
HI6006 Competitive Strategy Editing Service
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Visual presentation of the prices based on product from different brands
Price of Calvin Klein | Price of Zara | Price of Tommy Hilfiger | Price of Ralph Lauren |
94.27 | 47.53 | 60.73 | 63.77 |
70.71 | 48.87 | 87.92 | 97.55 |
60.48 | 58.19 | 47.07 | 64.47 |
61.83 | 32.64 | 96.78 | 149.02 |
73.78 | 54.35 | 83.47 | 29.53 |
82.79 | 51.75 | 92.30 | 95.28 |
65.26 | 75.09 | 35.89 | 131.19 |
48.48 | 61.67 | 118.92 | 54.59 |
78.00 | 31.28 | 98.56 | 100.66 |
81.44 | 42.92 | 72.89 | 111.94 |
128.17 | 72.02 | 49.06 | 32.78 |
31.00 | 28.04 | 14.38 | 80.25 |
37.42 | 60.34 | 79.39 | 89.56 |
64.86 | 59.32 | 130.20 | 115.84 |
63.02 | 34.98 | 41.18 | 22.08 |
84.65 | 56.49 | 47.57 | 38.35 |
84.74 | 25.07 | 29.02 | 41.41 |
86.82 | 36.22 | 44.24 | 62.18 |
50.59 | 70.48 | 87.10 | 77.11 |
41.17 | 39.91 | 60.49 | 45.88 |
57.74 | 46.55 | 72.93 | 60.77 |
29.09 | 48.50 | 98.93 | 41.05 |
63.03 | 52.36 | 52.82 | 94.84 |
60.27 | 66.86 | 123.02 | 89.81 |
56.75 | 38.36 | 15.42 | 74.94 |
44.24 | 51.88 | 56.38 | 73.80 |
109.63 | 47.16 | 103.36 | 60.55 |
101.15 | 54.65 | 105.25 | 44.30 |
100.12 | 66.34 | 86.78 | 42.48 |
75.55 | 73.45 | 124.03 | 59.07 |
Charts
One of the fundamental aspects in representing the data output on prices is through distribution. The price distribution arises because the rates will not yield the same at every time due to various macro and micro economic factors. Therefore, there exist a scattering of values around central tendency. The location, shape and variability of the distributions can be analysed using the following tools. (Freedman, 2010)
Location – Mean & median; Spread – Variance, Standard deviation and range and Shape – Skewness and Kurtosis
The descriptive statistics of prices
| Calvin Klein |
|
|
|
|
Location | Mean | 69.56889 |
| Median | 65.05888 |
Spread | Standard Deviation | 23.37837 |
| Sample Variance | 546.5483 |
| Range | 99.08802 |
Shape | Kurtosis | 0.111549 |
| Skewness | 0.401199 |
| Minimum | 29.08674 |
| Maximum | 128.1748 |
| Count | 30 |
| Zara |
|
|
|
|
Location | Mean | 51.10847 |
| Median | 51.8144 |
Spread | Standard Deviation | 13.86985 |
| Sample Variance | 192.3726 |
| Range | 50.02136 |
Shape | Kurtosis | -0.77666 |
| Skewness | -0.05338 |
| Minimum | 25.06978 |
| Maximum | 75.09114 |
| Count | 30 |
| Tommy Hilfiger |
|
|
|
|
Location | Mean | 73.86924 |
| Median | 76.16283 |
Spread | Standard Deviation | 32.29041 |
| Sample Variance | 1042.671 |
| Range | 115.8186 |
Shape | Kurtosis | -0.86718 |
| Skewness | -0.06482 |
| Minimum | 14.38371 |
| Maximum | 130.2023 |
| Count | 30 |
| Ralph Lauren |
|
|
|
|
Location | Mean | 71.50168 |
| Median | 64.12269 |
Spread | Standard Deviation | 31.26372 |
| Sample Variance | 977.4204 |
| Range | 126.9439 |
Shape | Kurtosis | -0.07787 |
| Skewness | 0.600018 |
| Minimum | 22.07529 |
| Maximum | 149.0192 |
| Count | 30 |
From the above descriptive analysis it is noted that the mean and median value of Calvin Klein is different, the mean states the average of the distribution whereas the median states the 50 percentile of the data or the mid value. However, it is noted that mean and median of Zara is almost same at 51, however Tommy Hilfiger and Ralph Lauren is also have a difference between the mean and median
While measuring the spread of the data, the standard deviation of Calvin Klein, Tommy Hilfiger and Ralph Lauren has highest value, this shows that the deviation of the values from the mean is huge. Therefore a large standard deviation means the observed value has a huge difference when compared with mean. Therefore the goal should be less standard deviation which is noted in Zara. Therefore, it can be stated that the mean value of Zara has more meaningful information . (Triola, 2014)
In order to analyse the shape of the data, skewness and kurtosis is applied, skewness will state the symmetry of the data. A data to be normally distributed should possess the value of skewness as 0. Similarly Kurtosis is used to measure the combined size of two tails of information, for a data to be normally distributed should have value equal to 3. In the descriptive analysis it is noted that Zara and Tommy Hilfiger has a skewness value almost 0, whereas Calvin Klein and Ralph Lauren have values more than 0, therefore these two products are asymmetrical and the prices may not have normal distribution.
