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The subjects of a study by Dugoff et al. (A-5) were 10 obstetrics and gynecology interns at the University of Colorado Health Sciences Center. The researchers wanted to assess competence in performing clinical breast examinations. One of the baseline measurements was the number of such examinations performed.
The following data give the number of breast examinations performed for this sample of 10 interns.
Intern Number No. of Breast Exams Performed
1 30
2 40
3 8
4 20
5 26
6 35
7 35
8 20
9 25
10 20
Source: Lorraine Dugoff, Mauritha R. Everett, Louis Vontver, and Gwyn E. Barley, "Evaluation of Pelvic and Breast Examination Skills of Interns in Obstetrics and Gynecology and Internal Medicine," American Journal of Obstetrics and Gynecology, 189 (2003), 655-658. 6.4 CONFIDENCE INTERVAL FOR THE DIFFERENCE BETWEEN TWO POPULATION MEANS 177 Construct a 95 percent confidence interval for the mean of the population from which the study subjects may be presumed to have been drawn.

Respuesta :

akiran007

Answer:

The 95 percent confidence interval for the mean of the population from which the study subjects may be presumed to have been drawn is (19.1269, 32.6730).

Step-by-step explanation:

Intern            No. of Breast

Number        Exams Performed               X²

1                         30                                  900

2                        40                                 1600

3                        8                                        64

4                        20                                   400

5                          26                                 676

6                           35                               1225

7                            35                               1225

8                            20                                400

9                             25                              625

10                                 20                        400

                                  ∑ 259                  ∑ 7515

Mean= X`= ∑x/n= 259/10= 25.9

Variance = s²= 1/n-1[∑X²- (∑x)²/n]

= 1/0[7515- (259)²/10]= 1/9[7515- 6708.1]

= 806.9/9=89.655= 89.66

Standard Deviation= √89.655= 9.4687

Hence

The value of t with significance level alpha= 0.05 and 9 degrees of freedom  is t(0.025,9)= 2.262

The 95 % Confidence interval is given by

x`±t(∝,n-1) s/√n

So Putting the values

25.9± 2.262( 9.4687/√10)

= 25.9 ±2.262 (2.9943)

= 25.9 ± 6.7730

= 25.9 +6.7730=32.6730

25.9 -6.7730= 19.1269

= 19.1269, 32.6730

The 95 percent confidence interval for the mean of the population from which the study subjects may be presumed to have been drawn is (19.1269, 32.6730).

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