Question

compare the means of two groups. the first group includes 20 people with a mean of 42 and a standard deviation of 3.1. the second group includes 24 people with a mean of 40 and a standard deviation of 3.2. what is the calculated value of t?

t = 2.09
t = 2.11
t = 2.14
t = 2.16

Explanation:

Step1: Calculate the pooled variance

The formula for pooled variance \(s_p^2=\frac{(n_1 - 1)s_1^2+(n_2 - 1)s_2^2}{n_1 + n_2-2}\)
Here, \(n_1 = 20\), \(s_1=3.1\), \(n_2 = 24\), \(s_2 = 3.2\)
\((n_1 - 1)s_1^2=(20 - 1)\times3.1^2=19\times9.61 = 182.59\)
\((n_2 - 1)s_2^2=(24 - 1)\times3.2^2=23\times10.24 = 235.52\)
\(n_1 + n_2-2=20 + 24-2=42\)
\(s_p^2=\frac{182.59+235.52}{42}=\frac{418.11}{42}\approx9.955\)

Step2: Calculate the t - statistic

The formula for t - statistic \(t=\frac{\bar{x}_1-\bar{x}_2}{\sqrt{s_p^2(\frac{1}{n_1}+\frac{1}{n_2})}}\)
Here, \(\bar{x}_1 = 42\), \(\bar{x}_2 = 40\)
\(\frac{1}{n_1}+\frac{1}{n_2}=\frac{1}{20}+\frac{1}{24}=\frac{6 + 5}{120}=\frac{11}{120}\approx0.0917\)
\(\sqrt{s_p^2(\frac{1}{n_1}+\frac{1}{n_2})}=\sqrt{9.955\times0.0917}\approx\sqrt{0.913}\approx0.956\)
\(t=\frac{42 - 40}{0.956}=\frac{2}{0.956}\approx2.09\)

Answer:

\(t = 2.09\)