Inferential Statistics and Findings

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BUSINESS RESEARCH PROJECT PART 3: SAMPLING AND DATA 2

BUSINESS RESEARCH PROJECT PART 3: SAMPLING AND DATA 7

Running head: 1

Business Research Project Part 3: Sampling and Data Collection Plan

Team C discussed the organization QuickMart Inc. and how employee satisfaction affects employee turnover rates specific to South Florida. Team C is narrowing the search to determine whether or not employee compensation is the most important component of job satisfaction. In order to determine the correlation that employee satisfaction has on employee turnover rates, Team C needs to collect and analyze data from current employees of QuickMart Inc.

For the past few decades, employee retention has been of interest to researchers and employers in various fields. Therefore, one may ask if there is a correlation in employee satisfaction based on turnover rates? Prior research has shown that job satisfaction is strongly and inversely associated with employee’s intention to leave an organization (Egan, Yang & Bartlett, 2004; Lambert, Hogan & Barton, 2001; MacIntosh & Doherty, 2010; Schwepker, 2001; Silverthorne, 2004). Wages could be one reason for QuickMart’s employee turnover.

On the other hand, given QuickMart can be viewed as a temporary “job” opposed to a career, therefore, employee turnover may not be related to employee satisfaction. Below is the statistics analysis:

RQ: Is there a relationship between employee satisfaction (DV) and compensation/pay (IV)?

The hypothesis statements are:

Ho: There is no relationship between employee satisfaction (DV) and compensation/pay (IV).

H1: There is a relationship between employee satisfaction (DV) and compensation/pay (IV).

Team C has identified the target population to be employees of QuickMart Inc., who reside in South Florida. Within this target population, Team C's focus is to determine what effect employee compensation has on employee satisfaction. The population and size refers to people or items with the characteristic one wishes to study or to survey, or to research and many times the research includes measurement of physical elements like population, or intangible elements like survey results. One problem facing Florida research is Walmart.com maps 23 super center stores with an average of 250 employees per store for the total population of 5,570 (Walmart, 2014). Random sampling allows segmentation of this population to create a sample that is an unbiased group of the larger population.

The sampling element that will be used is a survey broken down into 35 areas of job satisfaction and sub-divided into four topic areas; Career Development, Relationship with Management, Compensation and Benefits, and Work Environment. The survey depth is also part of the reliability achievement supporting the need for greater participation providing necessary data. Team C chose to use stratified sampling method to analyze the data which is the most efficient method providing for variability within strata that is minimized, variability within strata that is maximized, and the variables upon which the population is stratified are strongly correlated with the desired dependent variable.

Although many factors contributed to employee’s job satisfaction, only three have remained among the top five aspects most important to employee satisfaction:

1. Organization’s financial stability which contributes .55

2. Compensation/pay which contributes .54

3. Benefits which contributes .53

The study cumulated across 35 correlations from 596 independent samples, and results suggested that the pay-level was correlated .15 with the job satisfaction and .54 with pay satisfaction.

With any research, the processes require accurate and honest information received from trusted sources. Protecting these participants go beyond privacy disclosures into the survey content removing identifying information. Participant protection and the collected information is the plan's core. Information collected can and will be needed in the future. If any data is questioned, the background documentation supports current and future research models and business needs. Based on the researcher's need, the collection methods in this plan utilize electronic channels reaching respondents that are spread out allowing for faster response, reliable collection, and safe storage for future reference.

References

Walmart. (2014). Retrieved from http://www.glassdoor.com/Reviews/Walmart-Miami-Reviews-EI_IE715.0,7_IL.8,13_IM558.htm

Egan, T. M., Yang, B., & Bartlett, K. R. (2004). The effects of organizational learning culture

Average Hourly Rate for Walmart Employees. (n.d.). Retrieved from http://www.payscale.com/research/US/Employer=Wal-Mart_Stores%2c_Inc/Hourly_Rate

There Are Significant Business Costs to Replacing Employees. (2012). Retrieved from http://www.americanprogress.org/issues/labor/report/2012/11/16/44464/there-are-significant-business-costs-to-replacing-employees/

Appendix A

Customer Survey:

Employee Job Satisfaction:

