13.0k views
3 votes
A questionnaire to assess knowledge of communicable diseases was administered. There were a total of 36 questions to which respondents could answer "agree", "disagree", or "don’t know". Scores could range from 0 to 36. The mean score for U.S. study participants was 20.6 with a standard deviation of 5.8, while the mean score for Mexican study participants was 17.4 with a standard deviation of 5.8. The number of U.S. and Mexican study participants was 185 and 86, respectively. Supply the 90 percent confidence interval for the difference between the two population means.

User Jusid
by
4.6k points

1 Answer

5 votes

Answer:

Confidence interval: (1.95,4.45)

Explanation:

We are given the following information:

U.S Scores


\bar{x}_1 = 20.6, s_1 = 5.8, n_1 = 185

Mexico Scores


\bar{x}_2 = 17.4, s_2 = 5.8, n_2 = 86

Formula:

Degree of freedom =
n_1 + n_2 -2 = 185 + 86 -2 = 269

Confidence interval:


(\bar{x}_1 - \bar{x}_2) \pm t_(critical)\bigg(\sqrt{\displaystyle(s_1^2)/(n_1) + \displaystyle(s_2^2)/(n_2)}\bigg)

Putting the values, we get,


t_(critical)\text{ at}~\alpha_(0.10)\text{ and degree of freedom 269} = \pm1.6505


(20.6-17.4) \pm 1.6505\bigg(\sqrt{\displaystyle(33.64)/(185) + \displaystyle(33.64)/(86)}\bigg)= 3.2 \pm 1.25 = (1.95,4.45)

User Nilesh Verma
by
5.1k points