Thx! 3. 75 statistics teachers took part in a study. Each teacher self-reported

Question

Thx!

3. 75 statistics teachers took part in a study. Each teacher self-reported the number of years they had been teaching (x), and then each teacher was observed over several classes to determine how many mistakes the teacher made, on average, during a class (y). The results were: Scatterplot of Mistake Rates and Experience a. Explain why a linear regression is not appropriate here b. The researcher decided to try a transformation to the original variables and instead plotted log o y against log1o x (seen below). Explain why a linear regression is not appropriate for these transformed variables. Scatterplot of log Y vs log X 82 c. The researcher went ahead (!) and made the regression line for the plot in b. If the intercept and slope were 0.89 and-0.36 respectively, find what average error rate the model predicts for a teacher with 5 years of experience

Explanation / Answer

Answer:

a) First of all, as you can see from the first figure that the scatter plot looks curvilinear in nature instead for linear plot.

So, this is the reason that we will not go for linear regression in this case instead we can use quadratic regression in this case and comparative to linear regression, quadratic regression will fit to the data very well in this case because the scatter plot is curvilinear...

b) We use log transformation for variety of reasons :

i) As the y variable cannot take negative values so using the log transformation our prediction will be positive every time...

ii) the log-log transformation often makes the data easier to look at and sometimes normalizes the variance across observations.

But the reason that in this case why linear regression will not fit good when the model uses log log trasformation it is because the scatter plot is looks like a funnel shaped diagram and does not looks like that log(Y) and log(X) are linearly related because as the value of x increases the mouth of the funnel will keep on getting bigger and bigger and linear regression will give poor estimates at that time.

c) The information in the question is not sufficient to attempt this part

but to be rough if we estimate log(Y) using given information => log(Y) = 0.89 + (-0.36)*log(5) = 0.6383

and if we plot the point (log(x),log(y)) = (0.6989,0.6383) then this point will lie very close to the original data point and hence the error will be very small ...

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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