Exercise Problems 3

Case Study: Survey of Students

This is the same data set we considered in Exercise Problems 1. The data is in studentsurvey

Problem 1 What can you say about the relationship between Score and Gender?

Problem 2 What can you say about the relationship between Score and GPA? Find a 95% interval estimate for a student with a GPA of 2.5. Is this an interpolation or an extrapolation?

Problem 3 What can you say about the relationship between Score and Distance? Find a 99% interval estimate for a student who lives 1.5 miles from the school. Is this an interpolation or an extrapolation?

Problem 4 What can you say about the relationship between Score and Age? Find a 90% interval estimate for the mean score of 21 year old students. Is this an interpolation or an extrapolation?

attach(studentsurvey) 

Problem 1 What can you say about the relationship between Score and Gender?

In problem 1 of the Exercise Problems 2 we ran the ANOVA and found a statistically significant difference between the scores of males and females. Because there are just two groups there is no reason to run tukey, but rerunning the same command gives us a 95% confidence interval for the difference in scores:

oneway(Score, Gender) 

## p value of test of equal means: p = 0.0033 
## Smallest sd:  1.7    Largest sd : 2.2 
## A 95% confidence interval for the difference in group means is (0.3, 1.2)

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Problem2 What can you say about the relationship between Score and GPA?

In problem 3 of the Exercise Problems 2 we found a statistically significant correlation between Score and GPA. Let’s find a good model.

slr(Score, GPA)

## The least squares regression equation is: 
##  Score  = 3.12 + 1.33 GPA 
## R^2 = 11%

the residual vs fits plt and the normal plot looks good, so no problem with the assumptions. We find the model

\[ \text{Score} = 3.12 + 1.33 \text{ GPA} \]

Find a 95% interval estimate for a student with a GPA of 2.5. Is this an interpolation or an extrapolation?

slr.predict(Score, GPA, newx=2.5, interval="PI")

##  GPA  Fit Lower Upper
##  2.5 6.45  2.65 10.24

so a 95% prediction interval for a student with a GPA of 2.5 is (2.65, 10.24)

This is an interpolation because 2.5 is in the range of GPAs in the data set

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Problem 3 What can you say about the relationship between Score and Distance?

In problem 5 of the Exercise Problems 2 we used a log transform on Distance. Doing so again yields

slr(Score, log(Distance + 1))

## The least squares regression equation is: 
##  Score  = 6.2 + 0.042 log(Distance + 1) 
## R^2 = 0.05%

the residual vs fits plt and the normal plot looks good, so no problem with the assumptions. We find the model

Score = 6.2 + 0.042 log(Distance+1)

Find a 99% interval estimate for a student who lives 1.5 miles from the school. Is this an interpolation or an extrapolation?

slr.predict(Score, log(Distance+1), 
        newx=log(1.5+1), interval="PI", conf.level = 99)

##  log(Distance + 1)  Fit Lower Upper
##          0.9162907 6.24  0.94 11.54

so a 95% prediction interval for a student who lives 1.5 miles from the school is (0.94, 11.54)

Note newx=log(1.5+1) because we have the predictor log(Distance+1).

This is an interpolation because 1.5 is in the range of Distances in the data set.

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Problem 4 What can you say about the relationship between Score and Age?

Before we saw that observation #220 is an outlier and removed it. We do the same now. Then

slr(Age[-220], Score[-220])

## The least squares regression equation is: 
##  Age[-220]  = 20.132 - 0.033 Score[-220] 
## R^2 = 0.43%

Find a 90% interval estimate for the mean score of 21 year old students. Is this an interpolation or an extrapolation?

slr.predict(Score[-220], Age[-220], 
            newx=21, interval="CI", conf.level = 90)

##  Age[-220]  Fit Lower Upper
##         21 6.13  5.82  6.44

so a 90% confidence interval for the mean score of 21 year old students is (5.82, 6.44)

This is an interpolation because 21 is in the range of Ages in the data set.