Hypothesis testing with One Way ANOVA.Use hypothesis testing to answer the research questions, labeling each step and completing any post-tests.Researchers want to know if surroundings make a difference in ability to remember a list of words. They test this by randomly assigning subjects to three conditions: shopping mall setting, driving in a car setting (simulation) and in an outdoor setting. Here are the memory scores for each condition:Mall ConditionDriving ConditionOutdoor Condition8815771085126899101412101071288810

Name: Hypothesis testing with One Way ANOVA.Use hypothesis testing to answer
Uploaded: 2024-03-20T07:07:00.000Z
Channel: Bartleby
Description: Hypothesis testing with One Way ANOVA.Use hypothesis testing to answer the research questions, labeling each step and completing any post-tests.Researchers want to know if surroundings make a difference in ability to remember a list of words. They t

The one-way ANOVA compares the means between the groups you are interested in and determines whether any of those means are statistically significantly different from each other. Specifically, it tests the null hypothesis: where µ = group mean and k = number of groups. If, however, the one-way ANOVA returns a statistically significant result, we accept the alternative hypothesis (HA), which is that there are at least two group means that are statistically significantly different from each other. The following table is obtained: Mall Condition Driving Condition Outdoor Condition 8 8 15 7 7 10 8 5 12 6 8 9 9 10 14 12 10 10 7 12 8 8 8 10 Sum = 65 68 88 Average = 8.125 8.5 11 \sum_i X_{ij}^2 =∑iXij2= 551 610 1010 St. Dev. = 1.808 2.138 2.449 SS = 22.875 32 42 n = 8 8 8 The total sample size is N = 24N=24. Therefore, the total degrees of freedom are: df_{total} = 24 - 1 = 23dftotal=24−1=23 Also, the between-groups degrees of freedom are df_{between} = 3 - 1 = 2dfbetween=3−1=2, and the within-groups degrees of freedom are: df_{within} = df_{total} - df_{between} = 23 - 2 = 21dfwithin=dftotal−dfbetween=23−2=21 First, we need to compute the total sum of values and the grand mean. The following is obtained : Also, the sum of squared values is Based on the above calculations, the total sum of squares is computed as follows Similarly , The between sum of squares is computed . Now that sum of squares are computed, we can proceed with computing the mean sum of squares: Finally, with having already calculated the mean sum of squares, the F-statistic is computed as follows: ANOVA TABLE : Source of Variation SS df MS F Between Groups 39.08333 2 19.54167 4.236129 Within Groups 96.875 21 4.613095 Total 135.9583 23 Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: μ1 = μ2 = μ3 Ha: Not all means are equal The above hypotheses will be tested using an F-ratio for a One-Way ANOVA. Rejection Region Based on the information provided, the significance level is \alpha = 0.05α=0.05, and the degrees of freedom are df_1 = 2df1=2 and df_2 = 2df2=2, therefore, the rejection region for this F-test is R = \{F: F > F_c = 3.467\}R={F:F>Fc=3.467}. Test Statistics Decision about the null hypothesis Since it is observed that F = 4.236 > F_c = 3.467F=4.236>Fc=3.467, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p = 0.0285p=0.0285, and since p = 0.0285 < 0.05p=0.0285<0.05, it is concluded that the null hypothesis is rejected. Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that not all 3 population means are equal, at the \alpha = 0.05α=0.05 significance level.