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Q 12) Let's perform analysis on retention based on gender variable. You may find table completed in Question 5 useful for this analysis. At 90%
Q 12) Let's perform analysis on retention based on gender variable. You may find table completed in Question 5 useful for this analysis. At 90% confidence, test hypothesis that if a new option became available on campus for meals then male are less likely to stay with existing facility than female. Use subscript 1 for male and 2 for female. In the answers provide test statistics, p value and conclusion. | |||||||||||||||||||
Gender | Retention | The null hypothesis will be constructed so that currently, the males are just as or more likely to stay than the females. The alternative hypothesis is the males will leave. | |||||||||||||||||
Mean | SD | ||||||||||||||||||
Male | 3.046512 | 0.924623 | |||||||||||||||||
Female | 2.789474 | 1.228321 | |||||||||||||||||
Total | 2.967742 | 1.023779 | H0: (Male1) (Female2) | ||||||||||||||||
Ha: (Male1) > (Female2) | |||||||||||||||||||
1 to 5 rank with 1 stay, 5 leave | |||||||||||||||||||
Male 43 | Female 19 | This is a one tailed, left tailed test as the question is only asking for less retention visits. | |||||||||||||||||
2 | 1 | This is a test of two means with assumed normal distribution and equal variances. | |||||||||||||||||
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3 | 1 | The context of the comparison is to see if males are more likely to abandon this cafe if somewhere new opens. If the mean of the retention for males is higher than the mean of the female retention, this would indicate a greater probability of leaving. | |||||||||||||||||
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2 | 4 | t-Test: Two-Sample Assuming Equal Variances | |||||||||||||||||
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3 | 3 | Male 43 | Female 19 | ||||||||||||||||
3 | 3 | Mean | 3.046511628 | 2.789473684 | |||||||||||||||
4 | 3 | Variance | 0.854928018 | 1.50877193 | |||||||||||||||
4 | 3 | Observations | 43 | 19 | |||||||||||||||
3 | 2 | Pooled Variance | 1.051081191 | ||||||||||||||||
3 | 3 | Hypothesized Mean Difference | 0 | ||||||||||||||||
4 | 3 | df | 60 | ||||||||||||||||
4 | 1 | t Stat | 0.910110845 | ||||||||||||||||
1 | 3 | P(T<=t) one-tail | 0.183202895 | Calculated t value .9101 | Critical value 1.67 | ||||||||||||||
5 | t Critical one-tail | 1.670648865 | |||||||||||||||||
3 | P(T<=t) two-tail | 0.36640579 | Conclusion: | The calculated t value is not in the tail so, we can accept the null at alpha 0.05. | |||||||||||||||
3 | t Critical two-tail | 2.000297822 | The average of female retention is greater than male but, not enough to be staticially significant at 95% confidence level. | ||||||||||||||||
3 | Since the P value is .1832 < .05 the null hypothesis is accepted. | ||||||||||||||||||
3 | Stated planely, the retention rate of males would be about the same as females. | ||||||||||||||||||
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