Please take a few minutes Agree Somewhat Neither Somewhat Disagree

to tell us about your job: Agree Agree Disagree

or

Disagree

My job makes good use of my

skills and abilities:

I am satisfied that I am valued and

Important to my managers:

I am paid accurately for my job:

I am concerned about the future

of my career:

I am satisfied with the benefits

and growth potential:

I get updates and information

I need to do my job:

Overall, I am satisfied with my job:

Please add any additional comments regarding your satisfaction issues in this area:

I am a Male___ Female: ____Other:_____

My age group is: 16 – 21: __ 22 – 42 __ 43 –up __

I work in: Sales: ___ Management:___ Warehouse:____ Office:____

Appendix B:

Determining Sample Size with 95% Confidence and 5% margin of error:

Necessary sample size = (Z score)squared * StdDev*(1-stdDev)/(margin of error)sq

Zscore1.96STDDev0.5MOE0.05CL95%

3.84160.250.0025

384.16

Round up385

Week 1 Quiz

12) 24 0.06
76 0.19
176 0.44
72 0.18
52 0.13
400 1
0
13) 39 0.144
78 0.288
45 0.166
88 0.325
18 0.066
3 0.011
271
17) 2 4 1 1
3 9 2 4
5 25 4 16
6 36 5 25
3 9 2 4
Sigma 19 83 Sigma 2 14 50
19) 188.65
18) 2.1 237.83
2.4 430.86 430.9
3.2 456.09
3.4 459.15
3.6 1772.58
3.7 354.516
18.4 3.3
3.07
20) 41 22) 0 -1 1 23) 56 3136 Sample 93025
42 1 0 0 42 1764
43 3 1 0 0 41 1681 610.88
45 3 2 4 40 1600 87.27
46 4 3 9 36 1296
46 9 2.80 Sample 36 1296 S Variance 9.34
47 4 Mean 1.80 1.67 Stanton Deviation = Sample Squared 29 841
50 25 625
51 305 12239
51
51
51 41 1681 Sample 22201
52 6 27 729 388.83
55 23 529 77.77
56 22 484
58 21 441 S Variance 8.82
59 4 15 225
60 149 4089
63
64 3
1031
51.55 51 Skewed Right
24) 0 0 1764 25) X -2s to X+2S
3 9 X 140.05 S 19.74
4 16 71.6 2* 39.48
4 16 7.956 100.57 179.53
4 16 3.653 2.669 2.821
5 25 Number set greater than 1
6 36
7 49 (1- 1/2*2)X100%
9 81 75% within interval
0 0
42 248
26) Z = X-X/S Z=X-u/o 27) 28)
A 3 C -3 A 3.8461538462 C 1.0769230769
A 2 B 0.5 B -2 D 1 B 0.6923076923 D -0.5384615385
D -2.5 C 0
P Q N X !(N-X)
34) 0.01 0.38 35) A Q 0.4 (n/x)*P X q N-X 0.36 0.16 0.0576 24 2 24 48 0.5 0.0288
0.01 0.91 B P 0.6 (n/x)*P X q N-X 0.216 0.064 0.013824 720 6 6 36 20 0.2765
0.04 C Q 0.2 (n/x)*P X q N-X 0.8 0.04 0.032 6 1 2 2 3 0.0960
0.04 D Q 0.3 (n/x)*P X q N-X 1 0.0081 0.0081 24 1 24 24 1 0.0081
0.08 E P 0.4 (n/x)*P X q N-X 0.064 0.216 0.013824 720 6 6 36 20 0.2765
0.09 F Q 0.1 (n/x)*P X q N-X 0.81 0.1 0.081 6 2 1 2 3 0.243
0.11
0.15
0.12
0.09
0.09
0.06
0.04
0.03
0.02
0.01
0.01
1
36) Q 0.69 (n/x)*P X q N-X P Q N X !(n-x)
N 8 1 0.0513798374 0.0513798374 40320 1 40320 40320 1 0.0513798374
X 0
P 0.31
N 8
X 4 0.00923521 0.22667121 0.0020933562 40320 24 24 576 70 0.1465349358
37) U=np N= Sample P = Probablity 39) Z= X-U/O -2.75
U= 24(.15) -2.5
U= 3.6 -3
-3.25
-1.25
0=sqrt/npq N Sample P Probablity Q=1-p 1.25
0= 3.06
SQrt 3
40) z=x-u/o Z<1.14 subtract it from 0.5 to find z>1.14 0.373 0.127 41) Convert X Z=X-U/0 P(x>87) = P(Z>.58)
1.14 0.6944444444
-1.71 Z<-1.71 0.456 0.044 Subtract number is chart from .5 0.255 0.245
Z1 0.29 Area left z1 0.114 28.10%
Z2 1.43 Area left z2 0.424 0.31 X .98 0.45 Z Score 2.05
-3.43 P z<-3.43 = Pz>3.43 = .5 -.5=0
Z=X-u/0
2.05=X-82/8.6 X=100 9.5348837209 11.5848837209
43) 0=sqrt npq Interval u+/-30= np=+/-3sqrtnpq
880
29.66 u+/- 3o 1188.98
1011.02
45) u = 149 o= o/sqrt 100 1.2
Z= X-u/o
0.8333333333