Visual representation
| Business Style |
|
|
|
|
Location | Mean | 89.14395 |
| Median | 87.36749 |
Spread | Standard Deviation | 22.28638 |
| Sample Variance | 496.6829 |
| Range | 73.71693 |
Shape | Kurtosis | -0.94622 |
| Skewness | 0.274778 |
| Minimum | 56.48535 |
| Maximum | 130.2023 |
| Count | 30 |
| Sports style |
|
|
|
|
Location | Mean | 64.06367 |
| Median | 61.19693 |
Spread | Standard Deviation | 15.32209 |
| Sample Variance | 234.7665 |
| Range | 62.71523 |
Shape | Kurtosis | -0.6026 |
| Skewness | 0.34497 |
| Minimum | 36.21763 |
| Maximum | 98.93286 |
| Count | 30 |
| Casual style |
|
|
|
|
Location | Mean | 41.33898 |
| Median | 41.17424 |
Spread | Standard Deviation | 13.02517 |
| Sample Variance | 169.655 |
| Range | 50.87537 |
Shape | Kurtosis | -0.36824 |
| Skewness | -0.11966 |
| Minimum | 14.38371 |
| Maximum | 65.25908 |
| Count | 30 |
From the above descriptive analysis it is noted that the mean and median value of Business style and Sports style wears are different, the mean states the average of the distribution whereas the median states the 50 percentile of the data or the mid value. However, it is noted that mean and median of Casual style wear is almost same at 41. Therefore the mean and median is closely related for Casual style wear. (Boslaugh, 2012)
While measuring the spread of the data, the standard deviation of Business style has highest value with 22.28, this shows that the deviation of the values from the mean is huge. Therefore a large standard deviation means the observed value has a huge difference when compared with mean. Therefore the goal should be less standard deviation which is noted in Sports style and Casual style with SD of 15.32 and 13.02 respectively. Therefore, it can be stated that the mean value of Business style and Casual style has more meaningful information .
In order to analyse the shape of the data, skewness and kurtosis is applied, skewness will state the symmetry of the data. A data to be normally distributed should possess the value of skewness as 0. Similarly Kurtosis is used to measure the combined size of two tails of information, for a data to be normally distributed should have value equal to 3. In the descriptive analysis it is noted that Casual style has a skewness value almost 0, whereas Business style and Sports style have values more than 0, therefore these two style are asymmetrical and the prices may not have normal distribution.
Zara business | Zara sports | Zara casual |
58.19 | 54.35 | 47.53 |
75.09 | 51.75 | 48.87 |
72.02 | 61.67 | 32.64 |
60.34 | 42.92 | 31.28 |
59.32 | 36.22 | 28.04 |
56.49 | 48.50 | 34.98 |
70.48 | 52.36 | 25.07 |
66.86 | 51.88 | 39.91 |
66.34 | 47.16 | 46.55 |
73.45 | 54.65 | 38.36 |
| Zara business style |
|
|
|
|
Location | Mean | 65.85872 |
| Median | 66.60369 |
Spread | Standard Deviation | 6.859775 |
| Sample Variance | 47.05652 |
| Range | 18.60579 |
Shape | Kurtosis | -1.73364 |
| Skewness | -0.07044 |
| Minimum | 56.48535 |
| Maximum | 75.09114 |
| Count | 10 |
| Zara sports style |
|
|
|
|
Location | Mean | 50.14484 |
| Median | 51.8144 |
Spread | Standard Deviation | 6.984534 |
| Sample Variance | 48.78372 |
| Range | 25.45067 |
Shape | Kurtosis | 1.056389 |
| Skewness | -0.56811 |
| Minimum | 36.21763 |
| Maximum | 61.6683 |
| Count | 10 |
| Zara casual style |
|
|
|
|
Location | Mean | 37.32185 |
| Median | 36.6685 |
Spread | Standard Deviation | 8.370247 |
| Sample Variance | 70.06104 |
| Range | 23.8008 |
Shape | Kurtosis | -1.32792 |
| Skewness | 0.102239 |
| Minimum | 25.06978 |
| Maximum | 48.87058 |
| Count | 10 |
From the above descriptive analysis it is noted that the mean and median value of all style of products of Zara looks same, the mean states the average of the distribution whereas the median states the 50 percentile of the data or the mid value. However, it is noted that mean and median of all products of Zara is almost same.
