In: Statistics and Probability
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
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.