Santa Fe Grill Restaurant Survey - Essay Example

Santa Fe Grill Restaurant Survey College: Hypotheses The two entrepreneurs can be advised on the basis of the following hypothesesMen and women have the same reasons for choosing the restaurant Different distances taken to travel to the restaurant have the same influence on the decision of choosing the restaurant by the customers. Advertisement has no influence on the decision of choosing the restaurant by the customers Age group has an influence on the decision of choosing the restaurant by the customers Numbers of children below 18 years of age have an influence on the decision of choosing the type of restaurant. There is no significant difference in people’s opinions in regard to the dining out survey between the two restaurants Results and discussion Hypothesis 1 Men and women have the same reasons for choosing the restaurant In regard to the first hypothesis, the 2 samples have been used; the sample came from different people who are not matched. In this case the samples are independent from each other. Therefore, the appropriate test for testing this hypothesis is independent sample t-test, which is used to find out if there is a difference in the attitude to catalogue shopping between men and women. The null hypothesis: H0: µmen = µwomen Alternative hypothesis: µmen≠ µwomen In this case, µ is the mean number of the individuals/participants who have not been matched 2) This hypothesis is two tailed because it involves the phrase difference. 3) The α level: α = .05 4) Because the standard deviation is not known, we opt for t-test instead of z-score test. The t value is calculated using SPSS as in the next section: Table 1.Descriptive statistics of each group (men and women) Group Statistics X32 -- Gender N Mean Std. Deviation Std. Error Mean ID Male 293 228.93 129.651 7.574 Female 154 220.34 128.663 10.368 Table 1 provides a descriptive statistics for each of the two given groups .we have 293 male in the 0(X32) (N) and on average, they have 228.93 X32with standard deviation of 7.574. we also have 154 females in the 1(X32) (N) and on average, they have 128X32 with standard deviation of 10.368 X32.The last column shows the two groups Std error mean. Table 2.Independent sample test Basing on table 2, the p value of Levenes test is 0.816.In this case, this p value is greater than 0.05 the alpha. Therefore, we will have to use the middle row of the output (‘labeled Equal variances assumed.’).So we will have to assume that the variances are equal and we need to use the middle row of the output. The labeled column ‘‘t’’, provides a calculated t value. In this case the t value is 0.667 assuming equal variance. The labeled column df provides the degree of freedom related with the test. In this case we have 445 degrees of freedom. The labeled column sig (2 –tailed) provides p value related to the t test. In this case, the p value 0.816. If p ≤ α, then reject H0.Therefore, 0.816 is not equal or less than 0.05.So we fail to reject the null hypothesis. This means that we failed to observe a difference in the reasons to choosing the restaurant between men and women. Hypothesis 2. Different distances taken to travel to the restaurant have the same influence to the decision of choosing the restaurant by the customers. The one way ANOVA tests if any of the several means differ from each other. In this case, the there are 3 categories of distances that include less than 1 mile,1-5 miles and more than 5 miles. 1) The null hypothesis: H0: µless than 1 mile= µ1-5 miles= µmore than 5 miles Whereby µ represents mean opinion. 2) Alternative hypothesis: H1: not H0 3) The α level: α = .05 4) Determination of the statistical test, the opinion is the approx ratio scaled as well as there are 3 multiple groups. In this case the between subject ANOVA is the one appropriate. The appropriate statistics is calculated using SPSS as in the next section: Table 3.Descriptive statistics. As shown in table 3, for each dependent variable, the output shows the sample size, minimum, maximum standard deviation, standard error and confidence interval for every independent variable level (quasi) Test of Homogeneity of Variances ID Levene Statistic df1 df2 Sig. .766a 2 446 .466 a. Groups with only one case are ignored in computing the test of homogeneity of variance for ID. The homogeneity test of variances output tests the H0: σ2less than 1 mile= σ21-5 miles = σ2more than 5 miles In this case, the p value of 0.466 is greater than the 0.05; we fail to reject the null hypothesis. This means that there is limited evidence that the variances aren’t equal and the variance assumption homogeneity may fit. ANOVA ID Sum of Squares df Mean Square F Sig. Between Groups 43866.629 3 14622.210 .864 .460 Within Groups 7549845.871 446 16927.906 Total 7593712.500 449 The 2 value in the table above is the between groups