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Tiêu đề Case Study 1: Employee
Tác giả Tran Chau Giang, Cao Tien Phuc, Nguyen Cong Duy Khanh, Tran Bao Quan
Trường học National Economics University
Chuyên ngành Business Statistics
Thể loại Group Assignment
Năm xuất bản 2023
Thành phố Hanoi
Định dạng
Số trang 19
Dung lượng 1,2 MB

Nội dung

Correlation is significant at the 0.05 level 2-tailed.. The correlation between Joining Year and City is slightly positive, albeit weak, and is statistically significant with a significa

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NATIONAL ECONOMICS UNIVERSITY

Business School

BUSINESS STATISTICS Group Assignment

Group member : Tran Chau Giang – 11221833

Cao Tien Phuc – 11225128

Nguyen Cong Duy Khanh – 11223046

Tran Bao Quan- 11225409

Hanoi, July 5th 2023

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Case Study 1: Employee

1,

a, Do necessary data cleaning

Case Processing Summary

Cases

Age 1000 100,0% 0 0,0% 1000 100,0%

CT 1000 100,0% 0 0,0% 1000 100,0%

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After the coding process, the dataset is now clean Despite the presence of outliers in some variables as shown in the charts, the data can still be considered clear This clarity is attributed to the limited types of data representing the variables, which results in significantly high frequencies.

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Statistics

JoiningYea

r

PaymentTie

Gende r EverBenche d Experienc

LeaveOrNo

N Valid 1000 1000 1000 1000 1000 1000 1000 1000 1000

Missin

g

Mean 2015,1160 2,6930 26,494

0 ,62 ,10 3,1380 1,285

0 ,3560 1,780 0

Median 2015,0000 3,0000 26,000

0 1,00 ,00 3,0000 1,000

0 ,0000 2,000 0

Std

Deviation

1,84931 ,56309 2,5757

7 ,487 ,303 1,29408 ,5273

1 ,47905 ,8211 4

Minimum 2012,00 1,00 22,00 0 0 ,00 1,00 ,00 1,00

Maximum 2018,00 3,00 40,00 1 1 5,00 3,00 1,00 3,00

Descriptives

Statistic Std Error

95% Confidence Interval for Mean Lower Bound 2015,0012

Upper Bound 2015,2308

95% Confidence Interval for Mean Lower Bound 2,6581

Upper Bound 2,7279

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Range 2,00

95% Confidence Interval for Mean Lower Bound 26,3342

Upper Bound 26,6538

95% Confidence Interval for Mean Lower Bound ,59

95% Confidence Interval for Mean Lower Bound ,08

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Kurtosis 4,948 ,155

95% Confidence Interval for Mean Lower Bound 3,0577

Upper Bound 3,2183

95% Confidence Interval for Mean Lower Bound ,3263

Upper Bound ,3857

95% Confidence Interval for Mean Lower Bound 1,7290

Upper Bound 1,8310

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Upper Bound 1,3177

Percentiles

Percentiles

Weighted

Average(Definiti

on 1)

JoiningYear 2012,000

0 2013,000 0 2014,000 0 2015,000 0 2017,000 0 2017,000 0 2018,000 0

PaymentTie

r

1,0000 2,0000 2,0000 3,0000 3,0000 3,0000 3,0000

Age 24,0000 24,0000 25,0000 26,0000 27,0000 28,0000 30,0000

Experience 1,0000 2,0000 2,0000 3,0000 4,0000 5,0000 5,0000

EverBenche

d

,00 ,00 ,00 ,00 ,00 1,00 1,00

LeaveOrNot ,0000 ,0000 ,0000 ,0000 1,0000 1,0000 1,0000

Edu 1,0000 1,0000 1,0000 1,0000 1,0000 2,0000 2,0000

CT 1,0000 1,0000 1,0000 2,0000 3,0000 3,0000 3,0000

Tukey's Hinges JoiningYear 2014,000

0 2015,000 0 2017,000 0

PaymentTie

r

2,0000 3,0000 3,0000

Age 25,0000 26,0000 27,0000

Experience 2,0000 3,0000 4,0000

EverBenche

d

,00 ,00 ,00

Edu 1,0000 1,0000 1,0000

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Correlations

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Sig (2-tailed) ,024

Sum of Squares and Cross-products 3416,544 108,520

Sum of Squares and Cross-products 108,520 673,600

* Correlation is significant at the 0.05 level (2-tailed)

The correlation between Joining Year and City is slightly positive, albeit weak, and is statistically significant (with a significance level below 0.05) The Pearson Correlation coefficient lies between 0.1 and 0.3, indicating a modest but noticeable relationship between these two variables.

