The first answer is a and the second answer is d
Answer: It's me again ;) The answer is 778,095
Step-by-step explanation: Find 90% of 864,550.
Answer:
The 95% confidence interval estimate of the true population proportion of U.S. employers that were likely to require higher employee contributions for health care coverage is 0.52 +/- 0.0370
= (0.4830, 0.5570)
The margin of error M.E = 0.0370
Step-by-step explanation:
Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.
The confidence interval of a statistical data can be written as.
p+/-z√(p(1-p)/n)
p+/-M.E
Given that;
M.E = margin of error
Proportion p = 52% = 0.52
Number of samples n = 700
Confidence interval = 95%
z value (at 95% confidence) = 1.96
Substituting the values we have;
0.52 +/- 1.96√(0.52(1-0.52)/700)
0.52 +/- 1.96(0.0189)
0.52 +/- 0.0370
( 0.4830, 0.5570)
The 95% confidence interval estimate of the true population proportion of U.S. employers that were likely to require higher employee contributions for health care coverage is 0.52 +/- 0.0370
= (0.4830, 0.5570)
The margin of error M.E = 0.0370
The correct answer is the second option.
Hope this helps!
Answer:
Step-by-step explanation:
