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.

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

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

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

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.

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

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

The following formula is used to compute the level of significance

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

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

\(\begin{aligned}{c}df &= n - 2\\ &= 23 - 2\\ &= 21\end{aligned}\)

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.

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

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

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

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

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

Thus, the confidence interval becomes:

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

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).