4. A storeowner wishes to compare the average amount of money high school and co

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4. A storeowner wishes to compare the average amount of money high school and college students spend on music downloads. He randomly selects 10 students from three ferent student populations: high school, undergraduate, and graduate. The statistical assumptions required to perform a one-way ANOVA to compare the means of these three groups are provided below. reasonable based on the data. A partially completed ANOVA table is Sum of squares Source Groups Error Total DEMean square 3240 4450 What is the value of the pooled standard deviation? A) 10.9S B) 24.60 C) 56.92 D) 120 S. In a study on scholastic test scores of entering college freshmen, a random sample of colleges across the nation is selected and the average SAT Math score for the freshman class is recorded. The colleges are categorized according to their affiliation: public, private, or church. Does it appear that freshmen entering the three different types of schools do equally well on the SAT Math? Computer output is included below. Source Groups Error Total fsquares DF Mean squareFP-value 31953.1 63906.2 353440.2 417346.4 5.696 0.005 63 65 5610.2 One of the assumptions in ANOVA is that the population standard deviations are equal. Determine whether each of the following statements is true or false. A) We may use side-by-side boxplots to assess if this assumption of equal population standard deviations seems reasonable. As long as the ratio of the largest to the smallest sample standard deviation is C) D) B) greater than 2, then the assumption seems to be satisfied. An estimate for the common standard deviation in the three populations equals 74.90 We may use Normal quantile plots to determine if the assumption of equal population standard deviations is reasonable.

Explanation / Answer

Pooled Standard Deviation=sqrt(Mean sum of Square)=sqrt(605)=24.6

i.e option is B

2) Option B is the true

Groups 4450-3240=1210 3 different groups therefore df=3-1=2 mean sum of square=(1210/2)=605 Errors 3240 total df-groups df=9-2=7 3240/7=462.857 Total 4450 total 10 observations therefore df=10-1=9

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