Correlations

PaymentTier Experience

Sum of Squares and Cross-products 316,751 -2,634

Sum of Squares and Cross-products -2,634 1672,956

There is no correlation between Payment Tier and Experience (Sig > 0.05)

Correlations

Sum of Squares and Cross-products 6627,964 -11,790

Sum of Squares and Cross-products -11,790 277,775

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There is no correlation between Age and Education (Sig > 0.05)

Correlations

PaymentTier Edu

Sum of Squares and Cross-products 316,751 -51,505

Sum of Squares and Cross-products -51,505 277,775

There is a weak negative relationship between Payment Tier and Education (Sig

< 0.05, 0.1<| Pearson Correlation |< 0.3)

Correlations

EverBenched PaymentTier

Sum of Squares and Cross-products 91,596 4,314

Sum of Squares and Cross-products 4,314 316,751

There is no correlation between Ever Benched and Payment Tier (Sig > 0.05)

d,

Descriptives

Statistic Std Error

95% Confidence Interval for Mean Lower Bound 2015,0012

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Median 2015,0000

95% Confidence Interval for Mean Lower Bound 2,6581

Upper Bound 2,7279

95% Confidence Interval for Mean Lower Bound 26,3342

Upper Bound 26,6538

95% Confidence Interval for Mean Lower Bound ,59

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Minimum 0

95% Confidence Interval for Mean Lower Bound ,08

95% Confidence Interval for Mean Lower Bound 3,0577

Upper Bound 3,2183

95% Confidence Interval for Mean Lower Bound ,3263

Upper Bound ,3857

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Interquartile Range 1,00

95% Confidence Interval for Mean Lower Bound 1,7290

Upper Bound 1,8310

95% Confidence Interval for Mean Lower Bound 1,2523

Upper Bound 1,3177

With the confidence level of 95%, the confidence interval is from the lower bound to the upper bound on the descriptive tables.

2, Based on the human resource management perspective, here are some revised recommendations for businesses:

Since there's a positive correlation between the number of years an employee has worked and their city of residence, it's evident that those living farther away may exert more effort commuting, potentially impacting their loyalty Companies should consider implementing support policies for transportation or travel expenses, thereby boosting employee morale.

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Implementing a pay structure based on work experience is justified Employees with more experience should receive higher salaries compared to less experienced ones, fostering long-term commitment to the company.

Current wage structures do not reflect the educational levels of employees adequately For instance, individuals with doctorates may earn less than those with only bachelor's degrees, leading to a sense of unfairness Given the effort required to attain higher qualifications, it's crucial for businesses to align salaries with educational achievements to ensure long-term employee retention The observation that some employees, despite not being assigned work, receive equal or higher pay than their actively working counterparts, suggests a need for businesses to refine their salary distribution policies This will prevent disparities where more diligent workers are compensated less than those with less workload.

Case Study 2: Specialty Toys

1 Demand distribution for teddy bears is a normal probability distribution with confidence interval.

10000 < μ < 30000 = 1 - α = 90%

μ=x = 30000 + 2 10000 =20000

P (x < 30000) = 95%

⇔ P ( x −20000 σ < 30000 − 20000

σ ) = 95%

P (z < z ) = 95% ⇒ °

⇔ 30000 − 20000

σ = 6060 ⇔

2 Probability for the stock – out is probability for demand > order.

P (x > k) = P (z > k −20000 6060 ) = 1 – P (z < k −20000 6060 )

Order Probability [1 – P (z < k −20000

6060 )]

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28000 0,0934

3 Sell teddy bears for $24 and cost $16 per unit.

Profit per unit: 24 – 16 = $8 ⇒

All surplus inventory for $5 per unit

Loss in profit per unit: 5 – 16 = -11 ⇒

We have projected profit in the following table:

15000 8 x 10000 – 11 x 5000

= 25000

8 x 15000 = 120000

12000 0

18000 8 x 10000 – 11 x 8000

= -8000

8 x 18000 = 144000

14400 0

24000 8 x 10000 – 11 x 14000

= -74000

8 x 20000 – 11 x

4000 = 116000

19200 0

28000 8 x 10000 – 11 x 18000

= -118000

8 x 20000 – 11 x

8000 = 72000

22400 0

4 The order quantities meet 70%, and 30% chance of stock – out.

⇔ P (x < x ) = 70% °

P (z < ⇔ x ° −20000

6060 ) = 0,7

⇔ x ° −20000

6060 = 0,5244

x = 23177,864 ≈ 23178 ⇔ °

Profit if sales = 10000

P = 10000 x 8 – 11 x 13178 = -64958

Profit if sales = 20000

P = 20000 x 8 – 11 x 3178 = 125042

Profit if sales = 30000

P = 23178 x 8 = 185424

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5 Provide your own recommendation for an order quantity and note the associated profit projections Provide a rationale for your recommendation.

● According to the case and the above calculation, we get:

· underage cost: (Cu) = 24 – 16 = 8

· overage cost (Co)=16 – (16 – 5) = 5.

● According to the formula: Cu P(z)= Cu + Co

P(z)= 8 8 +5 = 0.62

● In the standard normal distribution function table, we find: P (z ≤ 0.31) = 0.5 + 0.12172 = 0.62172

● 0.62172 is the probability value closest to 0.62.

● According to the formula: z = x −μ

σ

● x= z *σ +μ = 0.31 × 6060 + 20000 = 21879

⇒ Hence, we would recommend Specialty Toys to have an order quantity of 21,879 because it will maximize the company’s profit.

Ngày đăng: 13/08/2024, 16:14