Uploaded by Seven7 Post

Worksheet 11 - Chapter 10 - key

advertisement
Worksheet 11 – Chapter 10 – Simple Linear Regression
Name:____________________
Section: __________________
1. You want to develop a model to predict the selling price of homes based on assessed
value. A random sample of 30 recently sold single-family houses in a small city is
selected to study the relationship between selling price (in thousands of dollars) and
assessed value (in thousands of dollars). The data are in the HOUSE file.
Using the steps shown in the notes, the flow chart from the notes & book, perform a simple
linear regression analysis of this data. Use DDXL to perform the initial analysis, then provide
interpretations within the context of this problem of the values found from DDXL.
If appropriate, predict the selling price for a house whose assessed value is $170,000 and
create a 95% confidence interval for that value.
1.
Hypothesize the deterministic component of the model
𝐸𝐸(𝑦𝑦) = 𝛽𝛽0 + 𝛽𝛽1 π‘₯π‘₯
2. Use the sample data to estimate the unknown parameters in the model
a. Plot the data to a scatter plot
Determine whether fitting a line to the data seems appropriate based on the graph.
The data appear to follow a linear pattern.
b. Find the slope and interpret
οΏ½1 = 1.78171
𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 𝛽𝛽
For every $1000 dollar increase in assessed value the asking price increases by 1.7817 (thousands of
dollars)
c. Find the y-intercept and interpret
οΏ½0 = βˆ’122.344
𝑦𝑦 βˆ’ 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 = 𝛽𝛽
There is not a valid interpretation of y-intercept for this problems a house assessment value of $0 is
in the scope of the data sampled or possible.
d. Prediction equation: 𝑦𝑦� = βˆ’122.344 + 1.78171(𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉)
3. Specify the probability distribution of the random error term and estimate the standard deviation of
this distribution
a. Check assumptions of probability of random error
i. The mean of Ξ΅ (random error) is 0
ii. The variance of Ξ΅ is constant
iii. The probability distribution of Ξ΅ is normal
iv. Ξ΅'s are independent of one another
b. Find the value of variability of random error and interpret
𝑠𝑠 = 3.475
𝑠𝑠 = (3.475)2 = 12.0756
We expect approximately 95% of the observed values of selling price to lie within 2(3.475) =
6.95 thousand dollars of their respective least squares predicted selling price value.
2
4. Statistically evaluate the usefulness of the model
a. Hypothesis Test for 𝛽𝛽1
Hypotheses:
𝐻𝐻0 : 𝛽𝛽1 = 0
𝐻𝐻𝐴𝐴 : 𝛽𝛽1 β‰  0
Assumptions: (See step 3a above)
Test:
Test Statistic: t = 18.7
p-value ≀ 0.0001
Summary: At the 5% significance level, my p-value is less than alpha therefore reject H0. There
is sufficient evidence to suggest that the population slope of the regression line predicting
selling price ($1000) from assessed value ($1000) is different from 0.
Our model is statistically useful
b. Calculate coefficient of correlation (r) and interpret
π‘Ÿπ‘Ÿ = √. 926 = 0.9623
There is a strong positive linear association between selling price ($1000) and assessed value
($1000).
c. Calculate coefficient of determination (r2) and interpret
π‘Ÿπ‘Ÿ 2 = .926
About 92.6% of the sample variation in selling price ($1000) can be explained by using assessed
value ($1000) to predict selling price in our linear model.
d. Is the model practically useful?
Since r2 is large and s is small relative to the possible values for y the model is practically useful.
5. Use the model for prediction, estimation
a. Use line for prediction
𝑦𝑦� = βˆ’122.344 + 1.78171(𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉)
𝑦𝑦� = βˆ’122.344 + 1.78171(170)
𝑦𝑦� = 180.5467 π‘‘π‘‘β„Žπ‘œπ‘œπ‘œπ‘œπ‘œπ‘œπ‘œπ‘œπ‘œπ‘œπ‘œπ‘œ 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑
b. Create Confidence Intervals for estimation and interpret
We are 95% confident that the true mean selling price value for a house whose assessed value is
$170,000 lies between $178,710 dollars and $182,390.
Download