For this assignment, we are to use the IBM SPSS software to input and analyze data set in which focuses on how sex, and feedback affects the self-efficacy scores that were measured via a psychological sound test , which had scores ranging from 0 to 100. This assignment focuses on the importance of utilizing the Analysis of Variance, otherwise known as ANOVA (Sullivan, 2010, p. 677). ANOVA is an inferential method that is used to test the equality of three or more population means. When using Analysis of Variance, the null hypothesis is always that the means of the different populations are equal. The alternative hypothesis is always that the mean of at least one population is different from the others (Sullivan, 2010, p. 677-678).
We created a frequency table and obtained the median, mode, standard deviation and variance for the self-efficacy variable. The results indicated that in regards to self-efficacy, the mean was at 74.65, median was 77.50, mode was 55.00, standard deviation was 15.31520, variance was at 234.555 and range was 47.00. This can be reaffirmed in the following table that was produced in IBM SPSS:
There are three variables that are being studied and they include sex, feedback and self-efficacy. There were two components of sex, either male or female and there were two components of feedback, which included no feedback and the informational feedback. The Self-efficacy score was measured in a scale of 0 to 100. From the analysis, we can observe that sex and feedback are independent variables and self-efficacy is the dependent variable. Due to the fact that we are measuring two independent variables, it necessitates the need to use ANOVA, otherwise known as the Analysis of Variance. The point here is to measure how sex and feedback conditions affect self-efficacy scores (Sullivan, 2010). Sex is measured in an ordinal scale, feedback is measured via nominal scale and self-efficacy is measured through an interval scale.
We are told that this data set represented the following study that states, “Participants were assigned to either the no feedback or informational feedback condition, where they were either not provided with feedback on a recent computer task (none) or were provided with detailed feedback on their performance and what they could do to improve (informational). Equal number of males and females were assigned to each condition. After exposure to the feedback condition, participants were asked to complete a self-efficacy test. It was expected that the informational group would have higher self-efficacy than those in the no feedback group. In addition, males were expected to have higher self-efficacy than the females. Lastly, females who received informational feedback were expected to have the highest self-efficacy scores.â€
The null hypothesis is Ho : u1 = u2 ; u3 = u4
The alternative hypothesis is H1: u1 ≠u2; u3 ≠u4In regards to the null and alternative hypothesis u1 represents informational group, u2 represents the no feedback group, u3 represents the male participants and u4 represents the female participants. We have to take into consideration that the null hypothesis is a statement of status quo or no difference and always contains a statement of equality. The null hypothesis is assumed to be true until we have evidence to the contrary (Sullivan, 2010, p. 506). The alternative hypothesis is a claim to be tested (Sullivan, 2010, p. 506-507).
We created a frequency table and obtained the median, mode, standard deviation and variance for the self-efficacy variable. The results indicated that in regards to self-efficacy, the mean was at 74.65, median was 77.50, mode was 55.00, standard deviation was 15.31520, variance was at 234.555 and range was 47.00. This is reiterated in the following ascertained frequency table:
The reason we ran a two-way ANOVA on this data is because two-way ANOVA allows us to compare the means differences between groups that have been split on two independent variables, which are called factors. A one-way ANOVA is used when we have one independent variable and one dependent variable, but when you have two independent variables and one dependent variable, it is preferred to use the two-way ANOVA (Sullivan, 2010). In this particular predicament, the two independent variables include sex and feedback conditions. The dependent variable is the self-efficacy score.
The interaction indicates that the performance of both men and women differed substantially in regards to the estimated marginal means and varied according to informational feedback and no feedback. From looking at the provided interaction graph, we can see that women faired much better in self efficacy scores when information feedback was used, and scored substantially lower when no feedback was used. Male performance generally improved when informational feedback was utilized as compared to no feedback. One also realizes that female performance was substantially higher than male performance when informational feedback was used. The mean score for men who were not given feedback was a 75.200, men’s mean score with informational feedback given was a 82.800. The mean score for women with no feedback was a 51.400, and women with informational feedback had a mean score of 89.00. This is reiterated in the following graph and table:
The Tukey Test is also known as the Honestly Significant Difference Tes or the Wholly Significance Test. This test is designed to compare pairs of means after the null hypothesis of equal means have been rejected. The Tukey test’s main objective is to be able to determine which population means differ significantly (Sullivan, 2010, p. 695). For this assignment, a Tukey is not used because there were fewer than three groups in the sex variable as well as the feedback variable.
The p value is <0.05, we have an F value of 24.667, and a significant value of 0.000, which is translated as p <0.05. Considering the substantial differences in men in men and women’s self efficacy scores in regards to no feedback and informational feedback, we will reject the null hypothesis and accept the alternative hypothesis of H1: u1 ≠u2; u3 ≠u4. In regards to the null and alternative hypothesis u1 represents informational group, u2 represents the no feedback group, u3 represents the male participants and u4 represents the female participants. We have to take into consideration that the null hypothesis is a statement of status quo or no difference and always contains a statement of equality. The null hypothesis is assumed to be true until we have evidence to the contrary (Sullivan, 2010, p. 506). The alternative hypothesis is a claim to be tested (Sullivan, 2010, p. 506-507).
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