Question

FCAT scores and poverty. In the state of Florida, elementary school performance is based on the average score obtained by students on a standardized exam, called the Florida Comprehensive Assessment Test (FCAT). An analysis of the link between FCAT scores and sociodemographic factors was published in the Journal of Educational and Behavioral Statistics (Spring 2004). Data on average math and reading FCAT scores of third graders, as well as the percentage of students below the poverty level, for a sample of 22 Florida elementary schools are summarized by the number given below. (x= percentage of students below poverty level, and y=math score ) n = 22 ??xi = 1292.7 ??yi = 3781.1 ??x2i =88668 ??yi2 =651612 ??xiyi =218292 (a) Propose a straight-line model relating math-score to percentage of students below poverty level. (b) Find the least-squares regression line fitting the model to the data. (c) Interpret the estimates for intercept and slope in the context of the problem. (d) Test whether the math score is negatively related to the percentage of students below the poverty level. (e) Construct a 99% confidence interval for the slope of the model, and interpret your result in the context of the problem.

Answer 1

model relating math-score to percentage of students below poverty level

let x = percentage of students below poverty level

y = math-score

a) model is

y = b_{0 +} b₁x

given n = 22 xi = 1292.7 yi = 3781.1 x_i² =88668 yi² =651612 xiyi =218292

xbar = xi /n = 1292.7/22 = 58.76

ybar = yi /n = 3781.1/22 = 171.87

b₁ = S_xy/S_xx

= 12710.12

= 1761.22

= -3882

b₁ = S_xy/S_xx = -3882/12710.12 = -0.305

b₀ = ybar - b₁*xbar = 171.87 - (-0.305)58.76 = 189.79

slpoe = b₁ = -0.305

Intercept = b₀ = 189.79

y = 189.79 - 0.305x

c) interpretation

intercept:- when there are no students below poverty line the math score intercept which does not have any meaning.

slope:- With increase in 1% of students below poverty line there is decrease in 0.305 times of math score.

d) hypothesis

H₀: b₁ =0

H: b₁ < 0

t =

MS_E = (S_yy - b₁S_xy)/(n-2) = (1761.22 - (-0.305)(-3882))/20 = 28.86

t = = -6.46

= 0.05

t_0.05,20 = 1.725

critera for rejection = t < -t_0.05,20

since the condition satisfies -6.46 < -1.725 we reject null hypothesis and we conclude that slope is negative.

e) 99% confidence interval

t_0.005,20 = 2.845

width of confidence interval = = 2.845* = 0.134

99% CI = (-0.305-0.134 , -0.305 + 0.134)

99% CI = (-0.439 , 0.171)