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Data are given for two variables 鈥淐hocolate Consumption鈥� and 鈥淣umber of Nobel Laureates鈥�.

Let x denote the variable 鈥淐hocolate Consumption鈥�.

Let y denote the variable 鈥淣umber of Nobel Laureates鈥�.

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).