Question 1

A nutritionist believes that obesity is more prevalent among American adults than it was in the past. He
discovers that in a study conducted in the year 1994, 380 of the 1630 randomly chosen adults were classified as
obese. However, in a more recent study, he finds 726 out of 2350 randomly chosen adults were classified as
obese. Construct a 95% confidence interval for the difference in proportions $\mathrm { P } _1 - \mathrm { P }_ 2$

Accepted Answer

The answer of A nutritionist believes that obesity is more...

Question 2

A random sample of 100 students at a high school was asked whether they would ask their father or mother for help with a homework assignment in science. A second random sample of 100 different students was asked the
Same question for an assignment in history. Forty-three students in the first sample and 47 students in the
Second sample replied that they turned to their mother rather than their father for help. Construct a 98%
Confidence interval for $\mathrm { p } _ { 1 } - \mathrm { p } _ { 2 }$
A)(-0.204, 0.124)
B)(-1.324, 1.521)
C)(-0.591, 0.762)
D)(-1.113, 1.311)

Accepted Answer

The confidence interval for the difference between two proportions $\mathrm { p } _ { 1 } - \mathrm { p } _ { 2 }$ can be calculated using the formula for the confidence interval of two proportions. Given the sample sizes (n1 = n2 = 100) and the number of successes (x1 = 43, x2 = 47), we first find the sample proportions $p_1 = \frac{x1}{n1} = 0.43$ and $p_2 = \frac{x2}{n2} = 0.47$. The standard error for the difference in proportions is calculated using the formula $\sqrt{\frac{p_1(1-p_1)}{n1} + \frac{p_2(1-p_2)}{n2}}$. For a 98% confidence interval, we use a Z-score that corresponds to the confidence level (approximately 2.33 for 98%). Plugging these values into the confidence interval formula gives us the interval (-0.204, 0.124), which suggests that the true difference in proportions could fall within this range with 98% confidence.

Quiz 8: Hypothesis Testing With Two Samples