degrees of freedom, 446 is within groups degrees of freedom,0.864 is the F ratio from the F column,0.460 is the p value and 16927.906 is within groups mean square estimate of variance. The last task now involves whether to reject null hypothesis. In this case the p value of 0.460 is seen to be greater than the 0.05.Therefore we fail to reject the null hypothesis. This means different distances have no influence on the decision of choosing the type of restaurant. Hypothesis 3 Advertisement has no influence to the decision of choosing the restaurant by the customers Group Statistics X31 -- Ad Recall N Mean Std. Deviation Std. Error Mean ID Do Not Recall Ads 307 234.10 131.827 7.524 Recall Ads 143 207.03 124.596 10.419 The appropriate test for testing this hypothesis is independent sample t-test, which is used to find out if there is a difference in the ability of recalling the advertisement in the last 60 days. The null hypothesis: H0: µYes = µNo Alternative hypothesis: µYes≠ µNo In this case, µ is the mean number of the individuals/participants who have not been matched The α level: α = .05 Since the standard deviation is not known, we opt for t-test instead of z-score test. Descriptive statistics of each group (Yes and No) Group Statistics X31 -- Ad Recall N Mean Std. Deviation Std. Error Mean ID Do Not Recall Ads 307 234.10 131.827 7.524 Recall Ads 143 207.03 124.596 10.419 In the table, we have 307 of individual who respondent “Yes” while 143 respondent “No” In regards to the table above, the p value of Levenes test is 0.437.In this case, this p value is greater than 0.05 the alpha. Therefore, we will have to use the middle row of the output (‘labeled Equal variances assumed.’).So we will have to assume that the variances are equal and we need to use the middle row of the output. The labeled column ‘‘t’’, provides a calculated t value. In this case the t value is .437 assuming equal variance .The labeled column df provides the degree of freedom related with the test. In this case we have 448 degrees of freedom. If p ≤ α, then reject H0: 0.437 is not equal or less than 0.05.so we fail to reject the null hypothesis. This means that we failed to observe a difference in the ability of recalling the advertisement for the last 60 days. Hypothesis 4 Age group has an influence on the decision to choosing the restaurant by the customers We have five age groups that include 18-25, 26-34, 35-49, 50-59, 60 and over The null hypothesis: H0: µ18-25= µ26-34= µ35-49 = µ50-59 = µ60 and over Alternative hypothesis: H1: not H0 The α level: α = .05 The between subject ANOVA is the one appropriate. Test of Homogeneity of Variances ID Levene Statistic df1 df2 Sig. .108a 4 442 .980 a. Groups with only one case are ignored in computing the test of homogeneity of variance for ID. The homogeneity test of variances output tests the H0: σ28-25= σ235-49 = σ250-59 = σ260 and over Basing on the results, the p value of 0.98 is greater than the 0.05; we fail to reject the null hypothesis. This means that there is limited evidence that the variances aren’t equal. ANOVA ID Sum of Squares df Mean Square F Sig. Between Groups 104061.751 7 14865.964 .877 .524 Within Groups 7489650.749 442 16944.911 Total 7593712.500 449 The value 7 is the between groups degrees of freedom, 442 is within groups degrees of freedom,0.877 is the F ratio from the F column,0.524 is the p value and 16944.911 is within groups mean square estimate of variance. We fail to reject the null hypothesis since the p value of 0.524, is seen to be greater than the 0.05. Hypothesis 5 Number of children below 18 years of age has an influence on the decision of choosing the type of restaurant. We have five age groups that include No Children at Home, 1-2 Children at Home and More Than 2 Children at Home The null hypothesis: H0: µ No Children at Home = µ1-2 Children at Home = µ More than 2 Children at Home Alternative hypothesis: H1: not H0 The α level: α = .05 The between subject ANOVA is used as well to test the hypothesis Test of Homogeneity of Variances ID Levene Statistic df1 df2 Sig. .297 2 447 .743 The homogeneity test of variances output tests the H0: σ2 No Children at Home = σ21-2 Children at Home = σ2 More than 2 Children at Home The p value of 0.743 is greater than the 0.05; we fail to reject the null hypothesis. This means that there is limited evidence that the variances aren’t equal and the variance assumption homogeneity fits. ANOVA ID Sum of Squares df Mean Square F Sig. Between Groups 7311.822 2 3655.911 .215 .806 Within Groups 7586400.678 447 16971.814 Total 7593712.500 449 The 2 is the between groups degrees of freedom, 447 is within groups degrees of freedom,0.215 is the F ratio from the F column,0.806 is the p value and 16971.814is within groups mean square estimate of variance. Hypothesis 6 There is no significant difference on people’s opinion in regards to