CHL Hospitality: Price Differentiation, Competition Analysis Case

This case study provides an action plan and spreadsheet analysis to address several key issues for Cunningham Holdings Limited's (CHL) Hospitality subsidiary. The analysis includes determining the average pricing of accommodations by brand, state, and location using pivot tables. Hypothesis testing, specifically ANOVA, is employed to evaluate price differentiation among accommodation brands, between states, and between locations. The study also investigates the impact of introducing the Comfort brand on internal competition, using t-tests to compare average prices with and without the Comfort brand. The results show statistical summaries and ANOVA outputs used to accept or reject null hypotheses about pricing differences. The findings from the analysis of 144 hotels, equally distributed across brands, states, and locations, indicate that while there are some observed average price variations, statistical tests are needed to confirm their significance. This detailed analysis helps CHL Hospitality understand its pricing strategies and competitive dynamics within the Australian hospitality market.

Part 1: Action plan
This section presents the action plan for solving each and every issue presented
Action plan for Issue 1:
This issue seeks to determine current average pricing of the accommodation by brands, states,
and locations
The following is the action plan to help solve issue 1;
i) Construct a Pivot Table from the data given (CHL Accommodation Data.xlsx)
ii) Restructure the Pivot Table fields to present the required information.
Action plan for Issue 2:
Determine whether price differentiation exists among the accommodation brands.
The following is the action plan to help solve issue 2;
i) Sort the data by BRAND. Copy and reorganize the data into three columns by brand
name.
Resort Cottage Classic
200.20 201.75 196.11
198.21 201.08 196.22
199.21 199.18 196.86
198.98 201.83 198.49
199.13 202.82 200.11
199.43 204.05 205.52
195.00 200.91 200.63
195.71 202.55 201.89
199.61 203.17 209.65
199.18 197.63 197.16
199.25 200.93 201.02
201.66 202.22 201.96
202.58 203.03 202.73
203.02 203.69 198.66
198.29 204.51 198.85
198.88 201.77 198.90
199.71 202.31 199.31
199.92 202.53 202.17
200.64 198.43 199.45
200.84 201.17 200.72
ii) Hypothesis testing procedure
a) Let u1 = average weekly rate for Resort brand u2 = average weekly rate for Cottage brand u3 =
average weekly rate for Classic brand
H0: u1 = u2 = u3
H1: at least one u is different
b) Level of significance (α) = 0.05

c) If p-value < α, Reject H0.
d) Using Excel data analysis add-inn, a One-Way ANOVA (Single factor) was ran
Action plan for Issue 3:
Determine whether price differentiation exists between states among the accommodation brands.
The following is the action plan to help solve issue 3;
i) Sort the data by STATE. Copy and reorganize the data into three columns by State name.
After sorting by state sort by brand and have the data as follows;

Brand NSW QLD VIC
1 200.2 199.71 199.18
1 198.21 199.92 199.16
1 199.21 200.64 199.38
1 198.98 200.84 199.55
1 199.13 198.58 199.55
1 199.43 199.20 202.50
1 195 199.87 195.66
1 195.71 199.99 195.77
1 199.61 199.45 202.49
1 199.18 204.08 200.26
1 199.25 204.69 197.84
1 201.66 205.99 198.36
1 202.58 206.12 200.58
1 203.02 199.05 200.65
1 198.29 199.49 207.01
1 198.88 202.07 207.01
2 201.75 201.77 203.93
2 201.08 202.31 207.20
2 199.18 202.53 202.65
2 201.83 198.43 198.14
2 202.82 201.17 198.43
iii) Hypothesis testing procedure
a) Three hypotheses are to be tested.
The first hypothesis will test the difference in average price for the three states
Let u1 = average weekly rate for NSW State u2 = average weekly rate for QLD State u3 =
average weekly rate for VIC State
H0: u1 = u2 = u3
H1: at least one u is different
The second hypothesis will test the difference in average price for the three brands
Let u1 = average weekly rate for Resort brand u2 = average weekly rate for Cottage brand
u3 = average weekly rate for Classic brand
H0: u1 = u2 = u3
H1: at least one u is different
The third hypothesis will test the significance of the interaction effect. The hypothesis is;
H0: There is no significant effect of interaction between state and brand on the
average price.
H1: There is significant effect of interaction between state and brand on the
average price.
b) Level of significance (α) = 0.05
c) If p-value < α, Reject H0.
d) Using Excel data analysis add-inn, ANOVA: Two-factor with replication was ran as
follows;

