Question

Let's use the data from Example 11.5 to estimate the difference in mean range of motion prior to treatment and the mean range of motion after ultrasound and stretch treatment for physical therapy patients. The data and the computed differences are shown in the accompanying table.

Subject	Range of Motion
	1	2	3	4	5	6	7
Pre-treatment	31	53	45	57	50	43	32
Post-treatment	32	59	46	64	49	45	40
Difference	−1	−6	−1	−7	1	−2	−8

We will use these data to estimate the mean change in range of motion using a 95% confidence interval, assuming that the 7 patients participating in this study can be considered as representative of physical therapy patients. The following boxplot of the 7 sample differences is not inconsistent with a difference population that is approximately normal, so the paired t confidence interval is appropriate.

A boxplot has a horizontal axis with values from −8 to about 1. The boxplot is also horizontal. The left whisker is approximately −8, the left edge of the box is approximately −7, the line inside the box is approximately, −2, the right edge of the box is approximately −1, and the right whisker is approximately 1.

The mean and standard deviation computed using the seven sample differences are -3.43 and 3.51, respectively. The t critical value for df = 6 and a 95% confidence level is 2.45, and therefore the confidence interval is

xd ± (t critical value) · sd √n	=	−3.43 ± (2.45)· 3.51 7
	=	−3.43 ± 3.25
	=	(−6.68, −0.18)

Based on the sample data, we can be 95% confident that the difference in mean range of motion is between -6.68 and -0.18. That is, we are 95% confident that the mean increase in range of motion after ultrasound and stretch therapy is somewhere between 0.18 and 6.68.

MINITAB output is also shown. MINITAB carries a bit more decimal accuracy, and reports a 95% confidence interval of (-6.67025, -0.18690).

Paired T-Test and CI: Pre, Post

Paired T for Pre – Post
	N	Mean	StDev	SE Mean
Pre	7	44.4286	9.9976	3.7787
Post	7	47.8571	10.8847	4.1140
Difference	7	−3.42857	3.50510	1.32480

95% CI for mean difference: (−6.67025, −0.18690)

T-Test of mean difference = 0 (vs not = 0): T-Value = −2.59    P-Value = 0.041

Suppose that for the 7 differences, the mean was −4.25, and the s_d was still 3.50510.

Calculate a 95% confidence interval for the this mean difference. (Round your answers to three decimal places.) ,

( ), ( )

Answer 1

From the given information,

The required 95% confidence interval is,

CI=-4.25 +- 2.45*(3.50510/squareroot(7))

CI= -4.25 +- 3.2458

CI= (-4.25-3.2458,-4.25+3.2458)

CI=(-7.4958,-1.0042)

Thank you.