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:

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

$t_{s} = -0.6033$

Answer 2

Following are the calculation to the given question:

The given data is as follows:

$\to \bold{x} \ \ \ 28 \ 31 \ 20\ 25\ 28\ 27\ 33\ 35} \\\\\to \bold{y } \ \ \ 26\ 27\ 26\ 25\ 29\ 32\ 33\ 34}$

Let $\mu_d$ is the mean difference of the two populations.

The difference between the two population is calculated as follows:

$\bold{x\ \ \ \ \ \ \ \ y \ \ \ \ \ \ \ \ d=x-y }$

$\bold{28 \ \ \ \ \ \ \ \ 26 \ \ \ \ \ \ \ \ 2 }\\\\bold{31 \ \ \ \ \ \ \ \ 27 \ \ \ \ \ \ \ \ 4 }\\\\bold{20 \ \ \ \ \ \ \ \ 26 \ \ \ \ \ \ \ \ -6 }\\\\bold{25 \ \ \ \ \ \ \ \ 25 \ \ \ \ \ \ \ \ 0}\\\\bold{28 \ \ \ \ \ \ \ \ 29 \ \ \ \ \ \ \ \ -1 }\\\\bold{27 \ \ \ \ \ \ \ \ 32 \ \ \ \ \ \ \ \ -5 }\\\\bold{33 \ \ \ \ \ \ \ \ 33 \ \ \ \ \ \ \ \ 0 }\\\\bold{35 \ \ \ \ \ \ \ \ 34 \ \ \ \ \ \ \ \ 1 }$

The mean of the differences is calculated as follows:

$\bar{d}= \frac{1}{n} \Sigma d$

  $=\frac{2+4-6+0-1-5+0+1}{8}\\\\=\frac{-5}{8}\\\\=-0.625$

Hence, the mean of the diferences is $\bar{d}=-0.625$

The standard deviation of the ditferences is calculated as follows:

$S_d=\sqrt{\frac{1}{n-1}\Sigma (d_1 -\bar{d})^2}$  $=\sqrt{ \frac{(2+0.625)^2+(4+0.625)^2 + (-6+0.625)^2 +(0+0.625)^2 + (-1+0.625)^2+ (-5+0.625)+(0+0.625)+(1+0.625)}{8-1}} \\\\=\sqrt{ \frac{(2.625)^2+(4.625)^2 + (-5.372)^2 +(0.625)^2 + (-0.372)^2+ (-4.372)^2+(0.625)^2+(1.625)^2}{7}} \\\\=\sqrt{ \frac{6.890+21.390 +28.858 +0.390 + 0.138+ 19.114+0.390+2.640}{7}} \\\\=\sqrt{ \frac{79.807}{7}} \\\\=\sqrt{ 11.401} \\\\=3.376$

Hence, the standard deviation of the differences is $s_d =3.376$

The test statistic is calculated as follows:

$t=\frac{\bar{d}-\mu_d}{\frac{s_d}{\sqrt{n}}}$

$=\frac{-0.625}{\frac{3.376}{\sqrt{8}}}\\\\=\frac{-0.625}{1.19}}\\\\=-0.5252$

Hence, the test statistic is $t=-0.5252$

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