Question

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).

Answer 1

Answer:

The two-sample t statistic for comparing the population means is -3.053.

Step-by-step explanation:

We are given that 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.

Let $\mu_1$ = average time taken to solve the puzzle in red-colored environment.

$\mu_2$ = average time taken to solve the puzzle in blue-colored environment.

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

Alternate Hypothesis, $H_A$ : $\mu_1\neq \mu_2$      {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. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$   ~ $t__n_1_-_n_2_-_2$

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

$\bar X_2$ = sample average time for solving the puzzles in the blue-colored environment = 15.84 seconds

$s_1$ = sample standard deviation for red-colored environment = 3.43 seconds

$s_2$ = sample standard deviation for blue-colored environment = 8.65 seconds

$n_1$ = sample of subjects in the red-colored environment = 21

$n_2$ = sample of subjects in the blue-colored environment = 21

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

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

                               =  -3.053

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