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HI6007 Statistics For Percent Frequency Distribution Of The Furniture Order

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Questions:

1. Missy had an assistant randomly select 50 recent orders that included furniture. The assistant recorded the value, to the nearest dollar, of the furniture portion of each order. The data collected is listed below (data set also provided in accompanying MS Excel file).
 
136 281 226 123 178 445 231 389 196 175
211 162 212 241 182 290 434 167 246 338
194 242 368 258 323 196 183 209 198 212
277 348 173 409 264 237 490 222 472 248
231 154 166 214 311 141 159 362 189 260
 
a. Prepare a frequency distribution, relative frequency distribution, and percent frequency distribution for the data set using a class width of $50.

b. Construct a histogram showing the percent frequency distribution of the furnitureorder values in the sample. Comment on the shape of the distribution.

c. Given the shape of the distribution in part b, what measure of location would be most appropriate for this data set? Shown below is a portion of a computer output for a regression analysis relating Y (demand) and X (unit price).
 
2. ANOVA
df SS
Regression 1 5048.818
Residual 46 3132.661
Total 47 8181.479
Coefficients Standard Error
Intercept 80.390 3.102
X -2.137 0.248
 
a. Determine whether or not demand and unit price are related. Use α = 0.05.
b. Compute the coefficient of determination and fully interpret its meaning.Be very specific.
c. Compute the coefficient of correlation and explain the relationship between demand and unit price.
 
3. The following are the results from a completely randomized design consisting of 3 treatments.
Source of Variation
Sum of
Squares
Degrees of
Freedom
Mean
Square F
Between Treatments 390.58
Within Treatments (Error) 158.40
Total 548.98 23
 
Using α = .05, test to see if there is a significant difference among the means of the three populations. The sample sizes for the three treatments are equal.
 
4. In order to determine whether or not the number of mobile phones sold per day (y) is related to price (x1 in $1,000), and the number of advertising spots (x2), data were gathered for 7 days. Part of the Excel output is shown below.
ANOVA
df SS MS F
Regression 40.700
Residual 1.016
Coefficients Standard Error
Intercept 0.8051
x1 0.4977 0.4617
x2 0.4733 0.0387
 
a. Develop an estimated regression equation relating y to x1 and x2.
b. At α = 0.05, test to determine if the estimated equation developed in Part a represents a significant relationship between all the independent variables and the dependent variable.
c. At α = 0.05, test to see if β1 and β2 is significantly different from zero.
d. Interpret slope coefficient for X2.
e. If the company charges $20,000 for each phone and uses 10 advertising spots, how many mobile phones would you expect them to sell in a day?

 

Answer:

Answer 1.

Answer 1. a) 

The frequency distribution, relative frequency distribution, percent frequency distribution of the data is given below-

Class

Class boundary

Mid -value

Frequency

Relative frequency

Percent frequency

100-149

99.5-149.5

125

3

0.06

6%

150-199

149.5-199.5

175

15

0.30

30%

200-249

199.5-249.5

225

14

0.28

28%

250-299

249.5-299.5

275

6

0.12

12%

300-349

299.5-349.5

325

4

0.08

8%

350-399

349.5-399.5

375

3

0.06

6%

400-449

399.5-449.5

425

3

0.06

6%

450-499

449.5-499.5

475

2

0.04

4%

Total

 

 

50

1.00

100%

b) The constructed histogram showing the percent frequency distribution of the furniture order values in the sample are-

The distribution of prices of furniture order indicates that the frequency distribution is positively skewed and skewed to the left. Its right tail is longer than the left tail.

c) Location of measure:

Location of measures summarize the numbers by a “Typical” value. Although the three most common measures of location are mean, median and mode, but here the best location measure is “Mode”. Basically, “Mode” refers the maximum number of occurrences of any value or any class. Here, the class of amount $150-$199 has highest frequency (30%). Therefore, most of the purchased furniture costs between $150 to $199.

Answer 2.

ANOVA:

 

df

SS

MS

F-statistic

P-value (Significance F)

Regression

1

5048.8818

5048.8818

74.13779

3.78764E-11

Residual

46

3132.661

68.10132609

 

 

Total

47

8181.479

174.0740213

 

 

 

 

Coefficients

Standard Error

t-statistic

p-value

Intercept

80.39

3.102

25.91553836

1.755E-29

X

-2.137

0.248

-8.61693548

3.106E-11

 

Coefficient of Variation

0.617103338

Adjusted R-square

0.608779497

Correlation of coefficient

0.785559252


The significant p-value of calculated t-statistic (-8.61693548) is 0.0 (3.106E-11) for the predictor variable unit price (X). The p-value is less than 0.05 (= α). Therefore, the demand (Y) and unit price (X) are significantly and linearly related to each other with 95% probability.

 

b) The coefficient of variation is found to be 0.6171. It interprets that the explanatory factor “Unit Price (X)” can explain 61.71% variability of the response variable “Demand (Y)”.

c) The correlation coefficient between unit price and demand is 0.78556. It indicates a strong, positive and direct correlation between these two variables. That is for the increment of the unit price, the demand also increases and vice versa. Similarly, for the decrement of the unit price, the demand also decreases and vice versa (Zou, Tuncali & Silverman, 2003).

Answer no. 3.

ANOVA:

 

 

No. of treatments =

3

 

 

SS

df

MS

F-statistic

P-value (Significance F)

Regression

390.58

2

195.29

25.89072

2.14826E-06

Residual

158.4

21

7.542857

 

 

Total

548.98

23

23.8687

 

 

The average values of three populations are equal to each other. The reason is that the F-statistic is 25.89072 according to the ANOVA table. The p-value of the significant F-statistic is calculated as 0.0. The p-value is less than 0.05 (=α). Therefore, the null hypothesis of inequality of averages of three populations are rejected with 5% level of significance. Hence, for the equal sample sizes of three treatments, the means of the three populations are found to be equal with 95% probability (Weisberg, 2005).

Answer 4.

ANOVA:

No. of observations =

7

No. of variables =

2

 

 

df

SS

MS

F-statistic

P-value (Significance F)

Regression

2

40.7

20.35

80.11811

0.000593174

Residual

4

1.016

0.254

 

 

Total

6

41.716

6.95266667

 

 

 

 

Coefficients

Standard Error

t-statistic

p-value

Intercept

0.8051

 

 

 

X1

0.4977

0.4617

1.07797271

0.322465

X2

0.4733

0.0387

12.2299742

1.82E-05

 

Multiple R-square

0.97564484

Adjusted R-square

0.96346725

Correlation coefficient ( r )

0.98774735

a) To determine whether that the number of mobile phones sold per day (Y) is associated or not to price (X1 in $000) and the number of advertising spots (X2), the previous tables are executed.

b) The estimated linear regression equation is –

Mobile phones sold per day (y) = 0.8051 + 0.4977*Price (x1) + 0.4733*Number of advertising spots (x2) (Park, 2011).

c) The calculated F-statistics is 80.11811 in this ANOVA table. Its p-value is 0.0005 that is less than 0.05. Therefore, it is 95% evident that simultaneously both the predictor variables have linear significant association with the response variable. Individually, price do not have statistically significance with dependent variable (0.322>0.05) and number of advertising spots have statistically significance with dependent variable (0.0<0.05).

At 5% level of significance, β1 (p-value = 0.322465) is significantly different from 0. However, β2 (p-value = 1.82E-05) is not significantly different from 0.  d

d) The slope of variable X2 (number of advertising spots) is 0.4733. It indicates that the predictor “number of advertising spots” has positive linear association with the dependent variable mobile phones sold per day. It also infers that for 1 unit increase or decrease in “number of advertising spots”, “mobile phones sold per day” increases or decreases by 0.4733 unit respectively (Chatterjee & Hadi, 2006).

e) If the company charges $20000 for each phone and uses 10 advertising spots, then the number of mobile phones sold per day is –

0.8051 + 0.4977*20 + 0.4733*10 = 0.8051 + 9.954 + 4.733 = 15.4921 = 15 (approx.).

 

References:

Chatterjee, S., & Hadi, A. S. (2006). Simple linear regression. Regression Analysis by Example, Fourth Edition, 21-51.

Park, S. H. (2011). Simple linear regression. In International Encyclopedia of Statistical Science (pp. 1327-1328). Springer Berlin Heidelberg.

Weisberg, S. (2005). Simple linear regression. Applied Linear Regression, Third Edition, 19-46.

Zou, K. H., Tuncali, K., & Silverman, S. G. (2003). Correlation and simple linear regression. Radiology, 227(3), 617-628.

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