Week 2 Quiz

1) N 93 Mean 25.4 SD 2.6 a 0.1 Sq rt N 9.6437
Mean +/- Z a/2 (s/sq rt n)
za/2 1.96 1.645 2.575
95% 90 99
s/sqrtn 0.27 0.27 0.27
Za/2*sqrt) 0.53 0.44 0.69
25.93 25.84 26.09
24.87 24.96 24.71
2) N 49 Mean 19.8 SD 11.9 Sq rt N 7
za/2 1.96
95%
s/sqrtn 1.70
Za/2*sqrt) 3.33
23.13
16.47
3) N 62 Mean 7.64 SD 8.91 Sq rt N 7.874
za/2 1.645
90%
s/sqrtn 1.13
Za/2*sqrt) 1.86
9.50
5.78
4) N 6 Mean 6 SD 3.16228 M+Sumt/n SQRT/M 2.44949
25 5
ta/2 4.032 2.797
99% 99%
s/sqrtn 1.2910 0.6325
ta/2 5.2053 1.7690
+ 11.21 7.77
- 0.79 4.23
5) N 10 Mean 2.8 SD 1.8 SQRT/M 1.89736 df 9 SQRT/N 3.1623
Ta/2 2.262
95%
s/sqrtn 0.5692059577
ta/2 1.2875438763 0.025
+ 4.888
- 2.312
6) Large Sample can be used when Sample is large enough if np>15 and nq >15 where q=1-p
NP = 100*0.50 = 100 Sample size NP = 100
q= 1-0..50 = 0.5 nq= 100
nq= 100*0..50 = 100 Both greater than 15
a 0.05 a2 0.025 n 100 p 0.5 q 100
za/2 1.96
PQ/n 0.5
sqrt/pq/n 0.7071 PQ/n 0.5
za/2/sqrt 1.385916
+ 1.69
- -1.09
7) np 200 0.5 100 8) p=x/n X 249500 N 325000
q 1 0.5 0.5 P 0.7677
nq 200 0.5 100 NP 423346.69
a (1-a)100% 95% 0.05 q 0.2323
Z 0.05 2 0.025 NQ 1399006.62
Z0.025 1.96 95% a 0.05 2 0.025
pq/n 0.00125 Sqrt 0.03536 z0.025 1.96
za/2 0.0693 pq/n 0.000001 sqrt 0.0007407706
+ 0.57 za/2 0.0015
- 0.43 + 0.7691
- 0.7662
10) 42 0.156 Red 12) 197.52 13) Z= X-U/o 16) A Convert X to Z score x 60 o 3.2
74 0.275 Blue 235.84 -1 z=X-u/o u 56
46 0.171 Yellow 443.87 2 1.25
89 0.331 Pink 457.06 3 P(x>60) P(z>1.25)
15 0.056 Orange 463.68 -2 Z>1.25 0.1056 P/62 miles to gallon 0.3944
3 0.011 Green 1797.97
269 1.0000 359.59 B: z=X-u/o -2.19 X 53
0.9554 P(<53) = (z<-2.19)
Z<-2.19 0.0143 P/52 miles gallon 0.4857
0.4263 C: Z=X-u/0 X 57 u 62
z1 0.31 0.1217
z2 1.88 0.4699 0.3482
P59<x<63) P(0.29<z<1.43)
P(z1<0.29) = 0.3264 z1
P(z2<1.43) = 0.424 z2
P(.029<z<1.43) 0.0976
P/between 59/63 is 0.310
0.003 D: z=x-u/o X 45
Z = -3.44
P(X<46) = P(z<-3.43)
0

