5. ANOVA calculations and rejection of the null hypothesis Aa Aa The following t

Question

5. 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 ina 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 Observations Sample Mean Squares (SS) 650 645 625 Treatment Private prep class High school prep class No prep class 40 40 40 87,750.00 97,500.00 107,250.00 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.) Sum of Squares (SS) Source Between treatments Within treatments df Mean Square (MS) ANOVA summary tables typically have a "Total" row not included in the partial table you just completed. Which of the following is a possible reason for including this row? The MStotal is used in the calculation of the F test statistic. O The SStotal is sometimes easier to calculate than SSbetween. Since SSwithin + SSbetween SStotal, you can use SStotal to calculate SSbetween The SStotal is used in the calculation of the F test statistic. The total sums of squares is the sometimes called the "error term."

Explanation / Answer

Ans:

SST=87750+97500+107250=292500

SSTR=[40*650^2+40*645^2+40*625^2]-[(650+645+625)*40]^2/120=14000

SSE=292500-14000=278500

ANOVA table:

Second option is correct.

SStotal sometimes is easier to calculate than SSbetween

SSwithin+SSbetween=SStotal

F statistic ratio of MSbwteen and MSwithin

F=MSbetween/MSwithin=2.941

When null hypothesis is true F-statistic is close to 0

When null hypothesis is fasle F-statistic large

If significance level is 0.05,F critical value=FINV(0.05,2,117)=3.074

As,F=2.941<3.074,We fail to reject null hypothesis.

(We reject null hypothesis for larger values of F)

SS df MS F p-value SSTR 14000 2 7000 2.941 0.0567 SSE 278500 117 2380.342 SST 292500 119

Related Questions

Navigate

• Browse (All)
• Browse 5
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.