The goal should be less standard deviation which is noted in Business style, Sports style and Casual style of Zara products are less. Therefore, it can be stated that the mean value of all products are closely related to the mean value.
In order to analyse the shape of the data, skewness and kurtosis is applied, skewness will state the symmetry of the data. A data to be normally distributed should possess the value of skewness as 0. Similarly Kurtosis is used to measure the combined size of two tails of information, for a data to be normally distributed should have value equal to 3. In the descriptive analysis it is noted that business style has a skewness value almost 0, whereas sports style and Sports style have values more than 0, therefore these two style are asymmetrical and the prices may not have normal distribution. (Witte, 2010)
A t-test is one of the statistical test which helps the researcher to determine whether the two set of data is significantly different from each other. A t test is one of the commonly implemented test analysis, which will follow the steps of the normal distribution. It should be noted that when the scaling term is unknown and it is replaced by an estimate which is available on the data then we can follow students t-test. (Sincich, 2012)
The first step is to formulate the hypothesis, the null hypothesis will state that there mean difference is the same or to put it in other words, there is no significant difference between the variables. The alternate hypothesis will be stated as there is a significant difference among the variables. The interpretation of the data can be stated if the test statistic value is greater than the significance value then null hypothesis is rejected and alternate hypothesis is accepted. (Freedman, 2010)
In the given problem, the objective is to test if there is a significant difference in average prices across these two gender targets. Therefore the following hypothesis is framed.
Null hypothesis: There is no significant difference between average prices and gender
Alternate hypothesis:There is a significant difference between average prices and gender
t-Test: Two-Sample Assuming Unequal Variances |
|
|
|
|
|
|
|
| Price | Gender |
|
Mean | 66.51207 | 1.5 |
|
Variance | 754.4582 | 0.252101 |
|
Observations | 120 | 120 |
|
Hypothesized Mean Difference | 0 |
|
|
df | 119 |
|
|
t Stat | 25.92355 |
|
|
P(T<=t) one-tail | 0.004 |
|
|
t Critical one-tail | 1.657759 |
|
|
P(T<=t) two-tail | 0.008 |
|
|
t Critical two-tail | 1.9801 |
|
|
From the above t-test analysis it can be noted that mean value of price is 66.51 and the gender is 1.5, the variance of prices are high when compared with the gender.
It is further noted that t statistic is 25.92 whereas the p value is 0.004, this shows that the mean of values between price changes and gender are same. Therefore, null hypothesis is accepted. Hence it is stated that there is no significant difference between average prices and gender.
A t-test is one of the statistical test which helps the researcher to determine whether the two set of data is significantly different from each other. A t test is one of the commonly implemented test analysis, which will follow the steps of the normal distribution. It should be noted that when the scaling term is unknown and it is replaced by an estimate which is available on the data then we can follow students t-test.
The first step is to formulate the hypothesis, the null hypothesis will state that there mean difference is the same or to put it in other words, there is no significant difference between the variables. The alternate hypothesis will be stated as there is a significant difference among the variables. The interpretation of the data can be stated if the test statistic value is greater than the significance value then null hypothesis is rejected and alternate hypothesis is accepted.
In the given problem, the objective is to test if there is a significant difference in average prices across brands. Therefore the following hypothesis is framed.
Null hypothesis:There is no significant difference between average prices and brands
Alternate hypothesis:There is a significant difference between average prices and brands
t-Test: Two-Sample Assuming Unequal Variances |
|
|
|
|
|
| Price | Brands |
Mean | 66.27881803 | 2.512605042 |
Variance | 754.2678226 | 1.25195841 |
Observations | 119 | 119 |
Hypothesized Mean Difference | 0 |
|
df | 118 |
|
t Stat | 25.30703038 |
|
P(T<=t) one-tail | 0.008 |
|
t Critical one-tail | 1.657869523 |
|
P(T<=t) two-tail | 0.001 |
|
t Critical two-tail | 1.980272226 |
|
From the above t-test analysis it can be noted that mean value of price is 66.27 and the brands is 1.5, the variance of prices are high when compared with the brands.
It is further noted that t statistic is 25.30 whereas the p value is 0.008, this shows that the mean of values between price changes and brands are same. Therefore, null hypothesis is accepted. Hence it is stated that there is no significant difference between average prices and brands.
A t-test is one of the statistical test which helps the researcher to determine whether the two set of data is significantly different from each other. A t test is one of the commonly implemented test analysis, which will follow the steps of the normal distribution. It should be noted that when the scaling term is unknown and it is replaced by an estimate which is available on the data then we can follow students t-test.
