27]4 years ago
6 0

Answer:

Step-by-step explaSuppose your instructor administers 3 different forms of a final exam. When scores are posted,

you see the observed mean scores for those 3 different forms—82, 66, and 60—are not the same.

Is this due to chance variation, or do the 3 exams not share the same level of difficulty?

Version 1: 65, 73, 78, 79, 86, 93, 100

Version 2: 39, 58, 63, 67, 69, 74, 92

Version 3: 39, 52, 62, 64, 66, 77

[Note: if there were only 2 different exams, a two-sample t test could be used.]

It would be unrealistic to expect 3 identical sample mean scores, even if the exams were all equally

difficult: in other words, there’s bound to be some variation among means in the 3 groups.

Also, of course there will be variation of scores within each group.

If the ratio of variation among groups to variation within groups is large enough, we will have

evidence that the population mean scores for the 3 groups actually differ. Picture two possible

configurations for the data, both of which represent 3 data sets with means 82, 66, and 60. Thus,

variation among means (from the overall mean of 70) would be the same for both configurations.

1. There could be large within-group variation, such as we see in Exams 1a, 2a, and 3a, in

which case the ratio among to within is small; in this case, the data could be coming from

populations that share the same mean.

2. There could be small within-group variation, such as we see in Exams 1b, 2b, and 3b, in

which case the ratio among to within is large; in this case, the population means probably

differ.nation:

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                                           Question # 1

Given the expression

6^2\div \:3+\left(5+3\cdot \:2\right)-2^3

Follow the PEMDAS order of operations

\mathrm{Calculate\:within\:parentheses}\:\left(5+3\cdot \:2\right)\::\quad 11

=6^2\div \:3+11-2^3

\mathrm{Calculate\:exponents}\:6^2\::\quad 36

=36\div \:3+11-2^3

\mathrm{Calculate\:exponents}\:2^3\::\quad 8

=36\div \:3+11-8

\mathrm{Multiply\:and\:divide\:\left(left\:to\:right\right)}\:36\div \:3\::\quad 12

=12+11-8

\mathrm{Add\:and\:subtract\:\left(left\:to\:right\right)}\:12+11-8\::\quad 15

=15

Therefore,

6^2\div \:3+\left(5+3\cdot \:2\right)-2^3=15

                                            Question # 2

Given the expression

\frac{4+3^2-15\div 5}{\left(2^4-5\cdot \:3\right)^2}

as

4+3^2-\frac{15}{5}

=4+9-\frac{15}{5}      ∵3^2=9

\mathrm{Add\:the\:numbers:}\:4+9=13

=-\frac{15}{5}+13

and

\left(2^4-5\cdot \:3\right)^2

=\left(16-5\cdot \:3\right)^2   ∵ 2^4=16

\mathrm{Multiply\:the\:numbers:}\:5\cdot \:3=15

=\left(16-15\right)^2

\mathrm{Subtract\:the\:numbers:}\:16-15=1

=1^2

\mathrm{Apply\:rule}\:1^a=1

=1

Thus the equation \frac{4+3^2-15\div 5}{\left(2^4-5\cdot \:3\right)^2}  becomes

=\frac{-\frac{15}{5}+13}{1}

\mathrm{Divide\:the\:numbers:}\:\frac{15}{5}=3

=\frac{-3+13}{1}

\mathrm{Apply\:rule}\:\frac{a}{1}=a

=-3+13

\mathrm{Add/Subtract\:the\:numbers:}\:-3+13=10

=10

Therefore,

\frac{4+3^2-\frac{15}{5}}{\left(2^4-5\cdot \:3\right)^2}=10

                                                       Question # 3

Given the expression

ab-c^2+2b

Putting a = 2, b = 4, and c = 1 in the expression

=\left(2\right)\left(4\right)-\left(1\right)^2+2\left(4\right)

Follow the PEMDAS order of operations

\mathrm{Calculate\:exponents}\:\left(1\right)^2\::\quad 1

=\left(2\right)\left(4\right)-1+2\left(4\right)

\mathrm{Multiply\:and\:divide\:\left(left\:to\:right\right)}\:\left(2\right)\left(4\right)\::\quad 8

=8-1+2\left(4\right)

\mathrm{Multiply\:and\:divide\:\left(left\:to\:right\right)}\:2\left(4\right)\::\quad 8

=8-1+8

\mathrm{Add\:and\:subtract\:\left(left\:to\:right\right)}\:8-1+8\::\quad 15

=15

Therefore,

ab-c^2+2b=\left(2\right)\left(4\right)-\left(1\right)^2+2\left(4\right)=15

                                                     Question # 4

Given the expression

4d^3+2e\div \:f+de

Putting d = 2, e = 3, and f = 6 in the expression

=4\left(2\right)^3+2\left(3\right)\div \:6+\left(2\right)\left(3\right)

Follow the PEMDAS order of operations

\mathrm{Calculate\:exponents}\:\left(2\right)^3\::\quad 8

=4\cdot \:8+2\left(3\right)\div \:6+\left(2\right)\left(3\right)

\mathrm{Multiply\:and\:divide\:\left(left\:to\:right\right)}\:4\cdot \:8\::\quad 32

=32+2\left(3\right)\div \:6+\left(2\right)\left(3\right)

\mathrm{Multiply\:and\:divide\:\left(left\:to\:right\right)}\:2\left(3\right)\div \:6\::\quad 1

=32+1+\left(2\right)\left(3\right)

\mathrm{Multiply\:and\:divide\:\left(left\:to\:right\right)}\:\left(2\right)\left(3\right)\::\quad 6

=32+1+6

\mathrm{Add\:and\:subtract\:\left(left\:to\:right\right)}\:32+1+6\::\quad 39

=39

Therefore,

4d^3+2e\div \:\:f+de=4\left(2\right)^3+2\left(3\right)\div \:6+\left(2\right)\left(3\right)=39

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dsp73

Answer:

Step-by-step explanation:

-5 i think hope that helps

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