Stats: Standard Deviation Essay

Submitted By JennaWisnicky1
Words: 534
Pages: 3

1.
a. The mean kindergarten reading score is 72.2312 with a standard deviation of 10.31780; the mean family SES is .0617 with a standard deviation of .77634. For these scores, SSx = 62170.96, SSy = 351.976, and SP=1671.971. The correlation is r=SP/√(SSx)(SSy) = 1671.971/√(62170.96)(351.976) = 0.357

B. Ŷ = a + bX
Intercept = a = (mean of Y) - b(mean of X) = 72.2312 - 4.75(.0617) = 71.938
This indicates that the predicted reading score for someone with an SES of zero is 71.938.

Slope = b = (covariance btwn X and Y)/(Variance of X) = 2.863/.603 = 4.75
The slope is b = 4.75, so for every additional SES point we can expect an increase of 4.75 points in reading score.

Predicted children’s reading score = Ŷ = 71.938 + 4.75(SES)

Standard error of estimate = √SSerror/n-2 = √54228.654/585-2 = 9.64451

Standard error of the slope = Sb = standard error of estimate/√SSx = 9.65441/√351.976 = 0.514

t = b-β/Sb = 4.75-0/0.514 = 9.240
Interpretation: reject Ho and conclude that the slope is significantly different from zero. There is a significant positive linear association between children’s kindergarten reading score and family SES such that children’s reading score is predicted to increase by around 4.75 points for every one point increase in SES.

c. Standard error of estimate = √ Y-Ŷ²/n-2 = √SSerror/n-2 = √54228.654/585-2 = 9.64451; means the average distance between the actual SES score and the predicted reading score is 9.65541 points. In other words, on average we expect that any given reading score might be “off” by about 9.64451 points.

D. Correlation = r = b(sx/sy) = 4.75(.77634/10.31780) = 0.357; the squared correlation = r² = (0.357)² = 0.128.
In correlation analysis this is interpreted to mean that 12.8% of the total variance in children’s reading scores is shared with the SES score and vice versa. In regression, it would be interpreted to mean that 12.8% of the total variance in…