Question

Confidence Interval for Mean Predicted Value Example 1 in this section illustrated the procedure for finding a prediction interval for an individual value of y. When using a specific value\((x_0}\)for predicting the mean of all values of y, the confidence interval is as follows:

\(\hat y - E < \bar y < \hat y + E\)

where

\[E = (t_(\frac(\alpha }(2}}}(s_e}\sqrt (\frac(1}(n} + \frac((n((\left( ((x_0} - \bar x} \right)}^2}}}((n\left( (\sum ((x^2}} } \right) - ((\left( (\sum x } \right)}^2}}}} \)

The critical value\((t_(\frac(\alpha }(2}}}\)is found with n - 2 degrees of freedom. Using the 23 pairs of chocolate/Nobel data from Table 10-1 on page 469 in the Chapter Problem, find a 95% confidence interval estimate of the mean Nobel rate given that the chocolate consumption is 10 kg per capita.

Short answer

The 95% confidence interval for the mean number of Nobel Laureates for the given value of chocolate consumption equal to 10 kgs is (17.1,26.0).

Step-by-step solution

Given information

Data are given for two variables “Chocolate Consumption” and “Number of Nobel Laureates”.

Let x denote the variable “Chocolate Consumption”.

Let y denote the variable “Number of Nobel Laureates”.

Referring to Example 1 of section 10-3, the provided information is:

Sample size, n = 23

\[(s_e} = (\bf(6}}(\bf(.262665}}\)

\(\hat y = - 3.37 + 2.49x\)

Formula of the confidence interval for predicting the mean of all values of y when specific value \[((\bf(x}}_(\bf(0}}}\) is known

The confidence interval for predicting the mean of all values of y by using the specific value\[(x_0}\)is as follows:

\(CI = \left( (\hat y - E,\hat y + E} \right)\)

where\[E = (t_(\frac(\alpha }(2}}}(s_e}\sqrt (\frac(1}(n} + \frac((n((\left( ((x_0} - \bar x} \right)}^2}}}((n\left( (\sum ((x^2}} } \right) - ((\left( (\sum x } \right)}^2}}}} \)

The critical value \[(t_(\frac(\alpha }(2}}}\) (also known as t multiplies) is found with n - 2 degrees of freedom and the significance level.

Predicted value at \(\left( ((x_0}} \right)\)

Substituting the value of\((x_0} = 10\)in the regression equation, the predicted value is obtained as follows:

\(\begin(array}(c}\hat y = - 3.37 + 2.49x\\ = - 3.37 + 2.49\left( (10} \right)\\ = 21.53\end(array}\)

Level of significance and degrees of freedom

The following formula is used to compute the level of significance

\[\begin(array}(c}(\rm(Confidence}}\;(\rm(Level}} = 95\% \\100\left( (1 - \alpha } \right) = 95\\1 - \alpha = 0.95\\ = 0.05\end(array}\)

The degrees of freedom for computing the value of the t-multiplier are shown below:

\(\begin(array}(c}df = n - 2\\ = 23 - 2\\ = 21\end(array}\)

The two-tailed value of the t-multiplier for level of significance equal to 0.05 and degrees of freedom equal to 21 is equal to 2.0796.

Value of \(\bar x,\)\(\sum (((\bf(x}}^(\bf(2}}}} \)and \(\sum (\bf(x}} \)

The following calculations are done to compute the \(\sum ((x^2}} \)and \(\sum x \):

The value of\(\bar x\)is computed as follows:

\(\begin(array}(c}\bar x = \frac((\sum x }}(n}\\ = \frac((134.5}}((23}}\\ = 5.847826\end(array}\)

Confidence interval

Substitute the values obtained from above to calculate the value of margin of error (E) as shown:

\[\begin(array}(c}E = (t_(\frac(\alpha }(2}}}(s_e}\sqrt (\frac(1}(n} + \frac((n((\left( ((x_0} - \bar x} \right)}^2}}}((n\left( (\sum ((x^2}} } \right) - ((\left( (\sum x } \right)}^2}}}} \\ = \left( (2.0796} \right)\left( (6.262665} \right)\sqrt (\frac(1}((23}} + \frac((23((\left( (10 - 5.847826} \right)}^2}}}((23\left( (1023.05} \right) - ((\left( (134.5} \right)}^2}}}} \\ = 4.44286\end(array}\)

Thus, the confidence interval becomes:

\(\begin(array}(c}CI = \left( (\hat y - E,\hat y + E} \right)\\ = \left( (21.53 - 4.44286,21.53 + 4.44286} \right)\\ = \left( (17.0871,25.9729} \right)\\ = \left( (17.1,26.0} \right)\end(array}\)

Therefore, 95% confidence interval for the mean number of Nobel Laureates for the given value of chocolate consumption equal to 10 kgs is (17.1, 26.0).