Answer to Question #213378 in Statistics and Probability for arwa

Question

Chan et al. (A-9) developed a questionnaire to assess knowledge of prostate cancer. There was 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 scores for Caucasian study participants was 20.6 with a standard deviation of 5.8, while the mean scores for African-American men was 17.4 with a standard deviation of 5.8. The number of Caucasian study participants was 185, and the number of African-Americans was 86.

Answer 1

Given,

Caucasian: $n_1=185,\ \ \bar x_1=20.6,\ \ S_1=5.8$

African American : $n_2=86,\ \ \bar x_2=17.4,\ \ S_2=5.8$

Poolesed Standard Deviation , $S_p=\sqrt{\dfrac{S_1^2(n_1-1)+S_2^2(n_2-1)}{n_1+n_2-2}}=\sqrt{\dfrac{5.8^2(184)+5.8^2(85)}{185+86-2}}=5.8$

Since, Sample is large, So we can assume that $\bar x_1-\bar x_2$ follow a normal distribution.

Now,

90% Confidence Interval for $\mu_1-\mu_2$ is

$\implies (\bar x_1-\bar x_2) \pm Z_{1-\frac{0.10}{2}}\times S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}\\\ \\\implies(20.6-17.4)\pm1.64\times 5.8\sqrt{\frac{1}{185}+\frac{1}{86}}\\\ \\\implies 3.2\pm 1.24\\\implies (1.96,\ 4.44)$

Practical: The CI doesn't include the value zero. Hence the mean score for Caucasian study participants and the mean scores for African- American men are significantly different at 0.10 significance level.

Probabilistic: The mean difference $\mu_1-\mu_2$ for Caucasian study participants and African-American men fall within the interval (1.96, 4.44) with 90% confidence.