A study of the effect of exposure to color (red or blue) on the ability to solve puzzles used 42 subjects. Half the subjects (21) were asked to solve a series of puzzles while in a red-colored environment. The other half were asked to solve the same series of puzzles while in a blue-colored environment. The time taken to solve the puzzles was recorded for each subject. The 21 subjects in the red-colored environment had a mean time for solving the puzzles of 9.64 seconds with standard deviation 3.43; the 21 subjects in the blue-colored environment had a mean time of 15.84 seconds with standard deviation 8.65.
The two-sample t statistic for comparing the population means has value _____ (± 0.001).

The 21 subjects in the red-colored environment had a mean time for solving the puzzles of 9.64 seconds with standard deviation 3.43; the 21 subjects in the blue-colored environment had a mean time of 15.84 seconds with standard deviation 8.65.

Let [tex]\mu_1[/tex] = average time taken to solve the puzzle in red-colored environment.

[tex]\mu_2[/tex] = average time taken to solve the puzzle in blue-colored environment.

So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu_1=\mu_2[/tex] {means that there is no difference in time taken to solve both the puzzles}

Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu_1\neq \mu_2[/tex] {means that there is difference in time taken to solve both the puzzles}

The test statistics that would be used here Two-sample t test statistics as we don't know about the population standard deviation;

T.S. = [tex]\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] ~ [tex]t__n_1_-_n_2_-_2[/tex]

where, [tex]\bar X_1[/tex] = sample average time for solving the puzzles in the red-colored environment = 9.64 seconds

[tex]\bar X_2[/tex] = sample average time for solving the puzzles in the blue-colored environment = 15.84 seconds

[tex]s_1[/tex] = sample standard deviation for red-colored environment = 3.43 seconds

[tex]s_2[/tex] = sample standard deviation for blue-colored environment = 8.65 seconds

[tex]n_1[/tex] = sample of subjects in the red-colored environment = 21

[tex]n_2[/tex] = sample of subjects in the blue-colored environment = 21

Also, [tex]s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }[/tex] = [tex]\sqrt{\frac{(21-1)\times 3.43^{2}+(21-1)\times 8.65^{2} }{21+21-2} }[/tex] = 6.58

So, test statistics = [tex]\frac{(9.64-15.84)-(0)}{6.58 \sqrt{\frac{1}{21}+\frac{1}{21} } }[/tex] ~ [tex]t_4_0[/tex]

= -3.053

The value of two-sample t test statistics is -3.053.