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dusya [7]
3 years ago
15

A study of the effects of color on easing anxiety compared anxiety test scores of participants who completed the test printed on

either soft yellow paper or on harsh green paper. The scores for five participants who completed the test printed on the yellow paper were 17, 19, 28, 21, and 1 8. The scores for four participants who completed the test on the green paper were 20, 26, 17, and 24. Using the .05 level, test the researcher's prediction that participants should have lower anxiety scores when taking the test on the yellow paper than when taking the test on the green paper. What is the research hypothesis
Mathematics
1 answer:
Pavlova-9 [17]3 years ago
8 0

Answer:

H0 : μYellow = μGreen

H1 : μYellow < μGreen

Step-by-step explanation:

Let :

Yellow paper = μYellow

Green paper = μGreen

To test the hypothesis :

The null hypothesis is will state that there is no difference in mean of anxiety scores obtained for test taken of yellow and green papers

H0 : μYellow = μGreen.

The alternative hypothesis is the opposite of the null and it is to the left, where we want to test if the anxiety score is lower when take on a yellow paper

H1 : μYellow < μGreen

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The personnel files of all eight employees at the Pawnee location of Acme Carpet Cleaners Inc. revealed that during the last six
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Answer:

a) For Pawnee:

Range = 11, Mean = 3.13, Mean deviation = 2.44

For Chickpee:

Range =  7, Mean = 4.25, Mean deviation = 1.59

b) Pawnee Location has fewer lost days.

c) variation of the Chickpee is less than Pawnee location

Step-by-step explanation:

Data given:

For Pawnee:

3 1 0 0 2 11 5 3

Max = 11

Min = 0

Range = Max - Min

Range = 11-0 = 11

Range = 11

For Chickpee:

3 6 8 4 4 5 3 1

Max = 8

Min = 1

Range = Max - Min

Range = 8 - 1

Range = 7

For mean of Pawnee:

Mean = (3 + 1 + 0 + 0 + 2 + 11 + 5 + 3)/8

Mean = 25/8

Mean = 3.125

For mean deviation, we find the difference from the mean of each point.

Note: all the negative values will be taken as 0.

Differences from the mean of each point =

(3-3.13)=0.13, (1-3.31) = 2.13 ,

(0-3.13)=0, (0-3.13)=3.13,

(2-3.13)=1.13, (11-3.13)=7.87,

(5-3.13)=1.87, (3-3.13)=0.13

So, our differences are:

(0.13, 2.13, 0, 3.13, 1.13, 7.87, 1.87, 0.13 )

Mean deviation =  Sum of all differences/8

Mean deviation = (0.13+2.13+3.13+3.13+ 1.13+ 7.87+ 1.87+ 0.13)/8

Mean deviation = 2.44

For mean of Chickpee:

Mean = (3+6+8+4+4+5+3+1)/8

Mean = 34/8

Mean = 4.25

Similarly, we need to find mean deviation for Chickpee, for that we need to find the differences first as done above for Pawnee.

Note: all the negative values will be taken as 0

Differences from the mean:

(3-4.25) = 1.25, (6-4.25) =1.75,

(8-4.25) = 3.75, (4-4.25)=0.25,

(4-4.25)=0.25, (5-4.25)= 0.75,

(3-4.25)=1.25, (1-4.25) =3.25

So, the differences are:

Differences = (1.25,1.75,3.75,0.25,0.25,0.75,1.25,3.25)

Mean deviation = (1.25+ 1.75+ 3.75+ 0.25+ 0.25+ 0.75+ 1.25+ 3.25)/8

Mean deviation = 1.59

b) which location has fewer lost days:

This can be found out by the total number of days of Pawnee and chickpee.

Pawnee = Sum of values = (3 + 1 + 0 + 0 + 2 + 11 + 5 + 3) = 25 days

Chickpee = Sum of values = (3+6+8+4+4+5+3+1) = 34 days

Hence, Pawnee Location has fewer lost days.

C) which location has less variation?

Formula for variation = ∑(\frac{x^{2} }{8}) - (mean^{2})

where x = values of the set.

For Pawnee:

mean^{2} = 3.13^{2} = 9.8

Sum of square of all the data points of Pawnee = 169

Variation = 169/8 - 98 = 11.325

Similarly,

For Chickpee:

mean^{2}  = 4.25^{2} = 18.06

Sum of square of all the data points of Chickpee = 176

Variation = 176/8 - 18.06

Variation = 22-18.06

Variation = 4

Hence, variation of the Chickpee is less than Pawnee location

8 0
3 years ago
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