Assume that you want to test the claim that the paired sample data come from a population for which the mean difference is $\mu _ { d } = 0$ Compute the value of the t test statistics. Round intermediate calculations to four decimal
Places as needed and final answers to three decimal places as needed. $$\begin{array} { | l | l l l l l | }
\hline x & 9 & 6 & 7 & 5 & 12 \
\hline y & 6 & 8 & 3 & 6 & 7 \
\hline
\end{array}$$
A) t = 0.415
B) t = 1.292
C) t = 0.578
D) t = 2.890

Question

Assume that you want to test the claim that the paired sample data come from a population for which the mean difference is $\mu _ { d } = 0$ Compute the value of the t test statistics. Round intermediate calculations to four decimal
Places as needed and final answers to three decimal places as needed. $$\begin{array} { | l | l l l l l | } 
\hline x & 9 & 6 & 7 & 5 & 12 \
\hline y & 6 & 8 & 3 & 6 & 7 \
\hline
\end{array}$$
A) t = 0.415
B) t = 1.292
C) t = 0.578
D) t = 2.890

Quizplus · Accepted Answer

We first need to calculate the differences between the paired sample data:

$$\begin{aligned}d_1 &= x_1 - y_1 = 9-6=3 \d_2 &= x_2 - y_2 = 6-8=-2 \d_3 &= x_3 - y_3 = 7-3=4 \d_4 &= x_4 - y_4 = 5-6=-1 \d_5 &= x_5 - y_5 = 12-7=5\end{aligned}$$

Next, we calculate the sample mean and standard deviation of the differences:

$$\begin{aligned} \bar{d} &= \frac{1}{n}\sum_{i=1}^{n}d_i = \frac{3-2+4-1+5}{5} = \frac{9}{5} = 1.8\ s_d &= \sqrt{\frac{1}{n-1}\sum_{i=1}^{n}(d_i-\bar{d})^2} = \sqrt{\frac{1}{4}(1.2^2+0.2^2+2.2^2+0.8^2+3.2^2)} \approx 1.517\end{aligned}$$

Finally, we calculate the t test statistic:

$$t = \frac{\bar{d}-\mu_d}{s_d/\sqrt{n}} = \frac{1.8-0}{1.517/\sqrt{5}} \approx 1.292$$

Since the alternative hypothesis is not specified, we use a two-sided test and look up the critical value for a t-distribution with 4 degrees of freedom at a significance level of 0.05, which is approximately 2.776. Since the calculated t statistic (1.292) is less than the critical value (2.776), we fail to reject the null hypothesis that the population mean difference is zero. Therefore, we conclude that there is not enough evidence to support the claim that the paired sample data come from a population for which the mean difference is not equal to zero.

Assume That You Want to Test the Claim That the Paired