Question

Is injecting medical cement effective in reducing pain for people who have suffered fractured vertebrae? A paper describes an experiment to compare patients who underwent vertebroplasty (the injection of cement) to patients in a placebo group who underwent a fake procedure in which no cement was actually injected. Because the placebo procedure was similar to the vertebroplasty procedure except for the actual injection of cement, patients participating in the experiment were not aware of which treatment they received.

All patients were asked to rate their pain at three different times—3 days, 14 days, and 1 month after the procedure. Summary statistics are given in the accompanying table.

	Pain Intensity
	Vertebroplasty Group		Placebo Group
	n = 78		n = 79
	mean	sd	mean	sd
3 days	4.2	2.8	3.9	2.9
14 days	4.3	2.9	4.5	2.8
1 month	3.9	2.9	4.6	3.0

1. part a Construct and interpret a 95% confidence interval for the difference in mean pain intensity 3 days after treatment for the vertebroplasty treatment and the fake treatment. (Use μ_{Vertebroplasty} − μ_Placebo. Round your answers to two decimal places.)

2.Construct and interpret 95% confidence intervals for the difference in mean pain intensity at 14 days after treatment. (Use μ_{Vertebroplasty} − μ_Placebo. Round your answers to two decimal places.) to

3.Construct and interpret 95% confidence intervals for the difference in mean pain intensity at 1 month after treatment. (Use μ_{Vertebroplasty} − μ_Placebo. Round your answers to two decimal places.) to

2.

An article describes a study in which 85 patients age 13 to 25 who were considered at-risk for mental illness were randomly assigned to one of two groups. Those in one group took four fish oil capsules daily. The other group took a placebo. After 1 year, 4% of those in the fish oil group and 26% of those in the placebo group had become psychotic.

Is it appropriate to use the two-sample z test of this section to test hypotheses about the difference in the proportions of patients receiving the fish oil and the placebo treatments who became psychotic?

p=

np=

Answer 1

1:

Here we have < =4.2, 51 =2.8, n =78 Iz =3.9, 5, -2.9, n =79 Level of significance: a =0.05 Confidence level: CL =0.95 Since we can assume that variances are equal so degree of freedom of t test will be: df = 1 +1 -2 =155 The pooled standard deviation is: -1) , + (n2-1)3 2 -2.8508 n +n - 2 Critical value: t, -1.975 The standard error is: SE =s, -+ = -2.8508*sqrt((1/78)+(1/79)) yn n = 0.455 Margin of error: ME = t - SE =0.8986 Confidence interval is: (1-7) ME =(4.2-3.9) + 0.8986 =(-0.6, 1.2) Excel function for critical value: ETINV(0.05,155) Hence, the required confidence interval is (-0.6,1.2).

Answer: (-0.60, 1.20)

2:

Here we have x =4.3, 5;=2.9, 1 =78 X=4.5, 5, 52.8, n =79 Level of significance: a =0.05 Confidence level: CL =0.95 Since we can assume that variances are equal so degree of freedom of t test will be: df = n +n - 2 =155 The pooled standard deviation is: (n-1)s?, +(n2 – 1)s, 2 -2.8501 n +n2 - 2 Critical value: t =1.975 The standard error is: 1 1 SE =,=+= Vn ne =2.8501*sqrt((1/78)+(1/79)) = 0.4549 Margin of error: ME = T, SE =0.8984 Confidence interval is: (Ę - Z) IME =(4.3-4.5) + 0.8984 =(-1.1, 0.7) Excel function for critical value: =TINV(0.05,155)

Answer: (-1.10, 0.70)

3:

Here we have X =3.9, S=2.9, 1 =78 12=4.6,53, n=79 Level of significance: a =0.05 Confidence level: CL =0.95 Since we can assume that variances are equal so degree of freedom of t test will be: df = 9, +n - 2 =155 The pooled standard deviation is: (n-1)s 1+(n2 - 1)S 2 -2.9507 ? V n +n -2 Critical value: t =1.975 The standard error is: SE = =+= yn n2 =2.9507*sqrt((1/78)+(1/79)) = 0.471 Margin of error: ME = t - SE =0.9302 Confidence interval is: (1-7) ME = (3.9-4.6) + 0.9302 =(-1.63, 0.23) Excel function for critical value: ETINV(0.05,155)

Answer: (-1.63, 0.23)

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