Huey and Dunham (1987) measured the running speed of fence lizards, Sceloporus merriam, in Big Bend National Park in Texas. Individual lizards were captured and placed in a 2.3-meter raceway, where their running speeds were measured. Lizards were then tagged and released. The data from the researchers is presented in modified form below. The lizard collections have occurred over three different years to see if the sprint speed of tagged lizards is changing over time.

Sprint Speed (m/s)

Lizard

Year One

Year Two

Year Three

1 1.43 1.37 1.60
2 1.56 1.30 1.71
3 1.64 1.36 1.83
4 2.13 1.54 1.92
5 1.96 1.82 1.09
6 1.89 1.79 2.06
7 1.72 1.72 1.86
8 1.80 1.80 1.78
9 1.87 1.87 2.04
10 1.61 1.88 2.13

The null hypothesis is that the mean of the lizard speed measurements are only different due to chance while the alternative hypothesis states that at least one year is different from the others.

[straight H subscript 0 : space straight mu subscript 1 equals space straight mu subscript 2 space equals space straight mu subscript 3 straight H subscript straight A : space at space least space one space straight mu subscript straight i space is space different space from space the space others]

Calculate the ANOVA table below.

Source of Variation Sum of Squares df Mean squares F-ratio [F subscript 0.05 left parenthesis 1 right parenthesis comma d f subscript g r o u p s end subscript comma d f subscript e r r o r end subscript end subscript]
Groups (treatments)
Error
Total

R2 =

Do the ANOVA results indicate that the null hypothesis (H0) can be rejected in favor of the alternative hypothesis (HA)? (yes or no)

Answers should be rounded to the nearest three decimal places where appropriate

applying one way ANOVA:

R² = SSR / SST

    = 0.133/1.807

     =0.0734

Source of VariationSS     df       MS       F   P-value    F0.05(2.27)

treatments               0.133      2      0.066      1.069   0.3574      3.354

error                       1.675     27     0.062    

Total                        1.807     29    

as p value is not significantly low:

Do the ANOVA results indicate that the null hypothesis (H0) can be rejected in favor of the alternative hypothesis (HA)? -No

Step-by-step explanation:

answer:   segregation violated the equal protection clause of the 14th amendment, which holds that no state can “deny to any person within its jurisdiction the equal protection of the laws.”

details/explanation:

brown v. board of education of topeka, decided by the us supreme court in 1954, ruled that all americans are entitled to the same civil liberties and protections in regard to access to education. until that decision, it was legal to segregate schools according to race, so that black students could not attend the same schools as white students.   an older supreme court decision, plessy v. ferguson (1896), had said that separate, segregated public facilities were acceptable as long as the facilities offered were equal in quality.   in the case of brown v. board of education, that standard was challenged and defeated.   segregation was shown to create inequality, and the supreme court unanimously ruled segregation to be unconstitutional.

the ruling was important in advancing civil rights because it affirmed that the 14th amendment applies to all rights and privileges of citizens, including access to   education.   this was being violated by states whose laws supported the segregation of schools.   section 1 of the 14th amendment reads as follows:

the ratio of grapes to blueberries she should use is 12: 3

step-by-step explanation:

if mandie is making 15 cups of fruit salad, the ratio of grapes to blueberries she should use is 12: 3.