Week 3 Quiz

4) x Mean 111 u 100 o sd 58 n sample 100 SQRTN 10
z=x-u/0sqrtn Table of Z Scores z0.05 = 1.65
z= 1.90
a. Rejection region requires a=0.05 upper tail z distribution >1.65 Null Hypothesis is rejected one tailed
b. z=x-u/0sqrtn Table of Z Score za/2 = 1.96 two tailed
z= 1.90
Rejection region requires a/2 in each tail z<-1.96 or z>1.96 Null Hypothesis is not rejected
c. alternative hypothesis in part A is more specific, easier to reject H0 than one in part B
5) X -1.8 U 0 o 12.5 n 220 sqrtn 14.83
z= x-u/0 p= s/sqrtn
-2.14 Do not reject H0 , there is insuffucient evidence to indicate that u does not equal 0
6) If value of a is less than the chosen p value, reject the null hypothesis, if not do not reject
7) If the P value is less than the A value reject the null hypothesis, if not then do not reject the null hypothesis
8) Sample Size N is large use: x 10.6 u 10 SD 2.9 sqrtn 7.07 http://www.mathsisfun.com/data/standard-normal-distribution-table.html
z= x-u/s/sqrtn
1.46 Use table 0.4279 P(z>1.46) 0.0721
2p(z>1.46) 0.1442
Areas of a Standard Normal Distribution Table
9) X 57.2 U 56 S 7.5 sqrtn 8.94 Two Tailed test larger than 30 sample size
1.43 Table 0.4236 P/2= 0.0764 One tailed
0.1528 Two tailed
10) Rejection region for lower one tailed test Ha is t<ta2 TA= N-1 a= 0.05 Rejection region is t<-ta
a. N 8 degrees of freedom N-1 7 T Table http://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf
t0.05= 1.895 from table
Rejection region = t<- 1.895
x 5.6 u 8 s 1.1 sqrtn 2.83
-6.171 Compare rejection region, if test statistic falls into rejection region, Ha is true, if not conclude sampling experiment does not provide sufficient evidence to reject at Ho at a level
b. Does not equal sign is a two tailed test a = 0.05 a/2 0.025 Degrees of Freedom 7
to.005= 2.365 from table Rejection Region |t| > 2.365
Test Statistic is -6.171 If test statistic falls into |t| range reject Ho, conclude alternative Ha is true, if statistic does not fall into range, conclude sampling experiment does not provide sufficient evidence to reject Ho at a level
c. P value test Move across rows for DF row 7 between .05 and .1 0.050<pvalue<0.100 One tail
P value test Move across rows for DF row 7 between2( .05) and 2(.1) 0.100<pvalue<0.200 Two tail
11) a) Does test fair or pass the null hypothesis
12) Is test upper tail, two tail, or lower tail Use T table 349
t<-ta DF 19 a= 0.1 443
t0.10 1.328 from table x 357.55 SD 118.52 570
t<- 1.328 u 423 SQRTN 4.47 283
-2.470 195
488
455
430
403
549
385
297
184
261
274
403
311
306
136
429
7151 357.55
14) Ho: P= 0.5 H2 P< 0.5
P= 0.285
15) za/2 = 0.025 1.96 from table 95% confidence level = 5% Standard Normal Distribution table
a)
(x1-x2)+/- za/2sqrt s2/2 divided by n1 - s 2/2 divided n2 x1 5274 x2 5248 n(sample) 410
x1-x2 26 (s1, s2)2 22500 36481 s1 150 s2 191