Action plan for Issue 4:
Determine whether price differentiation exists between locations among the accommodation
brands.
The following is the action plan to help solve issue 3;
i) Sort the data by LOCATION. Copy and reorganize the data into three columns by
Location name. After sorting by state sort by brand and have the data as follows;
Brand Metropolitan Cities Regional cities
1 200.20 198.21
1 199.21 199.13
1 198.98 199.43
1 198.29 204.08
1 198.88 199.18
1 199.71 202.50
1 199.92 197.84
1 200.64 200.65
1 200.84 195.00
1 198.58 195.71
1 199.20 199.18
1 199.87 201.66
1 204.69 199.99
1 199.05 199.45
1 199.49 198.36
1 202.07 200.58
1 199.16 199.61
1 199.38 199.25
1 199.55 202.58
1 199.55 203.02
iii) Hypothesis testing procedure
Three hypotheses are to be tested.
The first hypothesis will test the difference in average price for the two locations
Let u1 = average weekly rate for Metropolitan cities u2 = average weekly rate for regional
cities
H0 : μ1=μ2
H1 : μ1 ≠ μ2

The second hypothesis will test the difference in average price for the three brands
Let u1 = average weekly rate for Resort brand u2 = average weekly rate for Cottage brand
u3 = average weekly rate for Classic brand
H0: u1 = u2 = u3
H1: at least one u is different
The third hypothesis will test the significance of the interaction effect. The hypothesis is;
H0: There is no significant effect of interaction between location and brand on the
average price.
H1: There is significant effect of interaction between location and brand on the
average price.
e) Level of significance (α) = 0.05
f) If p-value < α, Reject H0.
Using Excel data analysis add-inn, ANOVA: Two-factor with replication was ran as
follows;
Action plan for Issue 5:
Determine if the introduction of the Comfort brand has increased internal competition among the
other accommodation brands.
The following is the action plan to help solve issue 2;

i) Sort the data by BRAND then by comfort. Copy and reorganize the data into six columns
by brand name and by whether comfort was introduced or not. We have a data as shown
below.
Classic
With Comfort Without Comfort With Comfort Without Comfort With Comfort Without Comfort
200.20 199.25 201.75 201.83 196.11 200.11
198.21 201.66 201.08 202.82 196.22 205.52
199.21 202.58 199.18 204.05 196.86 200.63
198.98 203.02 203.93 200.91 198.49 201.89
199.13 198.29 207.20 202.55 208.41 209.65
199.43 198.88 202.65 203.17 198.58 197.16
195.00 202.50 198.14 197.63 200.67 201.02
195.71 195.77 198.43 200.93 200.86 201.96
199.61 202.49 198.60 202.22 200.90 202.73
199.18 200.26 202.88 203.03 201.46 205.62
199.18 197.84 198.67 203.69 196.84 205.93
199.16 198.36 201.27 204.51 197.78 207.45
199.38 200.58 203.48 202.74 202.32 201.86
199.55 200.65 202.76 202.83 198.20 202.02
199.55 207.01 203.79 202.94 198.28 201.37
195.66 207.01 203.31 204.10 200.97 207.01
Resort Cottage
ii) Hypothesis testing procedure
a) Three hypotheses will tested
The first hypothesis will involve testing the difference in mean price for the
Resort brand for those with comfort and those without comfort. So we will have;
Let u1 = average weekly rate for Resort brand with comfort u2 = average weekly
rate for Resort brand without comfort
H0 : μ1=μ2
H1 : μ1 ≠ μ2
The second hypothesis will involve testing the difference in mean price for the
Cottage brand for those with comfort and those without comfort. So we will have;
Let u1 = average weekly rate for Cottage brand with comfort u2 = average weekly
rate for Cottage brand without comfort
H0 : μ1=μ2
H1 : μ1 ≠ μ2
The third hypothesis will involve testing the difference in mean price for the
Classic brand for those with comfort and those without comfort. So we will have;
Let u1 = average weekly rate for Classic brand with comfort u2 = average weekly
rate for Classic brand without comfort
H0 : μ1=μ2

