Question

Exercises 13–28 use the same data sets as Exercises 13–28 in Section 10-1. In each case, find the regression equation, letting the first variable be the predictor (x) variable. Find the indicated predicted value by following the prediction procedure summarized in Figure 10-5 on page 493.

Internet and Nobel Laureates Find the best predicted Nobel Laureate rate for Japan, which has 79.1 Internet users per 100 people. How does it compare to Japan’s Nobel Laureate rate of 1.5 per 10 million people?

Short answer

The regression equation is\(\hat y = - \;8.44 + 0.203x\).

Thebest predicted Nobel laureate rate for Japan, which has 79.1 internet users per 100 people, will be approximately 5.1 per 10 million people.

The predicted value is not very close to the given value of 1.5 per 10 million people.

Step-by-step solution

Given information

The given data depicts the number of internet users and Nobel laurates (per 10 million people).

State the equation for the estimated regression line

The formula for the estimated regression line is

\(\hat y = {b_0} + {b_1}x\),

where

\({b_0}\)is the y-intercept,

\({b_1}\)is the slope estimate,

\(x\)is the explanatory variable, and

\(\hat y\)is the response variable (predicted value).

Let X denote the number of internet users and Y denote the number of Nobel laureates.

Compute the slope and intercept estimates

The calculations required to compute the slope and intercept are as follows.

The number of observations in the sample are \(\left( n \right) = 6\).

The slope is computed as

\(\begin{array}{c}{b_1} = \frac{{n\left( {\sum {xy} } \right) - \left( {\sum x } \right)\left( {\sum y } \right)}}{{n\left( {\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}}}\\ = \frac{{6 \times 2301.16 - 399.7 \times 30.4}}{{6 \times 27987.71 - {{399.7}^2}}}\\ = 0.2028\end{array}\).

The intercept is computed as

\(\begin{array}{c}{b_0} = \frac{{\left( {\sum y } \right)\left( {\sum {{x^2}} } \right) - \left( {\sum x } \right)\left( {\sum {xy} } \right)}}{{n\left( {\sum {{x^2}} } \right) - {{\left( {\sum x } \right)}^2}}}\\ = \frac{{30.4 \times 27987.71 - 399.7 \times 2301.16}}{{6 \times 27987.71 - {{399.7}^2}}}\\ = - 8.4430\end{array}\).

Thus, the estimated regression equation is

\(\begin{array}{c}\hat y = {b_0} + {b_1}x\\ = - 8.44 + 0.203x\end{array}\).

Checking the model

Refer to exercise 21 of section 10-1 for the following result.

1) The scatter plot does not show an approximate linear relationship between the variables.

2) The P-value is 0.056.

As the P-value is greater than the level of significance (0.05), the null hypothesis is failed to be rejected.

Therefore, the correlation is not significant.

Referring to figure 10-5, the criteria for a good regression model is not satisfied.

Therefore, the regression equation cannot be used to predict the value of y.

For bad models, the best-predicted value of a variable is simply its sample mean of response variables.

Compute the prediction

The best-predicted number of Nobel laureate rate for Japan, which has 79.1 internet users per 100 people, is the sample mean of the response variable.

The sample meanis computed as

\(\begin{array}{c}\bar y = \frac{{\sum y }}{n}\\ = \frac{{\left( {5.5 + 9 + ... + 0.1} \right)}}{6}\\ = 5.1\end{array}\).

Therefore, the best predicted Nobel laureate rate for Japan, which has 79.1 internet users per 100 people, will be approximately 5.1per 10 million people.

The Nobel laureate rate of 1.5 per 10 million people is not close enough to the predicted value of 5.1 per 10 million people.

Thus, the two values are not comparable.