Chapter 25, Problem 15E

To determine

To write a $90\%$ confidence interval for the mean temperature difference between summer and winter in Europe and explain what your interval means.

Answer to Problem 15E

We are $90\%$ confident that the mean difference in temperature between summer and winter in Europe is between ${(-41.3251,-32.3415)}^0\text{F}$ .

Explanation of Solution

Now, the table of the mean high temperatures is given in the question. So, let us find the difference between the means of summer and winter as:

  

Now let us check the conditions for inference as:

Random condition: It is not satisfied but we assume that the measurements of wind speeds are representative for the population.

Independent condition: It is not satisfied because if the wind speed $6$ hours from now is very fast then the wind speed at this moment will very likely also be fast.

Normal condition: It is satisfied because the histogram of the differences symmetric and uni-model.

Thus, all the conditions are not met.

Now, it is also given that:

  $\begin{array}{l}n=12\\ c=90\%\end{array}$

The sample mean difference is calculated as:

  $\begin{array}{l}\overline{d}=\frac{-41-36-\mathrm{....}-55-47}{12}\\ =-36.8333\end{array}$

And the sample standard deviation of the difference is as:

  $\begin{array}{l}{s}_d=\sqrt{\frac{{(-41-(-36.8333))}^2+{(-36-(-36.8333))}^2+\mathrm{....}+{(-55-(-36.8333))}^2+{(-47-(-36.8333))}^2}{12-1}}\\ =8.6638\end{array}$

The degree of freedom is as:

  $\begin{array}{l}df=n-1\\ =12-1\\ =11\end{array}$

And therefore the t -value will be as:

  $t=1.796$

Thus, the confidence interval is then be as:

  $\begin{array}{l}\overline{d}-t_{\alpha /2}\times \frac{s_d}{\sqrt{n}}\\ =-36.8333-1.796\times \frac{8.6638}{\sqrt{12}}\\ =-41.3251\\ \overline{d}+t_{\alpha /2}\times \frac{s_d}{\sqrt{n}}\\ =-36.8333+1.796\times \frac{8.6638}{\sqrt{12}}\\ =-32.3415\end{array}$

Thus, we conclude that we are $90\%$ confident that the mean difference in temperature between summer and winter in Europe is between ${(-41.3251,-32.3415)}^0\text{F}$ .