143.86 sqrt(s&N) 11.99
23.5 49.5 + Confidence Levels
2.5 - Confidence Levels
b) Ho (u1-u2) = 0 Ha(u1-u2) does not equal 0
Z=(x1-X2)-(u1-u2)/sqrt( o1sq/n1 + 02sq/n2)
26 11.99
2.17 Two tailed test z<or= 2.17 + z>= 2.17
0.485 From table 0.015000
0.03
c) Upper Tail Test P(z>=2.17)
0.485 0.015
d) Z=(x1-X2)-(u1-u2)/sqrt( o1sq/n1 + 02sq/n2)
3 11.99
0.25 0.0987 from table 0.4013
0.8026
17) a. s sq2 over p = (n1-1)s sq2 over1+(N2-1)s sq2 over2/n1 +n2-2 Solve Mean X1 1.8 3.24 solve mean X2 2.8 7.84
Soilve for s1 sq 2 2.608 1.2 1.44 2.7 7.29
0.652 s12 S 5 3.2 10.24 S 4 3.4 11.56
1.3 1.69 2.8 7.84
Solve for X2 2.93 1.6 2.56 11.7 34.53 136.89
82.81 9.1 19.17 x2 2.93
Solve for s2 sq 2 0.3075 x1 1.82
0.1025 s22
Pooled Estimte of o 0.4165
b. u2>u1 a= 0.05 Ho: u1-u2=0 Ha: u1-u2<0
Test Statistic T -1.105 T= -2.55 DF 7 T distribution Table
sqrtPE 0.187425 1.895 from table
0.4329260907 Rejection: t<-ta T= -2.55 Ta =- 1.895
There is sufficient evidence to reject the null hypothesis Ho
c. 90% Confidence level u1-u2
n1= 5<30 n2= 4<30 a= 0.1 a/2= 0.05 DF 7
From Table Ta/2= 0.05 1.895
Confidence Interval -1.11 0.82 -0.28
-1.93
18) Interpret a<= .10 P= 0.113 If P greater than A fail to refect otherwise Ho
b. P 0.0565
Males Females
19) a. 95% a= 0.05 a/2= 0.025 N 130 116
From Tables Za/2= 1.96 X 39.19 38.47
s 6.76 6.88
0.72 45.6976 47.3344 0.7595751724
sqrt 0.8715360993
1.7082107547
2.428
-0.988
23) a. Confidence Coefficient 0.99 a= 0.01 Za/2= 1.64
n1 400 p1 0.68 n2 400 p2 0.56
0.12 0.00116 0.0340587727 0.056
0.176
0.064
b. n1 180 p1 0.31 n2 250 p2 0.24
0.07 0.0019179333 0.0437942158 0.072
0.142
-0.002
c. n1 100 p1 0.47 n2 120 p2 0.59
-0.12 0.0045068333 0.0671329527 0.110
-0.010
-0.230
24) a. p1= x1/n1 x1 21 n1 188
p1= 0.112
b. p2 x2/n2 x2 34 n2 137
p2= 0.248
Sampling distribution large sample p1-p2 n1-n2 n1,p1>= 15 n2,p2>=15
n1*p1 21.28 n2*p2 37.20
n1*q1 166.944 n2*q2 112.8 Above >=15 is true
c. 90% q1 0.888 q2 0.752 a 0.1 a/2 0.05 Za/2 1.645
-0.136 0.0018897202 0.0434709122 0.072
-0.065
-0.208
25) a. p1 boys 0.549 x1 745 n1 1358
b. p2 girls 0.701 x2 966 n2 1379
c. p1-p2 Point Estimate
-0.152 -0.150
d. 95% a= 0.05 0.025 q1=1-p1 q2=1-p2 N 2000
z= 1.96
-0.1519 0.0003344917 0.0182891152 0.036
-0.188
-0.116
0.41 0.55
26) Ho: p1-p2=0 Ha:p1-p2does not equal 0 a= 0.05 p1 41 n1 100 Test for large sample 41
p2 55 n2 218
z=(p1-p2)-0/sqrtpq(1/n1+1/n2) p 0.302 q 0.698
-0.14 0.0554460097 -2.525 n1q1 59
0.014587156 n2p2 119.9
0.210751157 n2pq2 98.1
0.00307426 Rejection Region z<-Za/2 or Z>Za/2
0.025 1.96 from table
http://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf http://www.mathsisfun.com/data/standard-normal-distribution-table.html

WK 3 95% Sample Size

Necessary sample size = (Z score)squared * StdDev*(1-stdDev)/(margin of error)sq
Zscore 1.96 STDDev 0.5 MOE 0.05 CL 95%
3.8416 0.25 0.0025
384.16
Round up 385