You will need to use an independent t-test to see if there is a difference between the means. To perform an independent t-test, you need to find the t-statistic and compare it with the significance value (in this case, 5%).

The t-statistic is given by the following formula:

T = (x̄1 - x̄2) / sqrt[s^2 * (1/n1 + 1/n2)]

The only value we don't have from that formula is s^2, or the pooled sample variance. To find that value, we need to sum the squares of the s values, 1.5 and 1.8—so (1.5^2 + 1.8^2) = 5.49.

T = (17 - 19) / sqrt[5.49 * (1/13 + 1/10)] = (-2)/0.9855 = -2.029.

For the given degrees of freedom (21), the 5% confidence interval is at a t-value of +/- 1.720743. This value is outside of that range, and therefore it is statistically significant.

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