Question

Umbrella Corporation is trying to determine whether a new drug improves mood in individuals with bipolar disorder. The company gave the drug to 260 individuals (Group 1) and a placebo (or sugar pill) to a separate sample of 247 individuals (Group 2). After one month on the drug, 192 of the patients in Group 1 reported improved mood, while 149 of the patients in Group 2 reported the same.

Calculate the lower bound of a 95% confidence interval for the true difference in the proportion of individuals who experienced improved mood. Take all calculations to four (4) decimal places.

Answer 1

The lower bound of the 95% confidence interval for the true difference in the proportion of individuals who experienced improved mood after taking a new drug for bipolar disorder is 0.0895.

Step-by-step explanation:

To calculate the lower bound of a 95% confidence interval for the true difference in the proportion of individuals who experienced improved mood after taking a new drug for bipolar disorder, we use the formula for the confidence interval of the difference between two proportions:

Z = Z-value from the standard normal distribution for the desired level of confidence
p1 = proportion of success in group 1
p2 = proportion of success in group 2
n1 = total number of observations in group 1
n2 = total number of observations in group 2

The formula for the standard error of the difference between two proportions is:
SE = sqrt(p1(1-p1)/n1 + p2(1-p2)/n2),
and the confidence interval is given by: (p1 - p2) ± Z × SE.

For Group 1 (drug group),
p1 = 192/260 = 0.7385,
and for Group 2 (placebo group),
p2 = 149/247 = 0.6032.

The standard normal Z-value for a 95% confidence interval (two-tailed) is typically 1.96 (Z=1.96).

Calculating SE, we get:
SE = sqrt(0.7385(1-0.7385)/260 + 0.6032(1-0.6032)/247) = sqrt(0.0021) = 0.0456.
Now, calculate the 95% CI:
Lower bound = (0.7385 - 0.6032) - 1.96 × 0.0456 = 0.0895.

The lower bound of the 95% confidence interval for the true difference in proportions is 0.0895.