A researcher wants to determine if there's a statistically significant difference in how busy someone is based on whether they identify as an early bird or a night owl. The data consists of scores representing busyness levels for each group, and we need to test for a difference between the two groups at a significance level of $\alpha = 0.05$. We will perform an independent samples t-test.

1. Problem Description

A researcher wants to determine if there's a statistically significant difference in how busy someone is based on whether they identify as an early bird or a night owl. The data consists of scores representing busyness levels for each group, and we need to test for a difference between the two groups at a significance level of $\alpha = 0.05$. We will perform an independent samples t-test.

2. Solution Steps

First, we calculate the sample statistics for each group (Early Bird and Night Owl).

Early Bird data: 25, 28, 29, 31, 26, 30, 22, 23, 26

Night Owl data: 26, 10, 20, 17, 26, 18, 12, 23

Calculate the sample mean for each group:

$n_1 = 9$ (Early Bird sample size)

$n_2 = 8$ (Night Owl sample size)

$\bar{x}_1 = \frac{25+28+29+31+26+30+22+23+26}{9} = \frac{240}{9} \approx 26.67$

$\bar{x}_2 = \frac{26+10+20+17+26+18+12+23}{8} = \frac{152}{8} = 19$

Calculate the sample standard deviation for each group:

$s_1^2 = \frac{\sum (x_i - \bar{x}_1)^2}{n_1 - 1}$

$s_1^2 = \frac{(25-26.67)^2 + (28-26.67)^2 + (29-26.67)^2 + (31-26.67)^2 + (26-26.67)^2 + (30-26.67)^2 + (22-26.67)^2 + (23-26.67)^2 + (26-26.67)^2}{9-1}$

$s_1^2 = \frac{2.78+1.77+5.43+18.75+0.45+11.09+21.81+13.47+0.45}{8} = \frac{75.99}{8} \approx 9.50$

$s_1 = \sqrt{9.50} \approx 3.08$

$s_2^2 = \frac{\sum (x_i - \bar{x}_2)^2}{n_2 - 1}$

$s_2^2 = \frac{(26-19)^2 + (10-19)^2 + (20-19)^2 + (17-19)^2 + (26-19)^2 + (18-19)^2 + (12-19)^2 + (23-19)^2}{8-1}$

$s_2^2 = \frac{49+81+1+4+49+1+49+16}{7} = \frac{250}{7} \approx 35.71$

$s_2 = \sqrt{35.71} \approx 5.98$

Calculate the pooled variance:

$s_p^2 = \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2}$

$s_p^2 = \frac{(9-1)(9.50) + (8-1)(35.71)}{9+8-2} = \frac{8(9.50) + 7(35.71)}{15} = \frac{76+249.97}{15} = \frac{325.97}{15} \approx 21.73$

Calculate the t-statistic:

$t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{s_p^2(\frac{1}{n_1} + \frac{1}{n_2})}}$

$t = \frac{26.67 - 19}{\sqrt{21.73(\frac{1}{9} + \frac{1}{8})}} = \frac{7.67}{\sqrt{21.73(0.111 + 0.125)}} = \frac{7.67}{\sqrt{21.73(0.236)}} = \frac{7.67}{\sqrt{5.129}} = \frac{7.67}{2.265} \approx 3.386$

Calculate the degrees of freedom:

$df = n_1 + n_2 - 2 = 9 + 8 - 2 = 15$

The critical t-value for a two-tailed test with $\alpha = 0.05$ and $df = 15$ is $t_{crit} = 2.131$. Since our calculated t-statistic $|3.386| > 2.131$, we reject the null hypothesis.

3. Final Answer

There is a statistically significant difference in the busyness levels between early birds and night owls at the $\alpha = 0.05$ level.