Attempts: Score: /10 2. ANOVA calculations and rejection of the null hypothesis

Question

Attempts: Score: /10 2. ANOVA calculations and rejection of the null hypothesis Aa Aa The following table summarizes the results of a study on SAT prep courses, comparing SAT scores of students in a private preparation class, a high school preparation class, and no preparation class. Use the information from the table to answer the remaining questions. Number of Sum of Treatment Observations Sample Mean Squares (SS) Private prep class 60 680 265,500.00 High school prep class 650 60 276,120.00 No prep class 302,670.00 using the data provided, complete the partial ANOVA summary table that follows. (Hint: T, the treatment total, can be calculated as the sample mean times the number of observations. G, the grand total, can be calculated from the values of T once you have calculated them.) Mean square (MS) Sum of Squares (SS) Source Between treatments Within treatments In some ANovA summary tables you will see, the labels in the first (source) column are Treatment, Error, and Total. which of the following reasons best explains why the within treatments sum of squares is sometimes referred to as the error sum of squares? Q The within-treatments sum of squares measures treatment effects as well as random, unsystematic

Explanation / Answer

Sol:

Find standard deviation of sampel:

s1=sqrt[265500/60-1]

In Anova F statistic is the ratio of the between

the value of F=6.604

p=0.002

Reject Null hypothsis

Conclusion:

There is sufficient evidence at 5% level of significance that means are diff.

SS n-1 sample varinace sample sd sample mean 265500 59 4500 67.08204 680 276120 59 4680 68.41053 650 302670 59 5130 71.62402 635

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