The first step is to formulate the hypothesis, the null hypothesis will state that there mean difference is the same or to put it in other words, there is no significant difference between the variables. The alternate hypothesis will be stated as there is a significant difference among the variables. The interpretation of the data can be stated if the test statistic value is greater than the significance value then null hypothesis is rejected and alternate hypothesis is accepted. (Freedman, 2010)
In the given problem, the objective is to test if there is a significant difference in average prices across styles. Therefore the following hypothesis is framed.
Null hypothesis: There is no significant difference between average prices and styles
Alternate hypothesis:There is a significant difference between average prices and styles
t-Test: Two-Sample Assuming Unequal Variances |
|
|
|
|
|
| Prices | Styles |
Mean | 66.27881803 | 2.008403361 |
Variance | 754.2678226 | 0.66942031 |
Observations | 119 | 119 |
Hypothesized Mean Difference | 0 |
|
df | 118 |
|
t Stat | 25.51697319 |
|
P(T<=t) one-tail | 0.003 |
|
t Critical one-tail | 1.657869523 |
|
P(T<=t) two-tail | 0.007 |
|
t Critical two-tail | 1.980272226 |
|
From the above t-test analysis it can be noted that mean value of price is 66.27 and the styles is 2.0, the variance of prices are high when compared with the styles.
It is further noted that t statistic is 25.5 whereas the p value is 0.003, this shows that the mean of values between price changes and styles are same. Therefore, null hypothesis is accepted. Hence it is stated that there is no significant difference between average prices and styles.
A t-test is one of the statistical test which helps the researcher to determine whether the two set of data is significantly different from each other. A t test is one of the commonly implemented test analysis, which will follow the steps of the normal distribution. It should be noted that when the scaling term is unknown and it is replaced by an estimate which is available on the data then we can follow students t-test.
The first step is to formulate the hypothesis, the null hypothesis will state that there mean difference is the same or to put it in other words, there is no significant difference between the variables. The alternate hypothesis will be stated as there is a significant difference among the variables. The interpretation of the data can be stated if the test statistic value is greater than the significance value then null hypothesis is rejected and alternate hypothesis is accepted.
In the given problem, the objective is to test if there is a significant difference in Zara average prices across Zara styles. Therefore the following hypothesis is framed.
Null hypothesis:There is no significant difference between Zara average prices and Zara styles
Alternate hypothesis:There is a significant difference between Zara average prices and Zara styles
Zara prices and styles can be grouped as follows
PRICE | STYLE |
47.53 | 3 |
48.87 | 3 |
58.19 | 1 |
32.64 | 3 |
54.35 | 2 |
51.75 | 2 |
75.09 | 1 |
61.67 | 2 |
31.28 | 3 |
42.92 | 2 |
72.02 | 1 |
28.04 | 3 |
60.34 | 1 |
59.32 | 1 |
34.98 | 3 |
56.49 | 1 |
25.07 | 3 |
36.22 | 2 |
70.48 | 1 |
39.91 | 3 |
46.55 | 3 |
48.50 | 2 |
52.36 | 2 |
66.86 | 1 |
38.36 | 3 |
51.88 | 2 |
47.16 | 2 |
54.65 | 2 |
66.34 | 1 |
73.45 | 1 |
t-Test: Two-Sample Assuming Unequal Variances |
|
|
|
|
|
| Zara prices | Zara styles |
Mean | 51.23195651 | 1.965517241 |
Variance | 198.7692613 | 0.677339901 |
Observations | 29 | 29 |
Hypothesized Mean Difference | 0 |
|
df | 28 |
|
t Stat | 18.78610969 |
|
P(T<=t) one-tail | 0.0014 |
|
t Critical one-tail | 1.701130908 |
|
P(T<=t) two-tail | 0.002 |
|
t Critical two-tail | 2.048407115 |
|
From the above t-test analysis it can be noted that mean value of Zara average price is 51.23 and the styles of Zara is 1.96, the variance of prices are high when compared with the styles.
It is further noted that t statistic is 18.8 whereas the p value is 0.001, this shows that the mean of values between price changes and styles are same. Therefore, null hypothesis is accepted. Hence it is stated that there is no significant difference between Zara average prices and Zara styles.
Boslaugh, Sarah (2012). Statistics in a Nutshell. Cengage Publishing
Freedman, David (2010). Statistics. 4th Edition. Cengage Publishing
Sincich, T. Terry (2012). Statistics. 12th Edition.
Triola (2014). Essentials of Statistics. 5th Edition. McGraw Hill
Witte, S. Robert. (2010). Statistics. 5th Edition. McGraw Hill