Question

In: Statistics and Probability

You conduct a study to determine the impact that noise volume in an office has on...

You conduct a study to determine the impact that noise volume in an office has on worker productivity (number of minutes worked in a half-hour). You obtain the following productivity scores

Condition 1 (Loud noise): 14, 15, 19, 17, 14, 18, 15

Condition 2 (Soft noise): 16, 18, 21, 21, 20, 18, 21

A) Combine all data into a frequency distribution table with simple frequency, relative frequency, cumulative frequency, and percentile.

B) What does the percentile of the score 17 tell us about the data?

C) What does the relative frequency of the score 22 tell us about the data?

D) What does the cumulative score of the 21 tell us about the data?

E) Summarize the central tendency for the scores for each condition by calculating the mean, median, mode, for each condition (show your work).

F) Summarize the variability for the scores of each condition by calculating the sample variance and sample standard deviation (show your work).

G) Interpret what the mean score and the standard deviation score in condition 2 (soft noise), tell us about the data in that condition

Solutions

Expert Solution

Solution:

Part A

Here, we have to construct the frequency distribution for the combined data for loud noise and soft noise. Required frequency distribution is given as below:

Score X

Frequency f

Relative Frequency

Cumulative frequency

Percentile

14

2

0.142857143

2

0.14285714

15

2

0.142857143

4

0.28571429

16

1

0.071428571

5

0.35714286

17

1

0.071428571

6

0.42857143

18

3

0.214285714

9

0.64285714

19

1

0.071428571

10

0.71428571

20

1

0.071428571

11

0.78571429

21

3

0.214285714

14

1

22

0

0

14

Total

14

1

(All calculations are done by using excel.)

Part B

The percentile of the score 17 is given as 0.4286 or we can say that the observation 17 is the 42.86th percentile or approximately 43rd percentile of the data. The percentile of the score 17 tell us about the data that there is 42.86% of the observations are less than 17.

Part C

The relative frequency of the score 22 is given as 0 which tell us about the data that there is no any observation equal to 22 or the frequency for productivity score is 0.

Part D

Cumulative frequency of the score 21 is given as 14 which means there are 14 observations are less than or equal to the score 21.

Part E

The mean, median, mode, etc. descriptive statistics for both conditions are given as below:

Mean = Xbar = ∑X/n

For loud noise,

∑X = 112, n = 7, Xbar = 112/7= 16

For soft noise,

∑X = 135, n = 7, Xbar = 135/7= 19.2857

Mode is the observation with highest frequency.

For loud noise, mode = 14

For soft noise, mode = 21

Median is the middle most observation when observations are increasing or decreasing order.

For loud noise, median = 15

For soft noise, median = 20

Descriptive summary is given as below:

Loud Noise

Soft Noise

Mean

16

Mean

19.28571429

Standard Error

0.755928946

Standard Error

0.746875578

Median

15

Median

20

Mode

14

Mode

21

Standard Deviation

2

Standard Deviation

1.97604704

Sample Variance

4

Sample Variance

3.904761905

Kurtosis

-1.55

Kurtosis

-0.864485425

Skewness

0.525

Skewness

-0.733171052

Range

5

Range

5

Minimum

14

Minimum

16

Maximum

19

Maximum

21

Sum

112

Sum

135

Count

7

Count

7

Part F

Formula for sample variance and sample standard deviation is given as below:

Sample variance = S^2 = ∑(X – Xbar)^2/(n – 1)

Sample standard deviation = S = sqrt[∑(X – Xbar)^2/(n – 1)]

Calculations tables are given as below:

Loud Noise X

(X - Xbar)

(X - Xbar)^2

14

-2

4

15

-1

1

19

3

9

17

1

1

14

-2

4

18

2

4

15

-1

1

Total

24

∑(X – Xbar)^2 = 24

Sample variance = S^2 = ∑(X – Xbar)^2/(n – 1) = 24/(7 – 1) = 24/6 = 4

Sample standard deviation = sqrt(4) = 2

Soft Noise X

(X - Xbar)

(X - Xbar)^2

16

-3.2857

10.79582449

18

-1.2857

1.65302449

21

1.7143

2.93882449

21

1.7143

2.93882449

20

0.7143

0.51022449

18

-1.2857

1.65302449

21

1.7143

2.93882449

Total

23.42857143

∑(X – Xbar)^2 = 23.42857143

Sample variance = S^2 = ∑(X – Xbar)^2/(n – 1) = 23.42857143/(7 – 1) = 3.904761905

Sample standard deviation = S = sqrt(3.904761905) = 1.97604704

Part G

Mean score for soft noise is more than loud noise and standard deviation for soft noise is less than loud noise.


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