A survey of working adults is described in an online article. The article states, "Workers with a college
education had a significantly increased chance of getting a pay raise in the past year, with 34 percent of the
group seeing an increase, compared with 19 percent of those with some college and 16 percent of those
with high school diplomas or less." Suppose that the three samples (college degree, some college, high
school diploma or less) were independently selected representative samples and that the sample size for
each sample was 400.
Use the given information and the five-step process for estimation problems (EMC ) to calculate a
95% confidence interval for the difference in the proportion who received raises for those with a
college degree and those with a high school diploma or less. (Use
College Degree − High School Diploma or Less. Enter your answer using interval notation. Round your
numerical values to four values.)
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Step 1: State the problem and identify the parameters.
The problem is to calculate a 95% confidence interval for the difference in the proportion who received raises for those with a college degree and those with a high school diploma or less.
Parameters:
P1 = proportion who received raises in the college degree group
P2 = proportion who received raises in the high school diploma or less group
Step 2: Formulate an analysis plan.
We will use the formula for the confidence interval for the difference in two independent proportions.
Step 3: Collect and summarize the data.
Given information:
Sample size for each group (n1) = 400
Sample size for each group (n2) = 400
Proportion who received raises in the college degree group (p1) = 0.34
Proportion who received raises in the high school diploma or less group (p2) = 0.16
Step 4: Perform calculations.
We will calculate the standard error (SE) and the margin of error (ME) using the formula:
SE = sqrt((p1(1-p1)/n1) + (p2(1-p2)/n2))
ME = 1.96 * SE (to calculate a 95% confidence interval)
SE = sqrt((0.34(1-0.34)/400) + (0.16(1-0.16)/400))
ME = 1.96 * SE
Step 5: Interpret the results.
The 95% confidence interval for the difference in the proportion who received raises for those with a college degree and those with a high school diploma or less is given by:
(p1 - p2) +/- ME
Substituting the values:
(0.34 - 0.16) +/- ME
Calculating the ME and rounding to four decimal places:
ME = 1.96 * SE = 1.96 * (sqrt((0.34(1-0.34)/400) + (0.16(1-0.16)/400))) = 0.0464
The 95% confidence interval is therefore:
(0.34 - 0.16) +/- 0.0464 = 0.18 +/- 0.0464
Interval notation: (0.1336, 0.2264)
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