SPSS data analysis in marketing field Free Essay Example, 2250 words

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 ÃŽ ± level: ÃŽ ± = . 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 â€Å"Yes† while 143 respondent â€Å"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. We will write a custom essay sample on SPSS data analysis in marketing field or any topic specifically for you Only $17.96 $11.86/pageorder now 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 ≠¤ ÃŽ ±, 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 ÃŽ ± level: ÃŽ ± = . 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: ÏÆ'28-25= ÏÆ'235-49 = ÏÆ'250-59 = ÏÆ'260 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.