Question

Referring to the table, suppose the researcher wants to obtain a 95% confidence interval estimate for the mean number of physical therapy programs received by patients who had exactly two muscle injuries. The confidence interval estimate is:

A researcher would like to analyze the effect of muscle injury on the number of physical therapy intervention programs received. She takes a random sample of 4 patients. For these 4 patients, she finds out how many times each had a muscle injury and how many physical therapy intervention programs they received. These data are presented in the table below:

Patient Number of Muscle Injuries Number of Physical Therapy Programs Received
1 1 4
2 2 6
3 1 3
4 0 1

The results of the simple linear regression are provided below.

Y= 1+2.5X;  = 0.5,  = 0.35355; 2-tailed p value = 0.0192 (testing B1)

		(4.7356, 7.2644)
		(7.2356, 9.7644)
		(4.1369, 7.8631)
		(6.6369, 10.3631)

Answer 1

Answer: (4.1369, 7.8631)

Using excel<data<data analysis<regression

Regression Analysis
	r²	0.962
	r	0.981
	Std. Error	0.500
	n	4
	k	1
	Dep. Var.	Physical Therapy
ANOVA table
Source	SS	df	MS	F	p-value
Regression	12.5000	1	12.5000	50.00	.0194
Residual	0.5000	2	0.2500
Total	13.0000	3
Regression output				confidence interval
variables	coefficients	std. error	t (df=2)	p-value	95% lower	95% upper
Intercept	1.0000	0.43	2.31	0.15	-0.86	2.86
Muscle Injury	2.5000	0.3536	7.071	.0194	0.9788	4.0212
Predicted values for: Physical Therapy
		95% Confidence Interval	95% Prediction Interval
Muscle Injury	Predicted	lower	upper	lower	upper	Leverage
2	6.000	4.1369	7.8631	3.1541	8.8459	0.7500