H1 : μ1 ≠ μ2
b) Level of significance (α) = 0.05
c) If p-value < α, Reject H0.
d) Using Excel data analysis add-inn, a t-test (Two-Sample assuming equal variances) was
ran
Part 2: Spreadsheet analysis
Results on issue 1
Summary by Brand (Average Price)
Row Labels Count of Brand Average of Price Brand No of Hotel Average Price
1 48 200.2704167 Resort 48 200
2 48 202.65625 Cottage 48 203
3 48 201.4627083 Classic 48 201
Grand Total 144 201.463125
Summary by State (Average Price)
Row Labels Count of State Average of Price State No of Hotel Average Price
1 48 201.02375 NSW 48 201
2 48 202.196875 QLD 48 202
3 48 201.16875 VIC 48 201
Grand Total 144 201.463125
Summary by Location (Average Price)
Row Labels Count of Location Average of Price Location No of Hotel Average price
1 72 201.5591667 Metropolitan Cities 72 202
2 72 201.3670833 Regional Cities 72 201
Grand Total 144 201.463125

Results on issue 2
Anova: Single Factor
SUMMARY
Groups Count Sum Average Variance
Resort 48 9612.98 200.2704 7.433966
Cottage 48 9727.5 202.6563 9.547049
Classic 48 9670.21 201.4627 11.5182
ANOVA
Source of Variation SS df MS F P-value F crit
Between Groups 136.6128 2 68.30641 7.190346 0.001063 3.060292
Within Groups 1339.463 141 9.499739
Total 1476.076 143
Results on issue 3

Anova: Two-Factor With Replication
SUMMARY NSW QLD VIC Total
Resort
Count 16 16 16 48
Sum 3188.34 3219.69 3204.95 9612.98
Average 199.2713 201.2306 200.3094 200.2704
Variance 4.333012 6.491673 10.41843 7.433966
Cottage
Count 16 16 16 48
Sum 3233.45 3260.43 3233.62 9727.5
Average 202.0906 203.7769 202.1013 202.6563
Variance 3.427366 18.31522 6.162158 9.547049
Classic
Count 16 16 16 48
Sum 3227.35 3225.33 3217.53 9670.21
Average 201.7094 201.5831 201.0956 201.4627
Variance 17.66901 8.619863 9.5774 11.5182
Total
Count 48 48 48
Sum 9649.14 9705.45 9656.1
Average 201.0238 202.1969 201.1688
Variance 9.708803 11.96402 8.897547
ANOVA
Source of Variation SS df MS F P-value F crit
Brand 136.6128 2 68.30641 7.231241 0.001039 3.063204
State 39.26861 2 19.63431 2.078581 0.129093 3.063204
Brand * State (Interaction) 24.98265 4 6.245661 0.661195 0.620017 2.438739
Within 1275.212 135 9.446015
Total 1476.076 143
Results on issue 4

Anova: Two-Factor With Replication
SUMMARY Metropolitan CitiesRegional citiesTotal
Resort
Count 24 24 48
Sum 4802.71 4810.27 9612.98
Average 200.1129 200.4279 200.2704
Variance 7.539274 7.600104 7.433966
Cottage
Count 24 24 48
Sum 4854.02 4873.48 9727.5
Average 202.2508 203.0617 202.6563
Variance 11.68394 7.482232 9.547049
Classic
Count 24 24 48
Sum 4855.53 4814.68 9670.21
Average 202.3138 200.6117 201.4627
Variance 12.24893 9.776745 11.5182
Total
Count 72 72
Sum 14512.26 14498.43
Average 201.5592 201.3671
Variance 11.25642 9.514677
ANOVA
Source of Variation SS df MS F P-value F crit
Brand 136.6128 2 68.30641 7.275512 0.000991 3.061716
Location 1.328256 1 1.328256 0.141476 0.707395 3.909729
Brand * Location (Interaction) 42.5169 2 21.25845 2.264299 0.10775 3.061716
Within 1295.618 138 9.388537
Total 1476.076 143
Results on issue 5

