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:

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

Level of significance and degrees of freedom

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.

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:

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

Confidence interval

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).