the dining out survey between the two restaurants The null hypothesis: H0: µ Joses Southwestern Cafe = µ Santa Fe Grill Alternative hypothesis: µ Joses Southwestern Cafe ≠ µ Santa Fe Grill This hypothesis is tow tailed because it involves the phrase difference. The α level: α = .05 The t value has been calculated as below: Group Statistics Mexican Restaurant Most Recently Eaten At . . . N Mean Std. Deviation Std. Error Mean ID Joses Southwestern Cafe 167 221.68 128.021 9.907 Santa Fe Grill 283 227.75 131.403 7.811 The p value of Levenes test is .633. It is seen to be greater than 0.05 the alpha. Therefore, we will have to use the middle row of the output (‘labeled Equal variances assumed.’).So we will have to assume that the variances are equal and we need to use the middle row of the output. The labeled column ‘‘t’’, provides a calculated t value. In this regard, t value is -.478 assuming equal variance (the sign is ignored for two tailed t test).s Since 0.478 is not equal or less than 0.05.so we fail to reject the null hypothesis. This means, there is no significant difference between the two restaurants. Competitive Advantage Advantages The market is not seen to be new as well as dilute.In this case,they learn from each other to gain experience. There is no much competition since there are no many competitors around. If there is,then they just offer general services/products. Area for improvement and Recommendations The landscape of consumer-controlled communication has been seen to emerge with a faster process of dynamic consumer learning (Thompson, 2002, p.23). Consequently, there is increased costs and decline in efficiency rates from traditional direct marketing activity (Brodie et al., 1997). Hence, this restaurant requires an integrated marketing communications approach which combines online channels, traditional media, affiliate partnerships, PR, social networks, people and products to be successful in its plan of entering into this highly competitive market (Griffin and Pustay 2010). The Restaurant will be required to coordinate their supply customers’ value-added information and marketing resources in order to build long-term relationships and create engagement. This will be achieved by applying brand (Service/ Product), audience, delivery and content (Tradingeconomics.com 2010). It will then be required to find the optimal mix based on our company and competitive position (Porter, 1998). Audience A major ingredient is to apply analytics and extensive data in order to understand the target segment. This will be through Creation of pen-portraits of the core customers that include lifestyle and demographic information which will help to optimize the media targeting strategy (Robinson et al., 1978). Brand The Restaurant’s service and product will need to have a unique selling proposition so that to provide a competitive advantage. It will be mandatory for the product to deliver superior value for the target audience as well as be differentiated clearly on service, features or price (Johnson et al., 2011). Content With the majority of the competitors employing the same linear marketing strategies to target the same consumer segments, content will be considered to be a point of competitive differentiation. When the Restaurant supplies value-added content it will provide a reason to consumers to engage with the brand as well as connect the company’s services or products (Grant, 2008). Delivery The Restaurant will then be required to connect digital media channels that are new with traditional channels for the creation of a push-pull system. Consumers will be required to engage with the media employed or else the company’s content delivery might be unsuccessful. To achieve this, the Restaurant will ensure the content is appropriately linked to the company brand positioning so that to create the emotional connection with the services or product. References Griffin, R. W. & Pustay, M. W 2010, International Business, 6th edn, Pearson, Boston, p500 Grant, R. M. 2008, Contemporary Strategy Analysis. Blackwell Publishing, Malden Brodie, R.J., Coviello, N.E., Brookes, R.W. and Little, V. (1997). Towards a paradigm shift in marketing: an examination of current marketing practices, Journal of Marketing Management, 13 (5), July, pp. 383-406. Robinson, S., Hichens, R. and Wade, D. 1978, The directional policy matrix-tool for strategic planning, Long Range Planning Journal, Vol. 11, pp.8-15. Johnson, G., Whittington, R., Scholes, K. (2011) Exploring Strategy. England: Prentice Hall Porter, M. E. 1998. Competitive Strategy: Techniques for Analyzing Industries and Competitors. New York: The Free Press. Thompson, J. 2002 Strategic Management, 4th Edition, London: Thomson. Tradingeconomics.com 2010, Free indicators for 231 countries, accessed on 22th Oct 2010, [online] available: http://www.tradingeconomics.com/ Read More