Question

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.

Answer 1

Answer:

Confidence interval: (1.95,4.45)

Step-by-step 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:

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

Putting the values, we get,

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

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