Question

The National Oceanic and Atmosphere Administration publishes temperature information of cities around the world in Climates of the World. A random sample of \(50\) cities gave the data on average high and low temperatures in January shown on the WeissStats in the site.

a. Obtain a scatterplot for the data.

b. Decide whether finding a regression line for the data is reasonable. If so, then also do parts (c)-(f)

c. Determine the interpret the regression equation for the data.

d. Identify potential outliers and influential observations.

e. In case a potential outlier is present, remove it and discuss the effect.

f. In case a potential influential observation is present, remove it and discuss the effect.

Short answer

Part a.

Part b. It can be noted that, there is no strong curvature present in the scatterplot, therefore it is reasonable to find the regression line to the data.

Part c. \(\hat{y}=-7.5692+0.9168x\)

Part d. There is no any potential outliers and influential observations.

Part e. Not applicable

Part f. Not applicable

Step-by-step solution

Part a. Step 1. Given information

The below table gives the average high and low temperatures of \(50\) cities.

Part a. Step 2. Graph

The below graph represents the given points and high temperature is on the horizontal axis and the low temperature is on the vertical axis.

Part b. Step 1. Explanation

It is reasonable to find the regression line for the data if there is no strong curvature present in the scatterplot.

It can be noted that, there is no strong curvature present in the scatterplot, therefore it is reasonable to find the regression line to the data.

Part c. Step 1. Explanation

Assume that, the response variable is low temperature and the predictor variable is high temperature of the cities.

The sample size \(n=50\).

Below are the necessary sums.

\(\sum x_{i}=2843\)

\(\sum y_{i}=2228\)

\(\sum x_{i}^{2}=181233\)

\(\sum x_{i}y_{i}=144636\)

To find \(s_{xy}\) and \(s_{xx}\):

\(s_{xx}=181233-\frac{2843^{2}}{50}=19580.02\)

\(s_{xy}=144636-\frac{(2843)(2228)}{50}=17951.92\)

To find the averages:

\(\bar{x}=\frac{2843}{50}=56.86\)

\(\bar{y}=\frac{2228}{50}=44.56\)

Hence the parameters are:

\(b_{1}=\frac{17951.92}{19580.02}\)

\(=0.9168\)

\(b_{0}=44.56-(0.9168)\times 56.86\)

\(=-7.5692\)

The regression equation to predict the low temperature \((y)\) from the high temperature \((x)\) is,

\(\hat{y}=-7.5692+0.9168x\)

From the regression equation, the low temperature is increase on average by \(0.9168\) if the high temperature is increase by \(1\).

Part d. Step 1. Concept Used

If a data point lies far from the regression line then it is an outlier.

If the removal of a point causes a considerable change in the regression equation then the point is called an influential observation. That is, the removal of a point causes a considerable change in the direction of the regression line.

Part d. Step 2. Calculation

The predicted values for the given data are summarized in the below table.

The below graph represents the given points and the fitted regression line.

From the plotted graph,

• All the points are closed to the regression line therefore there are no potential outliers in the dataset.

• The removal of a point does not cause any considerable change in the direction of the regression line therefore there are no potential influential observations.

Part e. Step 1. Explanation

Not applicable, because it was concluded that there are no outliers in part (d).

Part f. Step 1. Explanation

Not applicable, because it was concluded that there are no potential influential observation in part (d).