t-Test: Two-Sample Assuming Equal Variances
With Comfort Without Comfort
Mean 198.57125 201.009375
Variance 2.570718333 9.58512625
Observations 16 16
Pooled Variance 6.077922292
Hypothesized Mean Difference 0
df 30
t Stat -2.797199163
P(T<=t) one-tail 0.004456971
t Critical one-tail 1.697260887
P(T<=t) two-tail 0.008913943
t Critical two-tail 2.042272456
t-Test: Two-Sample Assuming Equal Variances
With Comfort Without Comfort
Mean 201.695 202.496875
Variance 6.52328 2.723369583
Observations 16 16
Pooled Variance 4.623324792
Hypothesized Mean Difference 0
df 30
t Stat -1.054810551
P(T<=t) one-tail 0.149966654
t Critical one-tail 1.697260887
P(T<=t) two-tail 0.299933308
t Critical two-tail 2.042272456
t-Test: Two-Sample Assuming Equal Variances
With Comfort Without Comfort
Mean 199.559375 203.245625
Variance 9.51760625 10.68253292
Observations 16 16
Pooled Variance 10.10006958
Hypothesized Mean Difference 0
df 30
t Stat -3.28070814
P(T<=t) one-tail 0.001313995
t Critical one-tail 1.697260887
P(T<=t) two-tail 0.002627991
t Critical two-tail 2.042272456
Resort
Cottage
Classic
Part 3: Communication

From the issue 1 results, it was observed that a total of 144 hotels were considered in the study
with 48 hotels from each of the three brands (Resort, Cottage and Classic). Also the sample was
equally distributed between the three states and also equally distributed between the two
locations. The average price for the cottage was $203 while for the Resort and Classic were
$200 and $201 respectively. Based on the State, it was established that QLD state had an average
of $202 while both NSW and VIC had average of $201 each. In terms of location, results showed
that Metropolitan cities had an average price of $202 while regional cities had an average of
$201. These results shows that there is no significant differences in the average prices based on
brand, state and even location. However, this needs to be confirmed by a hypothesis test.
From issue two results, we can see that the p-value is 0.001 (a value less than 5% level of
significance), we therefore reject the null hypothesis and conclude that at least one brand has
different mean. This confirms Oscar belief that price differentiation exists between the Resort,
Cottage, and Classic brands at 5% level of significance.
From the results presented on issue 3, considering the ANOVA, we can see that for the brand,
the p-value is 0.001 (a value less than 5% level of significance), we therefore reject the null
hypothesis and conclude that at least one brand has different mean. For the State, the p-value is
0.129 (a value greater than 5%), we therefore fail to reject the null hypothesis and conclude that
there is no significant difference in the average price based on the State. Lastly, considering the
interaction between brand and state, we see that the p-value is 0.620 (a value greater than 5%
level of significance), we fail to reject the null hypothesis and conclude that there is no
significant effect of the interaction between brand and state. Based on the results, I would like to
inform Oscar that there is no price difference across the three states among the accommodation
brands.
From the results presented on issue 4, considering the ANOVA, we can see that for the brand,
the p-value is 0.001 (a value less than 5% level of significance), we therefore reject the null
hypothesis and conclude that at least one brand has different mean. For the Location, the p-value
is 0.707 (a value greater than 5%), we therefore fail to reject the null hypothesis and conclude
that there is no significant difference in the average price based on the Location. Lastly,
considering the interaction between brand and location, we see that the p-value is 0.108 (a value
greater than 5% level of significance), we fail to reject the null hypothesis and conclude that

there is no significant effect of the interaction between brand and location. Based on the results I
don’t agree with Oscar’s view that price difference exist at different locations among the
accommodation brands.
Considering the results presented on issue 5, we can see that the p-value in the case of Resort
brand is 0.009 (a value less than 5% level of significance), we therefore reject the null hypothesis
and conclude that the mean average price differs for the resort brand based on whether they have
comfort or not. Hotels with comfort were significantly cheaper than those without comfort.
Next considering the Cottage brand, we can see that the p-value is 0.300 (a value greater than 5%
level of significance), we therefore fail to reject the null hypothesis and conclude that there is no
significant difference in the mean price for the hotel with comfort and those without comfort.
Lastly, considering the Classic brand, we can see that the p-value in the case of Classic brand is
0.003 (a value less than 5% level of significance), we therefore reject the null hypothesis and
conclude that the mean average price differs for the Classic brand based on whether they have
comfort or not. Hotels with comfort were significantly cheaper than those without comfort.
In conclusion, based on the above results we can conclude that introduction of the Comfort brand
has significantly increased internal competition among the accommodation brands.