Question

A random sample of 30 words from Jane Austen’s Pride and Prejudice had a mean length of 4.08 letters with a standard deviation of 2.40. A random sample of 30 words from Henry James’s What Maisie Knew had a mean length of 3.85 letters with a standard deviation of 2.26. Which of the following is a correct expression for the 95% confidence interval for the difference in mean word length for these two novels?

a. (4.08−3.85)±2.576(2.4030+2.2630)$\frac{305}{1526}=0.200=20.0\%(4.08-3.85)\pm 2.576(\frac{2.40}{\sqrt{30}}+\frac{2.26}{\sqrt{30}})$

b. (4.08−3.85)±2.045(2.4030+2.2630)$\frac{305}{1526}=0.200=20.0\%(4.08-3.85)\pm 2.045(\frac{2.40}{\sqrt{30}}+\frac{2.26}{\sqrt{30}})$

c. (4.08−3.85)±2.045(2.40229+2.26229)$\frac{305}{1526}=0.200=20.0\%(4.08-3.85)\pm 2.045\left(\sqrt{\frac{2.{40}^2}{29}+\frac{2.{26}^2}{29}}\right)$

d. (4.08−3.85)±2.045(2.40230+2.26230)$\frac{305}{1526}=0.200=20.0\%(4.08-3.85)\pm 2.045\left(\sqrt{\frac{2.{40}^2}{30}+\frac{2.{26}^2}{30}}\right)$

e. (4.08−3.85)±2.576(2.40230+2.26230)$\frac{305}{1526}=0.200=20.0\%(4.08-3.85)\pm 2.576\left(\sqrt{\frac{2.{40}^2}{30}+\frac{2.{26}^2}{30}}\right)$

Short answer

Option (d) is the correct answer

Step-by-step solution

Given information

$$\begin{gathered} {\overline{x}}_1=4.08 \\ {\overline{x}}_2=3.85 \\ {n}_1=30 \\ {n}_2=30 \\ {s}_1=2.40 \\ {s}_2=2.26 \\ \alpha =0.05 \end{gathered}$$

Explanation

Either a null hypothesis or an alternative hypothesis is asserted.

$$\begin{gathered} df=min(n_1-1,n_2-1) \\ =min(30-1,30-1) \\ =29 \end{gathered}$$

And the t-value will be as:

$t_{\alpha /2}=2.045$

Therefore, the confidence interval will be as:

$$\begin{gathered} ({\overline{x}}_1-{\overline{x}}_2)\pm t\times \sqrt{\frac{{s_1}^2}{n_1}+\frac{{s_2}^2}{n_2}} \\ =(4.08-3.85)\pm 2.045\times \sqrt{\frac{2.{40}^2}{30}+\frac{2.{26}^2}{30}} \end{gathered}$$

Therefore, this implies that option (d) is the correct option.