Question

Assume that you want to test the claim that the paired sample data come from a population for which the mean difference is μd = 0. Compute the value of the t test statistic. Round intermediate calculations to four decimal places as needed and final answers to three decimal places as needed.
x 28 31 20 25 28 27 33 35
y 26 27 26 25 29 32 33 34

Answer 1

Answer:

Test statistic, $t_{s} = -0.603$ (to 3 dp)

Step-by-step explanation:

Deviation, d = x -y

Sample mean for the deviation

$\bar{d} = \frac{\sum x-y}{n}$

$\bar{d} = \frac{(28-6) + (31-27)+(20-26)+(25-25)+(28-29)+(27-32)+(33-33)+(35-34)}{8} \\\bar{d} = -0.625$

Standard deviation: $SD = \sqrt{\frac{\sum d^{2} - n \bar{d}^2}{n-1} }$

$\sum d^{2} = (28-26)^2 + (31-27)^2 +(20-26)^2 +(25-25)^2 +(28-29)^2 +(27-32)^2 +(33-33)^2 +(35-34)^2\\\sum d^{2} = 63$

$SD = \sqrt{\frac{63 - 8 * (-0.625)^2}{8-1} }$

SD =2.93

Under the null hypothesis, the formula for the test statistics will be given by:

$t_{s} = \frac{ \bar{d}}{s_{d}/\sqrt{n} } \\t_{s} = \frac{- 0.625}{2.93/\sqrt{8} }$

$t_{s} = -0.6033$