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Chapter 10: Q.10.131 (page 443)

Anorexia Treatment. Refer to Exercise 10.125 and find a 90% confidence interval for the weight gain that would be obtained, on average, by using the family therapy treatment.

Short Answer

Expert verified

(-10.2918,-4.2282) is the data's 90% confidence interval.

Step by step solution

01

GIven Information

The given data is,

BeforeAfterBeforeAfter83.394.380.57528691.581.677.882.591.983.895.286.7100.382.195.579.676.777.690.787.39883.592.576.976.889.993.894.2101.68691.773.494.9

02

Explanation

Below is a table of contents.

BeforeAfterDifference=dd2BeforeAfterDifference=dd283.394.3-1112180.57525.328.098691.5-5.530.2581.677.83.814.4482.591.9-9.488.3683.895.2-11.4129.9686.7100.3-13.6184.9682.195.5-13.4179.5679.676.72.98.4177.690.7-13.1171.6187.398-10.7114.4983.592.5-98176.976.80.10.0189.993.8-3.915.2194.2101.6-7.454.768691.7-5.732.4973.494.9-21.5462.25-123.51716.85
03

Explanation

The sample mean is

d炉=鈭�dn

=-123.517

=-7.2647

The standard deviation is,

sd=鈭�di2-鈭�di2nn-1

=1716.85-(-123.5)21717-1

=7.1574

04

Explanation

The degree of freedom is,

dof=n-1

=17-1

=16

The level of significance's critical value is 卤1.746.

The confidence level is,

CI=d炉卤ta2sdn

=-7.2647卤(1.746)7.157417

=(-10.2918,-4.2282)

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Most popular questions from this chapter

H2:渭1<渭2

Define the phrase independent samples.

In each of exercise 10.13-10.18, we have presented a confidence interval for the difference,渭1-渭2, between two population means. interpret each confidence interval

90%CI from-10to-5


In each of Exercises 10.75-10.80, we have provided summary statistics for independent simple random samples from non populations. In each case, use the non pooled t-test and the non pooled t-interval procedure to conduct the required hypothesis test and obtain the specified confidence interval.

x炉1=15,s1=2,n1=15,x炉2=12,s2=5andn2=15.

a. Two-tailed test, 伪=0.05

b. 95%confidence interval.

Provide an example (different from the ones considered in this section) of a procedure based on a paired sample being more appropriate than one based on independent samples.
See all solutions

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